Physics 1: exam 2
centripetal force
*ac* always pointing towards the center (direction of acceleration)
CQ: You tie a rope to a tree and you pull on the rope with a force of 100N. What is the tension in the rope?
100 N -the tension in the rope is the force that the rope "feels" across any section of it (or that you would feel if you replaced a piece of the rope). Because you are pulling with a force of 100 N, that is the tension in the rope
what is T1 in terms of m and a?
F = ma T = ma so in order to have the greatest tension with the smallest mass, the acceleration of 1m will have to be the largest as well
Do scales measure your weight?
Scales actually measure the normal force!
Using Newton's 2nd law, the centripetal force is
the mass times the acceleration. Fc = (m)(ac) = (m)(v^2) / r
If the mass of a rope is zero, how much force is required to accelerate it?
∑F = ma = 0! Also, if we apply a force T at one end of a rope, by Newton's 3rd Law, there is a force -T exerted at the other end of the rope. -this means: acceleration of the rope breaks it, so you need a force in the rope to keep it from breaking = tension (T) -tension pulls in both directions and you have the same tensions along the whole entire rope = same force acting throughout
The work W0 accelerates a car from 0 to 50 km/h. Is the work required to accelerate the car from 50 km/h to 150 km/h equal to 2W0, 3W0, 8W0, or 9W0? Choose the best explanation from among the following: 1.) The increase in speed from 50 km/h to 150 km/h is twice the increase in speed from 0 to 50 km/h. 2.) The final speed is three times the speed that was produced by the work W0 3.) The work to accelerate the car depends on the speed squared.
*8W0* The work done on the car is equal to its change in kinetic energy, which in turn is proportional to thesquare of its velocity. Therefore, if the kinetic energy of the car is W0 when its speed is 50 km/h, its kinetic energy will be 9W0 when its speed is 150 km/h. We conclude that the work required to accelerate the car from 50 km/h to 150 km/his 9W0 - W0 = 8W0 *The final speed is three times the speed that was produced by the work W0*
12.) T/F: The force of static friction between two surfaces is independent of the area of contact between the surfaces.
*True* -fs = μsN -N = weight with no other forces present
18.) A person applies a constant force of 20 N to a rock of mass 1000 kg, for a total of 20 seconds. What is the work done by this person if the rock does not move at all by this applied force?
0J *A* -not moving (W = Fd --> W =F0 --> W = 0)
IP A 9.50 g bullet has a speed of 1.30 km/s 1.) What is its kinetic energy in joules? 2.) What is the bullet's kinetic energy if its speed is halved? 3.) If its speed is doubled?
1.) 9.5 g --> 0.0095 kg 1.3 km/s --> 1300 m KE = 1/2 mv^2 --> 1/2(0.0095)(1300^2) --> *KE = 8027.5 J* 2.) KE = 1/2 mv^2 --> 1/2(0.0095)(650^2) --> *KE = 2007 J* (energy is quartered) 3.) KE = 1/2 mv^2 --> KE = 1/2(0.0095)(2600^2) --> *KE = 32110 J* (energy is multiplied by 4)
Two blocks of ice, one four times as heavy as the other, are at rest on a frozen lake. A person pushes each block the same distance d. Ignore friction and assume that an equal force F⃗ is exerted on each block. 1.) Which of the following statements is true about the kinetic energy of the heavier block after the push? 2.) Compared to the speed of the heavier block, what is the speed of the light block after both blocks move the same distance d? 3.) Now assume that both blocks have the same speed after being pushed with the same force F⃗ . What can be said about the distances the two blocks are pushed?
1.) It is equal to the kinetic energy of the lighter block (The work-energy theorem states that the change in kinetic energy of an object equals the net work done on that object. The only force doing work on the blocks is the force from the person, which is the same in both cases. Since the initial kinetic energy of each block is zero, both blocks have the same final kinetic energy.) 2.) Twice as fast (Since the kinetic energy of the lighter block is equal to the kinetic energy of the heavier block, the lighter block must be moving faster than the heavier block) 3.) The heavy block must be pushed 4 times farther than the light block. (Because the heavier block has four times the mass of the lighter block, when the two blocks travel with the same speed, the heavier block will have four times as much kinetic energy. The work-energy theorem implies that four times more work must be done on the heavier block than on the lighter block. Since the same force is applied to both blocks, the heavier block must be pushed through four times the distance as the lighter block)
A 70 kg bicyclist rides his 8.8 kg bicycle with a speed of 16 m/s 1.) How much work must be done by the brakes to bring the bike and rider to a stop? 2.) How far does the bicycle travel if it takes 3.5 s to come to rest? 3.) What is the magnitude of the braking force?
1.) KE = -1/2mv^2 --> KE = -1/2(78.8)(16^2) --> *KE = -10086.4 J* 2.) average speed = Vf + Vi /2 --> average speed = (0 + 16)/2 --> average speed = 8 d = vt --> d = 8 x 3.5 --> *d = 28 m* 3.) W = Fd --> 10086.4 = F(28) --> *F = 360.2 N*
A road race is taking place along the track shown in the figure (Figure 1). All of the cars are moving at constant speeds. The car at point F is traveling along a straight section of the track, whereas all the other cars are moving along curved segments of the track 1.) Let v⃗ A be the velocity of the car at point A. What can you say about the acceleration of the car at that point? 2.) Let v⃗ C be the velocity of the car at point C. What can you say about the acceleration of the car at that point? 3.) Let v⃗ D be the velocity of the car at point D. What can you say about the acceleration of the car at that point? 4.) Let v⃗ F be the velocity of the car at point F. What can you say about the acceleration of the car at that point? 5.) Assuming that all cars have equal speeds, which car has the acceleration of the greatest magnitude, and which one has the acceleration of the least magnitude? 6.) Assume that the car at point A and the one at point E are traveling along circular paths that have the same radius. If the car at point A now moves twice as fast as the car at point E, how is the magnitude of its acceleration related to that of car E.
1.) The acceleration is perpendicular to v⃗ A and directed toward the inside of the track. 2.) The acceleration is perpendicular to v⃗ C and pointed toward the inside of the track. 3.) The acceleration is perpendicular to v⃗ D and pointed toward the outside of the track 4.) The acceleration is zero (car is going straight, not around a curve like the other cars) 5.) D has the greatest because its on the tightest curve, F has the least because its driving straight 6.) The magnitude of the acceleration of the car at point A is four times that of the car at point E.
to calculate the work done by various forces on an object that moves 160 meters to the right. 1.) 18 N of force applied at 0º 2.) Find the work W done by the 30-newton force pulling at 30º 3.) Find the work W done by the 12-newton force pulling at 180º 4.) Find the work W done by the 15-newton force pulling at 220º
1.) W = F(cosθ)d --> W = 18(1)160 --> W = 2880 = *(2 sig figs) 2900 J* 2.) W = F(cosθ)d --> W = 30(cos30)160 --> W = 4157 = *(2 sig figs) 4200 J* 3.) W = F(cosθ)d --> W = 12(cos180)160 --> W = -1920 = *(2 sig figs) -1900 J* 4.) W = F(cosθ)d --> W = 15(cos220)160 --> W = -1839 = *(2 sig figs) -1800 J*
To keep her dog from running away while she talks to a friend, Susan pulls gently on the dog's leash with a constant force given by F⃗ = (2.0 N )x + (1.5 N )y 1.) How much work does she do on the dog if its displacement is d⃗ = (0.24 m )x 2.) How much work does she do on the dog if its displacement is d⃗ = (0.30 m )y 3.) How much work does she do on the dog if its displacement is d⃗ = (-0.62 m )x + (-0.21 m )y
1.) W = Fd --> W = (2)(0.24) --> *W = 0.48 J* 2.) W = Fd --> W = (1.5)(0.30) --> *W = 0.45 J* 3.) W = Fd --> Wx = 2 x -0.62 --> Wx = -1.24 W = Fd --> Wy = 1.5 x -0.21 --> Wy = -0.315 Wx + Wy = Wtotal --> W total = -0.315 + -1.24 --> *W total = -1.555 J*
You pick up a 3.7 kg can of paint from the ground and lift it to a height of 1.2 m . 1.) How much work do you do on the can of paint? 2.) You hold the can stationary for half a minute, waiting for a friend on a ladder to take it. How much work do you do during this time? 3.) Your friend decides against the paint, so you lower it back to the ground. How much work do you do on the can as you lower it?
1.) W = mg --> W = 3.7 x 10 --> W = 37 kg W = Fd --> W = 37 x 1.2 --> *W = 44.4 J* 2.) W = Fd --> W = F(0) --> *W = 0* 3.) W = Fd --> W = 37 x -1.2 -> *W = -44 J*
Two blocks are connected by a string, as shown in the figure (Figure 1). The smooth inclined surface makes an angle of 35º with the horizontal, and the block on the incline has a mass of 5.7 kg. The mass of the hanging block is m = 4.0 kg 1.) Find the direction of the hanging block's acceleration. 2.) Find the magnitude of the hanging block's acceleration.
1.) downwards 2.) m x g - T = m x a --> T = -ma + mg T - M x g x sin 35 = Ma --> T = Ma /Mg sin35 -ma + mg = Ma/Mg sin35 --> a = g (m - Msin35) / (M + m) --> *a = 0.74 m/s^2*
In the figure we see two blocks connected by a string and tied to a wall. The mass of the lower block is 1.0 kg; the mass of the upper block is 2.0 kg; the angle of the incline is 31º 1.) Find the tension in the string connecting the two blocks. 2.) Find the tension in the string that is tied to the wall.
1.) since it's only the weight of the bottom mass pulling, the tension will balance it out because there is no acceleration (W1 = T1) W1x = Wsinθ --> W1x = mgsinθ --> W1x = (1.0)(9.81)sin31 --> W1x = *5.1 N = T1* 2.) W2x = Wsinθ --> W2x = mgsinθ --> W2x = (1+2)(9.81)sin31 --> W2x = *15 N = T2*
A block of mass m1 slides on a frictionless tabletop. It is connected to a string that passes over a pulley and suspends a mass m2. (Figure 1) Suppose m1 and m2 are both increased by a factor of 2. 1.) Does the acceleration of the blocks increase, decrease, or stay the same? 2.) Does the tension in the string increase, decrease, or stay the same?
1.) stay the same 2.) increase (tension will double)
CQ: Two tug-of-war opponents each pull with a force of 100 N on opposite ends of a rope. What is the tension in the rope?
100N -because tension is the same throughout the length of the rope -force = tension -whether the other end of the rope is pulled by a person, or pulled by a tree, the tension in the rope is still 100 N
Newton's 3rd law can be summarized as follows:
A physical interaction (e.g., gravity) operates between two interacting objects and generates a pair of opposite forces, one on each object. It offers you a way to test for real forces (i.e., those that belong on the force side of ∑F⃗ =ma⃗ )—there should be a 3rd law pair force operating on some other object for each real force that acts on the object whose acceleration is under consideration.
CQ: in which case does block m experience a larger acceleration? -case 1: there is a 10 kg mass hanging from a rope and falling -case 2: a hand is providing a constant downward force of 98 N assume massless ropes
Case 2 -In case 2 the tension is 98 N due to the hand. In case 1 the tension is less than 98 N because the block is accelerating down. Only if the block were at rest would the tension be equal to 98 N -case 2: ∑F = ma --> T = ma --> 98 = ma -case 1: T2 = ma (for the box on top) W - T2 = ma --> (9.8 x 10) - T2 = 10 a (for the hand) -tension is less in case 2
Problem solving tactics 3
Draw force diagrams and include acceleration of the object. -grav = gravity -app = applied -frict = friction -norm = normal *remember force is a vector --> -slide 9: there is a force applied to box A which is positioned next to box B. Box A has the applied force pushing it right as well as a force from box B pushing back at it. Box B has the force from box A pushing it right
example: Two 1 kg masses are sitting 1 m apart. What is the gravitational force between the two masses? What if the masses are 1 mm apart?
F = G (m1 m2 / r^2) F = 6.673 x 10^-11 (1x1/1^2) F = 6.673 x 10^-11 r = 0.001 m F = 6.673 x 10^-11 ((1x1/0.001^2) F = 6.673 x 10^-5 in the second case where they are closer together, the force becomes larger
You are pulling your little sister on her sled across an icy (frictionless) surface. When you exert a constant horizontal force of 120 N , the sled has an acceleration of 2.3 m/s^2 . If the sled has a mass of 7.0 kg , what is the mass of your little sister?
F = ma 120 = m(2.3) m = 52.2 52.2 - 7.0 = *45.17 kg*
When you take your 1900-kg car out for a spin, you go around a corner of radius 55 m with a speed of 19 m/s. The coefficient of static friction between the car and the road is 0.88. Assuming your car doesn't skid, what is the force exerted on it by static friction?
F = ma a = v^2/r F = mv^2/r F = 1900 x 19^2/55 F = 12470.91 N -static friction isn't needed in this problem because it doesn't skid
Consider the system shown in the figure. (Figure 1) Block A has weight 4.91 N and block B has weight 2.94 N . Once block B is set into downward motion, it descends at a constant speed. Assume that the mass and friction of the pulley are negligible. Calculate the coefficient of kinetic friction μ between block A and the table top.
F = μ N 2.94 = μ (4.91) *μ = 0.599* --> 0.600
Non-Equilibrium
If there is an imbalance in forces, there must be acceleration. If we have multiple unbalanced forces in a problem, we can add the forces and use Newton's 2nd Law to determined the acceleration of an object -have a net force = net acceleration
CQ: A block of mass (m) rests on the floor of an elevator that is accelerating upwards. What is the relationship between the force due to gravity and the normal force on the block?
Normal force > mass (mg) -The block is accelerating upward, so it *must* have a *net upward force*. The forces on it are N (up) and mg (down), so N must be *greater* than mg in order to give the net upward force
Fundamental
One of the three truly unique forces in nature; Gravitational force, Strong nuclear force, and the Electroweak force.
An ice cube is placed in a microwave oven. Suppose the oven delivers 115 W of power to the ice cube and that it takes 31200 J to melt it. How long does it take for the ice cube to melt?
P = W/t 115 = 31200/t t = 271.3 seconds t = 4.52 minutes
The average power is defined as:
P = work / time = W/t
7-1 Work Done by a Constant Force
The work can also be written as the dot product of the force and the displacement: W = Fvector • dvector --> W = Fd cosθ in picture on slide 5: - W = Wxd - W = Wcosθd -y is perpendicular to the incline = no work = perpendicular to the direction of motion = only x does work
CQ: A book is lying at rest on a table. The book will remain there at rest because
There is no net force on the book -there are forces acting on the book, but the only forces acting are in the y direction. Gravity acts downward, but the table exerts an upward force that is equally strong, so the 2 forces *cancel*, leaving no net force
Force is a vector
To work Newton's 2nd Law problems in multiple dimensions, we must treat forces in each direction independently. ∑Fx = m ax ∑Fy = m ay We find the x-components and the y-components of each force, add all of the like components, and find the acceleration.
A farmhand pushes a 23 kg bale of hay 3.9 m across the floor of a barn If she exerts a horizontal force of 91 N on the hay, how much work has she done?
W = Fd W = 91 * 3.9 W = 355 J
Children in a tree house lift a small dog in a basket 5.05 m up to their house. If it takes 201 J of work to do this, what is the combined mass of the dog and basket?
W = Fd --> 201 = F(5.05) --> F = 39.8 N F = mg --> 39.8 = m(10) --> *m = 3.98 kg*
20.) How much energy is needed to change the speed of a 1600 kg sport utility vehicle from 15.0 m/s to 40.0 m/s?
W = KEf - KE0 W = 1/2mvf^2 - 1/2mv0^2 1/2(1600)(40)^2 - 1/2(1600)(15)^2 = W W = 1100000 J --> *1.10 MJ* *C*
CQ: A hockey puck slides on ice at a *constant velocity*. What is the net force acting on the puck?
Zero
Let's use the acceleration from a force and look at kinematics of a force acting over a distance:
a = Ftotal/m 2ad = vf^2 - vi^2 2 (Ftotal/m)d = vf^2 - vi^2 (Ftotal)d = 1/2 mvf^2 - 1/2 mvi^2
CQ: You kick a smooth, flat stone out on a frozen pond. The stone slides, slows down, and eventually stops. You conclude that:
a net force acted on it all along (friction = net force)
The work done by a system is the force acting over some distance. If we rewrite our power equation:
average power (P) = w/t W = Fd d/t =v so... P = Fd/t = F(d/t) = F(average velocity, v) The work is the force acting over a distance, the power is the force acting at a speed.
Example: slide 7 All force vectors have the same magnitude, and undergo a displacement, x. Rank these four situations from most negative work to most positive work. The floor is frictionless.
box A: greater than 90º pull box B: 90º pull box C: about 45º pull box D: horizontal pull *A, B, C, D* -A is the only one with negative work, B is zero, and C and D are positive
The force acting at a right angle to the motion
does no work! Only the force acting in the direction of the motion does work!
Example: A boat is being pulled by two locomotives through a canal of length 2.00 km. The tension in the cable is 5.00x10^3 N, and θ = 20º What is the net work done on the boat by the two locomotives?
draw a picture: d = 2,000 m, T = 5 x 10^3 N, θ = 20º Ty = 0 Tx = Fx --> Fx = Tcosθ W = Fx • d --> W = (Tcosθ)d Wtotal = 2W Wtotal = 2 • (Tcosθ)d --> Wtotal = 2 • (5x10^3 cos20) 2000 --> *Wtotal = 1.88 x 10^7 J* -2 W above because you need the force of both (2) locomotives
Example: A 40 kg child is riding a sled down a flat, icy hill. The hill is angled 30º above the horizon. Starting from rest, after the child has traveled 100m, what will be the child's final speed?
draw a picture: in notes -m = 40, V0 = 0, Δd = 100, θ = 30º only force acting = Wx, so Wx = ma mg sin30 = ma (but m and m cancel out on each side) --so--> 9.81sin30 = a --> a = 4.9 m/s^2 V^2 = V0^2 + 2ad --> V^2 = 0 + 2(4.9)(100) --> V = 31.3 m/s --> *V = 70 mph*
According to Newton's 3rd law, the force on the (smaller) moon due to the (larger) earth is
equal in magnitude to, and in the opposite direction from, the force on the earth due to the moon.
work has units of
joules (J). -Force has units of N, displacement has units of m 1 J = 1 N•m
Mass has units of kg, velocity units of m/s, so kinetic energy has units of
kg•m2/s2 = N•m = J (joules!).
Any object moving with a velocity, v, has a quantity associated with it called the
kinetic energy.
A roller-coaster track has six semicircular "dips" with different radii of curvature. The same roller-coaster cart rides through each dip at a different speed. For the different values given for the radius of curvature R and speed v, rank the magnitude of the force of the roller-coaster track on the cart at the bottom of each dip. Rank from largest to smallest. To rank items as equivalent, overlap them. 1.) R=15m v=12m/s 2.) R=15m v=8m/s 3.) R=30m v=4m/s 4.) R=60m v=16m/s 5.) R=45m v=4m/s 6.) R=30m v=16m/s
largest 1 6 4, 2 are equal 3 5 smallest
inertia is measures by
mass in kilograms (kg)
4.) A fireman is sliding down a fire pole. As he speeds up, he tightens his grip on the pole, thus increasing the vertical frictional force that the pole exerts on the fireman. When this force equals the weight of the fireman, what happens?
motion with constant speed *B.) The fireman continues to descend, but with constant speed.* -if f > W in this case, then he would stop eventually
not moving vertically =
normal force (N) is equal to weight
for the previous example, what would be the normal force (force pushing on the sled N)?
normal force = positive direction and Wy in negative y direction --> N - Wy = 0 (equal to zero because no acceleration in y direction) --> N = Wy = mg cos30
CQ: Your little sister wants you to give her a ride on her sled. On level ground, what is the easiest way to accomplish this?
pulling her from the front when you push your sister, the force F is pushing down (in addition to mg), so the normal force is larger. when you pull your sister, the force F is pulling up, against gravity, so the normal force is lessened. Recall that the frictional force is proportional to the normal force
CQ: We just decided that the cart continues with *constant velocity*. What would have to be done in order to have the cart continue with constant acceleration?
push the cart continuously -In order to achieve a non-zero acceleration, it is necessary to maintain the applied force. The only way to do this would to be to continue pushing the cart as it moves down the track. This will lead us to a discussion of Newton's second law
Chapter 6
ramps
ramps
slide 1 put all forces acting on the picture: -angle θ -weight (arrow down) (W = mg) (weight is a vector, mass is scalar = weight has an arrow down) -weight is on an incline = choose the axis to label the incline (incline is x), so y is always perpendicular to x -then need to resolve/find components/weight of these 2 vectors (Wy and Wx) (W = Wy + Wx) the 2 angles labeled θ in the picture are equal, so (Wy = -Wcosθ) and (Wx = wsinθ) -has to be another force involved to keep it from moving in the y direction = normal force = need some length/magnitude as Wy (N = Wy)
7-2 Kinetic Energy and the Work-Energy Theorem
slide 11: -When positive work is done on an object, its speed increases; when negative work is done, its speed decreases. -left picture = displacement (d) and weight of the apple (mg) go down = W>0 -right picture = displacement (d) of the apple is up (because being thrown upwards) while weight (mg) is still down = W<0 -if velocity (v) increases over time, then kinetic energy increases with time -left picture: KEfinal - KEinitial = 1/2 mvf^2 - 1/2 mvi^2 ---> positive, because final speed is greater than initial speed
negligible
so small or unimportant as to be not worth considering; insignificant.
what is the truck doing 5 minutes after the fly hit it?
still moving at a constant speed
Newton's definition of force
that which causes the velocity of a object to change, hence it gives rise to acceleration.
The unit of force is
the Newton (N)
CQ: A box of weight 100 N is at rest on a floor where μs = 0.4. A rope is attached to the box and pulled horizontally with tension T = 30 N. Which way does the box move?
the box does not move -The static friction force has a maximum of mN = 40 N. The tension in the rope is only 30 N. So the pulling force is not big enough to overcome friction. -What happens if the tension is 35 N? What about 45 N?
CQ: In a baseball game, the catcher stops a 90-mph pitch. What can you say about the work done by the catcher on the ball?
the catcher has done negative work -The force exerted by the catcher is opposite the direction to the displacement of the ball, so work is negative. Or, using the definition of work (W = F (Δr)cosθ), because θ = 180º, then W<0. -Note that because the work done on the ball is negative, its speed decreases
what about the work done by the ball on the catcher?
the catcher is producing either zero work initially or negative work, and the ball makes it more negative?
Kinetic energy is
the energy of motion. Any moving object has kinetic energy, it does not matter if the object is moving in a straight line, in a circle, or any sort of zig-zagging path.
Our coefficients μs and μk depend on
the material. -Often, we are given these values, we can look them up, or we will determine them.
We have covered all of the rules governing
the motion of a system subject to a constant acceleration. -we didn't care what was producing the acceleration or how, now we do
Newton's definition of inertia
the natural tendency of an object to remain at rest or in motion at a constant speed along a straight line. The mass of an object is a quantitative measure of the inertia. -larger mass = larger inertia
The average power is
the rate at which work is done by a force.
net force
the total force -you could have 2 forces (force is a vector) pulling in different directions and the net force is the sum/reaction of all forces being applied -zero net force = balanced/forces cancel out
CQ: A box sits in a pickup truck on a frictionless truck bed. When the truck accelerates forward, the box slides off the back of the truck because
there is no net force acted on the box -generally, the reason that the box in the truck bed would move with the truck is due to friction between the box and the bed. *If there is no friction, there is no force to push the box along with the truck, so it remains at rest.* The truck accelerated away, essentially leaving the box behind
CQ: you and a friend can each pull with a force of 20 N. If you want to rip a rope in half, what is the best way?
tie the rope to a tree and have both of you pull from the same end -apply max force -take advantage of the fact that the tree can pull with almost any force (until it falls down, that is). You and your friend should team up on one end, and let the tree make the effort on the other end
Practice exam
vvvvv
The unit of power is the
watt (W) 1 W = 1 J/s 1 horsepower = 1 hp = 746 W -work per second = J/s = 1 watt (W) -how much energy is used in 1 second
If a force acts at an angle to the direction of motion,
we can think of breaking the force into two parts, the part in the direction of the motion and the part at a right angle to the motion. -work can be positive, negative, or zero
When we have multiple forces,
we sum the contribution from each force together. With work we do the exact same thing, the total work is the sum of the work done by each force. -ex: previous cable and locomotives example (had 2 forces = add them together/multiply by 2 to get total force on the boat)
Suppose a rocket launches with an acceleration of 29.0 m/s^2 What is the apparent weight of an 93-kg astronaut aboard this rocket?
weight = mass x acceleration due to gravity weight = 93 x 9.81 weight = 912.33 N weight = mass x acceleration of rocket weight = 93 x 29 weight = 2697 2697 + 912.33 = apparent weight *apparent weight = 3609.33*
equilibrium means acceleration =
zero
The work done may be positive, zero, or negative, depending on the angle between the force and the displacement:
θ < 90º = positive work θ = 90º = zero work θ > 90º = negative work -positive = force and displacement pointing in the same direction -zero = because there is no horizontal movement -negative = motion going in one direction, but force acting in the opposite direction of motion = always does negative work
CQ: Antilock brakes keep the car wheels from locking and skidding during a sudden stop. Why does this help slow the car down?
μs > μk so static friction is better -static friction is always > kinetic friction -static friction is greater than sliding friction, so by keeping the wheels from skidding, the static friction force will help slow the car down more efficiently than the sliding friction that occurs during a skid
Total work
Work is the product of a force acting over a displacement. What can we do if we have multiple forces?
can friction ever do positive work?
Yes -consider the case of a box on the back of a pickup truck from earlier. If the box moves along with the truck, then it is actually the force of friction that is making the box move
Since force involves acceleration, and acceleration is a vector, is force a vector?
Yes! ∑F = ma
Example: (a) What is the magnitude of the centripetal acceleration of an object on Earth's equator due to the rotation of Earth? (b) What the period of rotation of the Earth has to be such that objects on the equator have a centripetal acceleration with a magnitude of 9.8 m/s2? radius of Earth = 6.37x10^6 m T = 24 h = 86,400 s
a.) V = 2πr / T --> V = 2π(6.37x10^6) / 86400 --> V = 463.2 ac = V^2 / r --> ac = (4.63.2)^2 / 6.37x10^6 --> *ac = 0.034 m/s^2* b.) ac = V^2 /r --> 9.81 = V^2 / 6.37x10^6 --> V = 7901 m/s V = 2πr / T --> 7901 = 2π(6.37x10^6 ) / T --> T = 5065 seconds = *84.4 min*
CQ: A box is being pulled across a rough floor at a constant speed. What can you say about the work done by friction?
friction does negative work -friction is acting in the opposite direction to the displacement, so the work is negative. Or, using the definition of work (W = F (Δr)cosθ), because θ = 180º, then W < 0 ------> pull and displacement ↑ N ↓ mg <------ f -opposite direction of motion
W>0 if force
has component in direction of motion (θ < 90º) -work is positive when the angle it is pulled at is less than 90º (the highest number of work is given to the angle closest to 0 = more horizontal = stronger)
W<0 if force
has component opposite direction of motion (θ > 90º) -work is negative when the angle it is pulled at is greater than 90º (the more negative the number is = the closer the number of work is to 180 in the opposite direction of movement)
what is the force that usually keeps the book on the seat?
inertia
The force keeping the rope spread out is called the
tension T
Weight
the special name we give to the force a very large object exerts on a much smaller object. Since it is a force, weight has units of N. It is also a vector (though we often forget about the direction). Above the surface of the earth, the weight of an object is given as: W = G (ME m / r^2) = mg -e; earth exerts force onto me = weight (mg) -weight = gravitational constant (G)
Dynamics is
the study of the root causes the accelerations in kinematics.
*period, T* of the motion is
the time it takes to travel once around a circle. -Since the distance traveled around the circle is 2πr, the speed of an object in uniform circular motion is: V = 2πr / T -period = time it takes to go around one full circle -if change in distance = V x T and the distance traveled is the circumference of the circle, then 2πr = V x T --> V = 2πr / T
10.) A 40.0-kg crate is being raised by means of a rope. Its upward acceleration is 2.00 m/s2. What is the force exerted by the rope on the crate?
*A* 472 N force = tension W = 40 x 9.8 F = ma T-W T = ma + W
3.) An object of weightW is in free-fall close to the surface of Earth. What is the force that the object exerts on Earth?
*C.) a force equal to W* -because of Newton's 3rd law = action and reaction
There are two general types of forces:
-Fundamental -Non-fundamental
13.) A packing crate slides down an inclined ramp at constant velocity. Thus we can deduce that
-no net upward force -no downward force (no acceleration) -friction force acting *D.) a frictional force is acting on it.*
7.) A net force of 125 N is applied to a certain object. As a result, the object accelerates with an acceleration of 24.0 m/s2. The mass of the object is
A.) 5.20 kg
9.) A person has a mass of 45 kg. How much does she weigh?
B.) 440 N
17.) Can work be done on a system if there is no motion?
B.) No, since a system which is not moving has no energy.
6.) A block of mass M slides down a frictionless plane inclined at an angle θ with the horizontal. The normal reaction force exerted by the plane on the block is directed
B.) perpendicular to the plane.
The work-energy theorem
Both work and energy have the same units, but they are more intrinsically linked. The work-energy theorem states that the work done on a system is equal to the change in kinetic energy. Mathematically: W = KEf - KE0 W = mvf^2 - 1/2 mv0^2 -work = final minus initial
A puck attached to a string undergoes circular motion on an air table. If the string breaks at the point indicated in the figure (Figure 1), is the subsequent motion of the puck best described by path A, B, C, or D?
C (velocity pointing this way)
5.) You ride on an elevator that is moving downward with constant speed while standing on a bathroom scale. The reading on the scale is
C.) equal to your true weight, mg
2.) A golf club hits a golf ball with a force of 2400 N. The golf ball hits the club with a force
C.) exactly 2400 N
Newton's 1st law
Consider a body on which no net force acts. -If a body is at rest it will remain at rest. -If the body is moving it will continue to move at constant velocity
Consider a scale on an elevator going down:
Consider a scale on an accelerating elevator: If there is a net downward acceleration, a person's weight in the elevator will appear to be smaller than it is. N - mg = -ma N = mg - ma -we call this special normal force the "apparent weight" -normal force is less
Chapter 5
Dynamics
1.) An object is moving with constant velocity. Which of the following statements is true? A) There are no forces acting on the object. B) There is no frictional force acting on the object. C) A constant force is being applied in the direction of motion. D) A constant force is being applied in the direction opposite of motion. E) The net force on the object is zero.
E.) The net force on the object is zero. -no acceleration
15.) What type of acceleration does an object moving with constant speed in a circular path experience?
E.) centripetal acceleration
Equilibrium and motion
Equilibrium does *not* mean the object must be stationary! Static equilibrium means stationary, equilibrium means there is no acceleration. Examples of static equilibrium: This lecture hall Examples of dynamic (moving) equilibrium: An airplane traveling at constant speed -dynamic equilibrium = moving with constant velocity (so acceleration = zero) -forces balance out = equilibrium
8.) A 3.0-kg and a 5.0-kg box rest side-by-side on a smooth, level floor. A horizontal force of 32 N is applied to the 5.0-kg box pushing it against the 3.0-kg box, and, as a result, both boxes slide along the floor. How large is the contact force between the two boxes?
F - F2 = m1a F1 = m2a (F1 and F2 are equal) --> F - m2a = m1a F = a (m1 + m2) --> a = 32/8 --> 4m/s F1 = (3)(4) --> *F1 = 12N* *A*
In a grocery store, you push a 13.1-kg shopping cart horizontally with a force of 11.2 N. If the cart starts at rest, how far does it move in 3.00 s?
F = ma 11.2 = (13.1)a a = 0.855 X = V0t + (1/2)at^2 X = 0 + 1/2(0.855)(3)^2 X = 3.85 m
Newton's 2nd Law says that:
Ftotal = ma
14.) In Figure 6-1, the block of massm is at rest on an inclined plane that makes an angleθ with the horizontal. The force of static frictionf must be such that
Fx = Fcosθ F = mg f = mgcosθ *B* -because initially at rest (no speed)
19.) A constant force of 20 N is applied to an object of mass 8.0 kg at an angle of 25° with the horizontal. What is the work done by this force on the object if it causes a displacement of 2.0 m along the horizontal direction?
Fx = Fcosθ Fx = 20cos25 Fx = *36J* *C*
11.) A 40.0-kg suitcase is being pulled along the ground by means of a strap which exerts a force of 10.0 N at an angle of 43.0° above the horizontal. What is the normal force exerted by the ground?
Fy = 10sin40 --> Fy = 6.43 Fy and N are upward forces weight is a downward force so.... N + Fy = W W = mg --> W = 392 N = 392 - 6.43 --> *N = 385* *B*
Newton's 3rd Law
If one object is exerting a force on a second object, then the second object is also exerting a force back on the first object. The two forces have exactly the same magnitude but act in opposite directions. F B->A = negative F A->B -force one pushes on force 2 with a magnitude and force 2 pushes back with the same magnitude --> <--
kinetic friction
If the object is moving, the frictional force is called the kinetic friction -Kinetic friction is usually smaller than static friction. -The form is similar: fk = μkN
Consider a scale on an accelerating elevator:
If there is a net upward acceleration, a person's weight in the elevator will appear to be greater than it is. N - mg = ma N = ma + mg -you have a greater weight going up in the elevator because you need more of a force to prevent vertical motion and to balance - normal force - weight (mg) = mass x acceleration -weight = mass x gravity
Newton's Gravitational Constant
In MKS units (SI units), the gravitational constant has a value of: G = 6.673 x 10^-11 N•m^2/kg^2 On the surface of the earth: F = G (ME m / RE^2) = mg G (ME/RE^2) = 6.673 x 10^-11 (5.98 x 10^24 / (6.38 x 10^6)^2) G (ME/RE^2) = 9.80 N = (9.80 kg•m/s^2 ME = mass of the earth -m = object on the surface of the earth -r = radius -this is why gravity = 9.80 -know G gets smaller when moving away from the center of the earth because of the radius of the earth (RE)
Non-fundamental
Just about everything else will boil down to the electromagnetic forces from interactions of atoms and molecules.
The kinetic energy is defined as:
KE = 1/2 mv^2 -kinetic energy = only when moving (velocity dependent)
When Skylab reentered the Earth's atmosphere on July 11, 1979, it broke into a myriad of pieces. One of the largest fragments was a 1770-kg lead-lined film vault, and it landed with an estimated speed of 120 m/s. What was the kinetic energy of the film vault when it landed?
KE = 1/2mv^2 KE = 1/2(1770)(120^2) KE = 12744000 J KE = 12.7 MJ A megajoule (MJ) is 1,000,000 J
Force F1 does 5 J of work in 10 seconds, force F2 does 3 J of work in 5 seconds, force F3 does 6 J of work in 18 seconds, and force F4 does 25 J of work in 125 seconds. Rank these forces in order of increasing power they produce. Rank forces from largest power they produse to smallest. To rank items as equivalent, overlap them.
Largest F2 F1 F3 F4 Smallest P = W/t
CQ: Is it possible to do work on an object that remains at rest?
No -F•d = W
Work & Kinetic Energy
Say we apply some force, F, along some distance, d. We have just done work. Work is the application of a force along a displacement.
CQ: Three blocks of mass, 3m, 2m, and 1m, are connected by strings and pulled with constant acceleration (a). What is the relationship between the tension in each of the strings? 3m ---- 2m ---- 1m ---->
T1 > T2 > T3 -T1 pulls the whole set of blocks along, so it must be the largest. T2 pulls the last two masses, and T3 only pulls the last mass a ----------->
16.) T/F: Work done by a force on an object can be a positive or a negative quantity
True
If we apply the force at some angle to the displacement, the work done by us is given as:
W = (Fcosθ)d -any force that goes along the direction of movement/motion = work -Fcosθ = Fx -d = distance traveled/displacement -if pulling horizontal = only worried about work in the x-axis [ ] --> F
A 57-kg packing crate is pulled with constant speed across a rough floor with a rope that is at an angle of 37º above the horizontal. If the tension in the rope is 142 N, how much work is done on the crate by the rope to move it 6.1 m?
W = F(cosθ)d W = 142(cos37)6.1 W = 691.8 J
Chapter 7
Work & Kinetic Energy
Traveling in circles
It is sometimes more useful to talk about the time an object travels, rather than in terms of position and velocity vectors. Circular motion is one of those situations. We can define the *period, T* of the motion
static friction
If the object is at rest, this is the force of static friction fsMAX = μsN -friction that holds the object in place
Equilibrium
*If an object is in equilibrium, there is no acceleration* ∑Fx = m ax = 0 ∑Fy = m ay = 0 If we have multiple forces, the condition of equilibrium means that the net force is zero.
Example: A 20.0 kg sled is being pulled across a horizontal surface at a constant velocity. The pulling force has a magnitude of 80.0 N and is directed at an angle of 30.0º above the horizontal. Determine μk.
-draw picture: in notes Fx = Fcosθ Fy = Fsinθ weight = mg (arrow down) Normal force (arrow up, equal to weight) so Fx = max Fy = may (constant velocity means zero acceleration, so F = 0) -friction force opposes the motion (fk to the left and fx to the right) -Newtons 2nd law along x: Fx - Fk = 0 -Newton's 2nd law along y: Fy - weight + normal force = 0 fk=fx --> fx = cosθ --> fx = 80cos30 --> fx = 69.3 N N = W-Fy --> N = mg-Fsinθ --> N = (20)(9.81) - 80sin30 --> N = 156.2 N μk = fk/N --> μk = 69.3/156.2 --> *μk = 0.444*
1 Newton of force is the force needed to give an acceleration of
1 m/s^2 to an object of 1 kilogram. [N] = [kg m/s2]
Two forces, F⃗ 1 and F⃗ 2, act at a point, as shown in the picture. (Figure 1) F⃗ 1 has a magnitude of 9.60 N and is directed at an angle of α = 56.0º above the negative x axis in the second quadrant. F⃗ 2 has a magnitude of 6.00 N and is directed at an angle of β = 52.2º below the negative x axis in the third quadrant. 1.) What is the x component Fx of the resultant force? 2.) What is the y component Fy of the resultant force? 3.) What is the magnitude F of the resultant force? 4.) What is the angle γ that the resultant force forms with the negative x axis? In this problem, assume that positive angles are measured clockwise from the negative x axis
1.) -cos(56) = x1/9.6 --> x1 = -5.368 -cos(52.2) = x2/6 --> x2 = -3.677 Fx = x1 + x2 --> *Fx = -9.05 N* 2.) sin(56) = y1/9.6 --> y1 = 7.959 sin(52.2) = y2/6 --> y2 = 4.741 Fy = y1 - y2 --> *Fy = 3.2 N* 3.) F^2 = Fx^2 + Fy^2 F^2 = 9.05^2 + 3.2^2 *F = 9.6 N* 4.) tanγ = Fy/Fx tanγ = 3.2/9.05 *γ = 19.6º*
A force of magnitude 7.50 N pushes three boxes with masses m1 = 1.30kg, m2 = 3.20kg, and m3 = 4.90kg, as shown in the figure (Figure 1). (Ignore friction.) 1.) Find the magnitude of the contact force between boxes 1 and 2 2.) Find the magnitude of the contact force between boxes 2 and 3.
1.) 1.3 + 3.2 + 4.9 = 9.4 kg total a = 7.50/9.40 --> a = 0.798 F = ma --> (8.10 kg)(0.798 m/s²) --> *F = 6.46 N* 2.) F = (4.90)(0.798) --> *F = 3.91 N*
In the situation shown in the figure, a person is pulling with a constant, nonzero force F⃗ on string 1, which is attached to block A. Block A is also attached to block B via string 2, as shown. For this problem, assume that neither string stretches and that friction is negligible. Both blocks have finite (nonzero) mass. 1.) correctly describe the relationship between the accelerations of blocks A and B? 2.) How does the magnitude of the tension in string 1, T1, compare with the tension in string 2, T2?
1.) Both blocks have the same acceleration. 2.) T1 > T2
An object acted on by three forces moves with constant velocity. One force acting on the object is in the positive x direction and has a magnitude of 7.0 N ; a second force has a magnitude of 5.0 N and points in the negative y direction. 1.) Find the magnitude of the third force acting on the object. 2.) Find the direction of the third force acting on the object. ( º from the + x-direction)
1.) F3^2 = Fx^2 + Fy^2 --> F3^2 = 7^2 + 5^2 --> *F3 = 8.6 N* 2.) tanθ = Fy/Fx -> tanθ = 5/7 --> θ = 35.5 --> 180 - 35.5 = *144.5*
When you push a 1.82-kg book resting on a tabletop, it takes 2.27 N to start the book sliding. Once it is sliding, however, it takes only 1.55 N to keep the book moving with constant speed. 1.) What is the coefficient of static friction between the book and the tabletop? 2.) What is the coefficient of kinetic friction between the book and the tabletop?
1.) Fs = μ m g --> 2.27 = μ(1.82)(9.81) --> μ = 0.127 2.) Fk = μ m g --> 1.55 = μ(1.82)(9.81) --> μ = 0.0868
IP A 37-kg crate is placed on an inclined ramp. When the angle the ramp makes with the horizontal is increased to 29º, the crate begins to slide downward. 1.) What is the coefficient of static friction between the crate and the ramp? 2.) At what angle does the crate begin to slide if its mass is doubled?
1.) Fx = Fcosθ --> Fx = 37cos29 --> Fx = 32.36 Fy = Fsinθ --> Fy = 37sin29 --> Fy = 17.94 F = Fy/Fx --> F = 17.94/32.36 --> *F = 0.55* 2.) mass doesn't matter = *29º*
A 6-kg bucket of water is being pulled straight up by a string at a constant speed. 1.) What is the tension in the rope? 2.) At a certain point the speed of the bucket begins to change. The bucket now has an upward constant acceleration of magnitude 3 m/s^2. What is the tension in the rope now? 3.) Now assume that the bucket has a downward acceleration, with a constant acceleration of magnitude 3 m/s^2. Now what is the tension in the rope?
1.) The answer is about 60 N. Since the bucket is not accelerating, the only force acting on it is gravity (6 kg x 10 m/s^2 = 60 N) 2.) T = mg + F --> T = (6)(10) + (6 x 3) --> *T = 78 N* 3.) Since the bucket is accelerating downwards, the rope is not fixed and only a part of the gravitational force is creating tension. Just subtract the downward acceleration from g and multiply by the mass. This will give you about 42 N. 10 - 3 = 7 x 6 = *42 N*
*Mark each of the following statements as true or false*. If a statement refers to "two objects" interacting via some force, you are not to assume that these two objects have the same mass. 1.) Every force has one and only one 3rd law pair force. 2.) The two forces in each pair act in opposite directions. 3.) The two forces in each pair can act on the same object or on different objects. 4.) The two forces in each pair may have different physical origins (for instance, one of the forces could be due to gravity, and its pair force could be a normal contact force). 5.) The two forces of a 3rd law pair always act on different objects. 6.) Given that two objects interact via some force, the accelerations of these two objects have the same magnitude but opposite directions. (Assume no other forces act on either object.)
1.) True 2.) True 3.) False 4.) False 5.) True 6.) False
You push two identical bricks across a tabletop with constant speed, v, as shown in the figure. In case 1, you place the bricks end to end; in case 2, you stack the bricks one on top of the other 1.) Is the force of kinetic friction in case 1 greater than, less than, or equal to the force of kinetic friction in case 2? 2.) what can you say about normal force and friction with both cases
1.) equal to 2.) The normal force is the same in the two cases, and friction is independent of surface area.
A skateboarder on a ramp is accelerated by a nonzero net force. For each of the following statements, state whether it is always true, never true, or sometimes true. 1.) The skateboarder is moving in the direction of the net force 2.) The acceleration of the skateboarder is at right angles to the net force. 3.) The acceleration of the skateboarder is in the same direction as the net force. 4.) The skateboarder is instantaneously at rest.
1.) sometimes true. (for example, if the skateboarder is moving down the ramp the motion is in the same direction as the net force. On the other hand, if the skateboarder moves up the ramp, the direction of motion is opposite to the direction of the net force. In general, there is no reason for the direction of motion to be in the same direction as the net force) 2.) never true (the acceleration of an object is always in the direction of the net force) 3.) always true (look at 2) 4.) sometimes true (again, the fact that an object accelerates in a certain direction tells you nothing about its direction of motion, or whether it is instantaneously at rest,an example would be a skateboarder coasting upward on a ramp. At the skateboarder's highest point, he or she is instantaneously at rest, though still accelerating in a direction pointing down the ramp)
Chadwick now needs to push the piano up a ramp and into a moving van. (Figure 2) The piano slides up the ramp without friction. Is Chadwick strong enough to push the piano up the ramp alone or must he get help? To solve this problem you should start by drawing a free-body diagram. 1.) Determine the object of interest for this situation 2.) Now draw the free-body diagram of the piano in this new situation
1.) the piano 2.) weight arrow straight down, normal force arrow goes up and to the left, and pushing force goes up and to the right
Suppose that you are asked to solve the following problem: Chadwick is pushing a piano across a level floor (see the figure). (Figure 1) The piano can slide across the floor without friction. If Chadwick applies a horizontal force to the piano, what is the piano's acceleration? 1.) Determine the object of interest for the situation described in the problem introduction. 2.) Identify the forces acting on the object of interest. 3.) free-body diagram you have drawn for the piano
1.) you should draw a free-body diagram for the piano 2.) gravitational force acting on the piano (piano's weight), force of the floor on the piano (normal force), and force of Chadwick on the piano 3.) normal force arrow up, weight arrow down, and pushing force going right
Inertial reference frames
According to Newton's 1st Law, if there is no net force, then an object at rest remains at rest and an object moving at a constant velocity remains at a constant velocity. Can we tell the difference between being at rest or moving at a constant velocity? We call a reference frame (coordinate system) that remains stationary or a reference frame moving at a constant velocity inertial reference frames. If our frame of reference is accelerating, it is not an inertial reference frame!
force.
Any sort of push or pull that causes or alters motion is called a force -force gives rise to acceleration
A 26 kg suitcase is being pulled with constant speed by a handle that is at an angle of 30º above the horizontal. If the normal force exerted on the suitcase is 180 N, what is the force F applied to the handle?
Forces on the suicase are 1. Weight Mg downward 2. Applied force F1 3. Normal force Fn = 180 N upward 4. Frictional force (horizontal and opposite to direction in which the suitcase moves) The suitcase is being pulled horizontally. Therefore net vertical force = 0 F1 x sin(θ) + Fn - Mg = 0 F1 x sin30 + 180 - 26(9.81) = 0 *F1 = 150 N*
Normal forces
If objects are constantly subject to gravitational forces, why am I not falling through the floor right now? According to Newton's 1st Law, if an object is at rest, there must be no net force. If gravity is exerting a force on me, and I am in contact with the floor, the floor is exerting a force on me. From Newton's 3rd Law, the force of the floor on me is equal to the force of me on the floor. This reaction force is the *normal force* -there has to be a force pushing upwards to balance out gravity pulling = normal force -normal force and weight have to balance each other out/be equal in order for there to be no vertical movement
Ropes and tension
Many times we will use a rope to transfer force from one object to another. We usually make several assumptions about the rope: 1) the rope's mass is negligible (has no mass) 2) the rope does not change length as a force is exerted on the rope
Newton's 2nd Law
Newton observed that the relationship between force and mass was linear and concluded: ∑F = ma a = ∑F/m -When forces act on an object, it will cause the object to accelerate in the direction of the net force (the sum of the forces). The amount of acceleration is given by the net force divided by the object's mass. -force produced acceleration -∑F is net force
Example: Two blocks are sitting on a smooth table. Block A has a mass of 5 kg. Block B has a mass of 10 kg. You push on block A with a force of 15 N. What is the acceleration of block A? Is it different from block B?
Newton's 2nd Law for block A: mA x a = Fapplied - Force of BA Newton's 2nd Law for block B: mB x a = Force of AB Fapplied = (mA + mB) x a --> a = 15/15 --> *a = 1 m/s^2) *No, it is the same acceleration for B because its the same force* B pushes on A and A pushes on B = forces are the same magnitude in opposite directions (Newton's 3rd law)
The rules of forces were worked out over 300 years ago. These rules are known as
Newton's Laws.
CQ: You put your book on the bus seat next to you. When the bus stops suddenly, the book slides forward off the seat. Why?
No net force acted on the book -Newton's first law -book was moving before -motion relative to the bus -the book was initially moving forward (because it was moving on the bus). When the bus stopped, the book *contiueds moving forward*, which was it *initial state of motion*, and therefore it slid forward off the seat
CQ: A block of mass (m) rests on the floor of an elevator that is moving upward at a constant speed. What is the relationship between the force due to gravity and the normal force on the block?
Normal force = mass (mg) -the block is moving at a constant speed, so it must have no net force on it. The forces on it are N (up) and mg (down), so N = mg. Just like the block at rest on a table
Uniform Circular Motion
Say that an object is traveling with a constant (uniform) speed along a circular path. The object is accelerating at a constant magnitude by constantly changing direction. -This is what we call uniform circular motion. -The force causing the circular motion is the *centripetal force* -uniform because the magnitude of velocity/speed is always constant/tangent to the trajectory -*if* vector velocity is changing = acceleration
Fundamental Force: Gravity
So far we have mentioned a gravitational acceleration of 9.80 m/s2 downward arising from the gravitational force. Why this value of 9.80 m/s2? Newton deduced that all object with mass exert an attractive force on each other. If an object with mass m1 and another object with mass m2 are a distance r apart, the magnitude of the attracting force is: F = G (m1 x m2/r^2) -if there are 2 objects of 2 masses, they interact/there is a force between the 2 -r is the distance between 2 objects/radius
A man attempts to pick up his suitcase of weight ws by pulling straight up on the handle. However, he is unable to lift the suitcase from the floor. Which statement about the magnitude of the normal force n acting on the suitcase is true during the time that the man pulls upward on the suitcase?
The magnitude of the normal force is equal to the magnitude of the weight of the suitcase minus the magnitude of the force of the pull.
Now assume that the man of weight wm is tired and decides to sit on his suitcase. Which statement about the magnitude of the normal force n acting on the suitcase is true during the time that the man is sitting on the suitcase?
The magnitude of the normal force is equal to the sum of the magnitude of the man's weight and the magnitude of the suitcase's weight.
How do we measure force?
The most common means of measuring a force is through the use of a spring scale. Springs exert different forces as they are extended or compressed. We can measure the length of the spring to calibrate the force. (We will do this later in the course)
Period and Frequency (f)
The period is the time that is takes to travel once around a circle. The frequency is the number of times traveled around the circle each second. We define these quantities as: T = 2πr / V frequency (f) = 1/T
Acceleration
The position vector has a constant magnitude as does the velocity vector. They are changing vector quantities because the direction changes. What direction is the acceleration? ac = change in velocity / change in time = V^2 / r -ac is centripetal acceleration -in the picture (slide 16) the 2 radii that meet are equivalent to the acceleration. they create an angle θ, so the distance between the 2 accelerations is the velocity
A 93-kg water skier floating in a lake is pulled from rest to a speed of 10 m/s in a distance of 24 m . What is the net force exerted on the skier, assuming his acceleration is constant?
V^2 = V0^2 + 2ax 10^2 = 0 + 2a(24) a = 2.083 F = ma F = 93 x 2.083 F = 193.719 N
Friction
When an object in the real world is in contact with another object, there is a force countering any applied force. Experiments show that the force is independent of the contact area and the magnitude of the force is proportional to the magnitude of the normal force. -friction is a vector -μ = coefficient of friction
Example: A 40 kg child is sitting in a sled down a flat, grassy hill. The hill is angled 30º above the horizon. What must the coefficient of friction be to keep the sled from moving?
Wy = 40cos30 --> Wy = 34.6 Wx = 40 sin30 --> Wx = 20 ∑Fx = ma ∑Fy = ma (both equal to zero because acceleration is zero so force is zero, means there is no movement in either direction Wx - fs = 0 --> Wx = fs N-Wy = 0 --> Wy = N fs = μ N --> 20 = μ(34.6) --> *μ = 0.577* μ can also be found using tan30 in this case
Example: What is the tension in the rope and the acceleration of the blocks? picture: slide 11: top weight is W1 = 422 N and bottom weight is W2 = 185 N
both accelerations are equal ∑F = ma (top block) --> only force acting is tension (T) --> T = ma ∑F = ma --> W2 - T = ma m = w/g --> m1 = 43 and m2 = 18.9 W2 = (m1 + m2) a --> 185 = (43 + 18.9) a --> *a = 2.99 m/s^2* T = m1a --> T = (43)(2.99) --> *T = 129N*
change in velocity =
change in acceleration -force produces acceleration
net force needs to be
constant -if there is acceleration, then there is a net force
CQ: A very large truck sits on a frozen lake. Assume there is *no friction* between the tires and the ice. A fly suddenly smashes against the front window. What will happen to the truck?
it moves backward at a constant speed (because a ****ing tiny ass fly can move a big ass truck now) -when the fly hit the truck, it exerted a force on the truck (only for a fraction of a second). So, in this time period, the truck accelerated (backward) up to some speed. After the fly was squashed, it no longer exerted a force, and the truck simply continued moving at a constant speed
CQ: Consider a cart on a horizontal, frictionless table. Once the cart has been given a push and released, what will happen to the cart?
it will continue with a constant velocity -no friction -no net force = no acceleration
An example: Pulleys (Equilibrium) If block 1 weighs 55 N, block 2 weighs 35 N and block 3 weighs 28 N, what is the magnitude of the table's normal force on block 1? picture: block 1 and two are on the left side resting on a table and block 3 is by itself hanging
normal force and the weight of block 1 and 2 are acting on the table -nothing moves = everything in equilibrium ∑Fy = may = 0 (shouldn't have a net force on y = balance) W1 + W2 - W3 - N = 0 --> Normal force on the table (N) = 62 N -W3 is equal to tension -N is subtracted along with W3 because they are acting in opposite directions of W1 and W2
What is the normal force if the elevator is in free fall downward?
normal force would be zero
When an object rests on a surface, there is always a force perpendicular to the surface; we call this the
normal force, denoted by n⃗
Example: A 5.7 kg block is on a 35º incline connected by a mass-less rope over an ideal pulley to a vertically hanging mass. What must be the mass of the block for the system to be in equilibrium?
slide 12 m1 = 5.7 kg, θ = 35º, and m2 = ? -on m2 (mass of the block on the side) the only 2 forces acting are tension (T) up and weight (W2) down -on m1 (mass of the block on the incline) the only 2 forces acting are tension (T) up at the angle and weight (W1) down Wx = W sinθ Wy = W cosθ W-T = 0 --> W = T --> m2g = T T-Wx = 0 --> T = Wx --> Wx = m1g sinθ since T is equal to mg and equal to Wx --> m2g = m1g sinθ --(g cancels out)--> m2 = m1 sinθ --> m2 = 5.7sin35 --> *m2 = 3.3 kg*
pulley example with tension (same weight on both sides of pulley)
slide 9 (left side picture) -have a pair of forces in both directions -have the same tension pulling up and the same weight pulling down (in the one on the powerpoint they are connected to the same box on both sides) = T + T - W = 0 --> 2T = W --> T = W/2 (because no motion = static equilibrium)
pulley example with tension (weight on one side)
slide 9 (right side picture) -if you were pulling on the other side of the rope that is empty with a constant speed = net effect the same when pulling the rope with a constant velocity and when jest holding the rope still because acceleration = 0 --> ∑Fy = ma = 0 --> T - W = 0 --> T = W (have the same magnitude but opposite directions because they balance out)
