Physics 2053C Exam 2 Review
In uniform circular motion, which of the following quantities are constant? instantaneous velocity speed the magnitude of the net force centripetal acceleration
-speed -the magnitude of the net force
Conversion between force units
1 pound = 4.45 N 1 N = 0.225 pound Correspondence between mass and weight, assuming g = 9.80 m/s^2 1 kg = 2.20 lbs 1 lbs = 0.454 kg = 454 grams
A 1.2 kg book lies on a table. The book is pressed down from above with a force of 15 N. What is the normal force acting on the book from the table below?
A 1.2 kg book lies on a table. The book is pressed down from above with a force of 15 N. What is the normal force acting on the book from the table below? book is in static equilibrium ASSESS The magnitude of the normal force is larger than the weight of the book. From the table's perspective, the extra force from the hand pushes the book further into the atomic springs of the table. These springs then push back harder, giving a normal force that is greater than the weight of the book.
Example 5.4 Tension in towing a car A car with a mass of 1500 kg is being towed at a steady speed by a rope held at a 20° angle from the horizontal. A friction force of 320 N opposes the car's motion. What is the tension in the rope?
A car with a mass of 1500 kg is being towed at a steady speed by a rope held at a 20° angle from the horizontal. A friction force of 320 N opposes the car's motion. What is the tension in the rope? Known theta = 20 degrees Mass = 1500 kg Force = 320 N We have to find tension = T Okay so, the car is moving in a straight line at a constant speed so it is in dynamic equilibrium and must have a net force = ma = 0. There are three contact forces acting on the car: tension force, friction, normal force, AND the long-range force of gravity Normal force is straight up weight is straight down friction is pointing in the opposite direction that the car is traveling in tension is point in the direction of the tow truck SOLVE This is still an equilibrium problem, even though the car is moving, so our problem-solving procedure is unchanged. With four forces, the requirement of equilibrium is Σ Fx = nx + Tx + fx + wx = max = 0 Σ Fy = ny + Ty + fy + wy = may = 0 We can again determine the horizontal and vertical components of the forces by "reading" the free-body diagram. The results are shown in the table. With these components, Newton's second law becomes T cosθ − f = 0 n + T sinθ − w = 0 The first equation can be used to solve for the tension in the rope: T=(F/cos(theta)) = 320N / cos(20) = 340 N It turned out that we did not need the y-component equation in this problem. We would need it if we wanted to find the normal force .
A 50-kg student (mg = 490 N) gets in a 1000-kg elevator at rest and stands on a metric bathroom scale. As the elevator accelerates upward, the scale reads A. > 490 N B. 490 N C. < 490 N but not 0 N D. 0 N
A. > 490 N
You are riding in an elevator that is accelerating upward. Suppose you stand on a scale. The reading on the scale is A. Greater than your true weight. B. Equal to your true weight. C. Less than your true weight.
A. Greater than your true weight.
The drag force pushes opposite your motion as you ride a bicycle. If you double your speed, what happens to the magnitude of the drag force? A. The drag force increases. B. The drag force stays the same. C. The drag force decreases.
A. The drag force increases.
Newton's 2nd Law
Acceleration is produced when a force acts on a mass. The greater the mass (of the object being accelerated) the greater the amount of force needed (to accelerate the object).
A cyclist is rounding a 21-m radius curve at 11 m/s . What is the minimum possible coefficient of static friction between the bike tires and the ground?
Angular acceleration is v²/r = (11m/s)²/21m ≈ 5.76m/s² acceleration due to g is 9.81m/s². 5.76m/s² ÷9.8m/s² ≈ .585
Two boxes are suspended from a rope over a pulley. Each box has weight 50 N. What is the tension in the rope? A. 25 N B. 50 N C. 100 N D. 200 N
B. 50N
A box with a weight of 100 N is at rest. It is then pulled by a 30 N horizontal force. Does the box move? A. Yes B. No C. Not enough information to say
B. No 30 N < fs max = 40 N
Two identical satellites have different circular orbits. Which has a higher speed? A. The one in the larger orbit B. The one in the smaller orbit C. They have the same speed.
B. The one in the smaller orbit
A box on a rough surface is pulled by a horizontal rope with tension T. The box is not moving. In this situation: A. fs > T B. fs = T C. fs < T D. fs = µsmg E. fs = 0
B. fs = T Newton's First Law
In general, the coefficient of static friction is A. Smaller than the coefficient of kinetic friction. B. Equal to the coefficient of kinetic friction. C. Greater than the coefficient of kinetic friction.
C. Greater than the coefficient of kinetic friction.
A satellite orbits the earth. A Space Shuttle crew is sent to boost the satellite into a higher orbit. Which of these quantities increases? A. Speed B. Angular speed C. Period D. Centripetal acceleration E. Gravitational force of the earth
C. Period
A box is being pulled to the right at steady speed by a rope that angles upward. In this situation: A. n > mg B. n = mg C. n < mg D. n = 0 E. Not enough information to judge the size of the normal force
C. n < mg
A 50-kg student (mg = 490 N) gets in a 1000-kg elevator at rest and stands on a metric bathroom scale. Sadly, the elevator cable breaks. What is the reading on the scale during the few seconds it takes the student to plunge to his doom? A. > 490 N B. 490 N C. < 490 N but not 0 N D. 0 N
D. 0 N
Which of these objects is in equilibrium? A. A car driving down the road at a constant speed B. A block sitting at rest on a table C. A skydiver falling at a constant speed D. All of the above
D. All of the Above
A box is being pulled to the right over a rough surface. T > fk, so the box is speeding up. Suddenly the rope breaks. What happens? The box A. Stops immediately. B. Continues with the speed it had when the rope broke. C. Continues speeding up for a short while, then slows and stops. D. Keeps its speed for a short while, then slows and stops. E. Slows steadily until it stops.
E. Slows steadily until it stops.
Finding the force to slide a sofa Carol wants to move her 32 kg sofa to a different room in the house. She places "sofa sliders," slippery disks with µk = 0.080, on the carpet, under the feet of the sofa. She then pushes the sofa at a steady 0.40 m/s across the floor. How much force does she apply to the sofa?
F = fk = µkn = µkmg = (0.080)(32 kg)(9.80 m/s2) = 25 N
When you are on a swing, and at the lowest point of your motion, is your apparent weight greater than, less than, or equal to your true weight? Explain.
In a circular motion there must be a centripetal force directed towards the center. At the bottom of the swing trajectory we can pretend you are in uniform circular motion, so the net force points up toward the center of the circle. Therefore, the swing must exert a larger upward force on you than the earth does in the downward direction. So your apparent weight is greater than your true weight.
Why is it impossible for an astronaut inside an orbiting space station to go from one end to the other by walking normally?
In an orbiting station, after one foot pushes off there isn't a force to bring the astronaut back to the "floor" for the next step.
Newton's 1st Law
Isaac Newton's first law of motion, also known as the law of inertia, states that an object at rest will stay at rest and an object in motion will stay in motion with the same speed and direction unless acted upon by unbalanced force.
Example 5.1 Forces supporting an orangutan An orangutan weighing 500 N hangs from a vertical rope. What is the tension in the rope?
PREPARE The orangutan is at rest, so it is in static equilibrium. The net force on it must then be zero. FIGURE 5.1 first identifies the forces acting on the orangutan: the upward force of the tension in the rope and the downward, long-range force of gravity. These forces are then shown on a free-body diagram, where it's noted that equilibrium requires Net Force = 0 Because the tension (T) vector points straight up, in the positive y-direction, its y-component is Ty = T. Because the weight vector (W) points straight down, in the negative y-direction, its y-component is wy = − w. This is where the signs enter. With these components, Newton's second law become T − w = 0 This equation is easily solved for the tension in the rope: T = w = 500 N ASSESS It's not surprising that the tension in the rope equals the weight of the orangutan. That gives us confidence in our solution.
Anjay's mass is 70 kg. He is standing on a scale in an elevator that is moving at 5.0 m/s. As the elevator stops, the scale reads 750 N. Before it stopped, was the elevator moving up or down? How long did the elevator take to come to rest?
PREPARE The scale reading as the elevator comes to rest, 750 N, is Anjay's apparent weight. Anjay's actual weight is w = mg = (70 kg)(9.80 m/s2) = 686 N SOLVE We can read components of vectors from the figure. The vertical component of Newton's second law for Anjay's motion is ΣFy = n − w = ma(y) n is the normal force, which is the scale force on Anjay, 750 N. w is his weight, 686 N. We can thus solve for ay a(y) = (n-w)/m = (750 - 686) / 70 kg = +0.91 m/s^2 The acceleration is positive and so is directed upward, exactly as we assumed—a good check on our work. The elevator is slowing down, but the acceleration is directed upward. This means that the elevator was moving downward, with a negative velocity, before it stopped
Variation in your apparent weight is desirable when you ride a roller coaster; it makes the ride fun. However, too much variation over a short period of time can be painful. For this reason, the loops of real roller coasters are not simply circles like (Figure 1) . A typical loop is shown in (Figure 2) . The radius of the circle that matches the track at the top of the loop is much smaller than that of a matching circle at other places on the track.
Reason: The radius of the loop decreases as the carts enter and exit the loop. The centripetal acceleration is smaller for larger radius loops and larger for smaller radius loops. This means the centripetal acceleration increases from a minimum at the entry to the loop to a maximum at the top of the loop and then decreases as the cars exit the loop. This prevents a sudden change of acceleration, which can be painful. This also limits the largest accelerations to the top of the loop, so that riders only experience the maximum acceleration for a portion of the trip. Assess: This is reasonable. If the cars entered a small radius loop directly, the centripetal acceleration would increase suddenly.
A car with a mass of 1500 kg is being towed by a rope held at a 20° angle to the horizontal. A friction force of 320 N opposes the car's motion. What is the tension in the rope if the car goes from rest to 12 m/s in 10 s?
T= (m x a(x) + f ) / cos(20) T= (1500kg x 1.2 m/s^2 + 320N) = 2300 N
An elevator is suspended from a cable. It is moving upward at a steady speed. Which is the correct free-body diagram for this situation?
Tension in the string and Weight are the same
Newton's 2nd Law
The BIG Equation. Newton's second law of motion can be formally stated as follows: The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object.
Newton's 3rd Law
These two forces are called action and reaction forces and are the subject of Newton's third law of motion. Formally stated, Newton's third law is: For every action, there is an equal and opposite reaction.
Two satellites have circular orbits with the same radius, one with more mass than the other. Which has a higher speed?
They have the same speed
Chapter 6 Learning Goal
To learn about motion in a circle, including orbital motion under the influence of a gravitational force.
Definition of Apparent Weight
W(app)= magnitude of supporting contact forces
What is the orbital period of a spacecraft in a low orbit near the surface of Mars? Spacecraft have been sent to Mars in recent years. Mars is smaller than Earth and has correspondingly weaker surface gravity. On Mars, the free-fall acceleration is only 3.8m/s2 . What is the orbital period of a spacecraft in a low orbit near the surface of Mars? Assume the radius of the satellite's orbit is about the same as the radius of Mars itself, rMars = 3.37×106m .
apply Keplers third law as T^2 = 4pi^2 R^3/GM as gravity g = GM/R^2 GM = gR^2 so T^2 = 4pi^2*R^3/(gR^2) T^2 = 39.4384 * 3.37e6/(3.8) T^2 = 34.97 *10^6 T = 5914.02 secs T = 5914.04/60, = 98.534 min T = 98.534/60 = 1.64 Hrs
Drag Force
comparable to frictioin
Planet X has free-fall acceleration 8 m/s2 at the surface. Planet Y has twice the mass and twice the radius of planet X. On Planet Y
g = 4 m/s^2
The mass of the sun is 1.99×1030kg and its distance to the Earth is 1.50×1011m. What is the gravitational force of the sun on the earth?
http://www.chegg.com/homework-help/questions-and-answers/part-mass-sun-199-1030kg-distance-earth-150-1011m-gravitational-force-sun-earth-part-b-mas-q6117147
A rod is free to slide on a frictionless sheet of ice. One end of the rod is lifted by a string. If the rod is at rest, which diagram in FIGURE 5.3 shows the correct angle of the string?
if the string is at an angle there must be friction in order to hold the rod in place. since the rod is on a frictionless surface the piece of sting that's straight up and the rod at an angle is actually at equilibrium
The box is sitting on the floor of an elevator. The elevator is accelerating upward. The magnitude of the normal force on the box is
n > mg
normal force
normal force is the component, perpendicular to the surface (surface being a plane) of contact, of the contact force exerted on an object by, for example, the surface of a floor or wall, preventing the object from falling.
In Equilibrium...
the sums of the x- and y-components of the force are 0
A car drives over the top of a hill that has a radius of 51 m . What maximum speed can the car have without flying off the road at the top of the hill?
v = sqrt(r x g) = v = sqrt(51 x 9.8) = 22.25 m/s
A roller coaster car is going over the top of a 14-m-radius circular rise. At the top of the hill, the passengers "feel light," with an apparent weight only 60 % of their true weight. How fast is the coaster moving?
v=(g x r x 0.4)^1/2 v=(9.8 x 14m x 0.4m) v=7.41 m/s
The passengers in a roller coaster car feel 50% heavier than their true weight as the car goes through a dip with a 40 m radius of curvature. What is the car's speed at the bottom of the dip?
v=(gr/2)^1/2 v=(9.8 x 40m / 2)^1/2 14 m/s
A wind turbine has 12,000 kg blades that are 35 m long. The blades spin at 21 rpm . If we model a blade as a point mass at the midpoint of the blade, what is the inward force necessary to provide each blade's centripetal acceleration?
w = 21*2π/60 = 2.199 rad/sec Fi = (m*R/2)*w² = 1015476 N when the blade is horizontal ans = 1015476 N
how can you find weight from mass? (what is the equation?)
w=mg
A car drives over the top of a hill that has a radius of 48 m. What maximum speed can the car have without flying off the road at the top of the hill?
when we have: mg = mv2/R -- the car will not fly over => v = √(gR) = √(9.8m/s2*48m) =21.69m/s so when the v<= 21.69 m/s, the car will not fly over, and otherwise it will.
Weightlessness
• A person in free fall has zero apparent weight. • "Weightless" does not mean "no weight." • An object that is weightless has no apparent weight.
Rolling Friction
• A wheel rolling on a surface experiences friction, but not kinetic friction: The portion of the wheel that contacts the surface is stationary with respect to the surface, not sliding. • The interaction between a rolling wheel and the road can be quite complicated, but in many cases we can treat it like another type of friction force that opposes the motion, one defined by a coefficient of rolling friction µr : fr = µrn
Normal Forces
• An object at rest on a table is subject to an upward force due to the table. • This force is called the normal force because it is always directed normal, or perpendicular, to the surface of contact. • The normal force adjusts itself so that the object stays on the surface without penetrating it.
Looking Ahead: Circular Motion
• An object moving in a circle has an acceleration toward the center, so there must be a net force toward the center as well.
Gravity on Other Worlds
• If we use values for the mass and the radius of the moon, we compute g(moon) = 1.62 m/s2. • A 70-kg astronaut wearing an 80-kg spacesuit would weigh more than 330 lb on the earth but only 54 lb on the moon.
Gravity on Other Worlds
• If you traveled to another planet, your mass would be the same but your weight would vary. The weight of a mass m on the moon is given by w=m x g(moon)
Kinetic Friction
• Kinetic friction, unlike static friction, has a nearly constant magnitude given by fk = µkn where µk is called the coefficient of kinetic friction.
Mass and Weight
• Mass and weight are not the same thing. • Mass is a quantity that describes an object's inertia, its tendency to resist being accelerated. • Weight is the gravitational force exerted on an object by a planet: weight = -mg
Dynamics and Newton's Second Law
• The essence of Newtonian mechanics can be expressed in two steps: • The forces acting on an object determine its acceleration acceleration = net force / mass • The object's motion can be found by using in the equations of kinematics. • Thus Newton's second law, , is Σ F(x) = ma(x) Σ F(y) = ma(y) (These two are Newton's 2nd Law in Component Form)
Apparent Weight
• The only forces acting on the man are the upward normal force of the floor and the downward weight force: n = w + ma wapp = w + ma • Thus wapp > w and the man feels heavier than normal
Looking Ahead: Apparent Forces
• The riders feel pushed out. This isn't a real force, though it is often called centrifugal force; it's an apparent force.
Chapter 6 Preview Looking Ahead: Gravity and Orbits
• The space station appears to float in space, but gravity is pulling down on it quite forcefully. • You'll learn Newton's law of gravity, and you'll see how the force of gravity keeps the station in orbit.
Apparent Weight
• The weight of an object is the force of gravity on that object. • Your sensation of weight is due to contact forces supporting you. • Let's define your apparent weight wapp in terms of the force you feel: W(app)=
Equilibrium
• We say that an object at rest is in static equilibrium. • An object moving in a straight line at a constant speed (acceleration = 0) is in dynamic equilibrium. • In both types of equilibrium there is no net force acting on the object:
