Exam 2 Conceptual Class Questions

Three blocks of mass 3m, 2m, and m are connected by strings and pulled with constant acceleration a. What is the relationship between the tension in each of the strings?

1) T1 > T2 > T3 T1 pulls the whole set of blocks along, so it must be the largest. T2 pulls the last two masses, but T3 only pulls the last mass.

If you push with force F on either the heavy box (m1) or the light box (m2), in which of the two cases below is the contact force between the two boxes larger? in case A, there is force acting on m2 first (smaller bxo) then M1 right after (bigger) in case B, force switches and acts on taller box first then small

1) case A The acceleration of both masses together is the same in either case. But the contact force is the only force that accelerates m1 in case A (or m2 in case B). Since m1 is the larger mass, it requires the larger contact force to achieve the same acceleration

consider 2 identical blocks, one resting on a flat surface, and the other resting on an include for which case is the normal force greater? photo A depicts a rectangular block with a cube right in the middle resting. photo B depicts a block on a right triangle, looks like its sliding down it. 1. case a 2. case b 3. both the same (N=mg) 4. both the same ( 0<N<mg) 5. both the same (N=0)

1, case a in case a we know that N=W. in case b due to the angle of the incline, N<W. in fact we can see that N=W costheta

a plant is hung from wire as shown. whats the tension in each wireif the plant weights 20N? draw the free body diagram for the point where wires meet. take the horizontal direction for x, and the vertical direction for y and count the number of forces that have componenets in each direction.

2 x and 3 y components Strategy: • Clearly we are going to need FBDs • This plant is at rest, so all the forces are going to balance since • After we have FBDs, our 2nd Law equations will solve this problem for us. • Since we have forces with angles we are going to have to use components

Your little sister wants you to give her a ride on her sled. On level ground, what is the easiest way to accomplish this?

2) pulling her from the front In Case 1, the force F is pushing down (in addition to mg), so the normal force is larger. In Case 2, 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.

In the game of tetherball, the struck ball whirls around a pole. In what direction does the net force on the ball point?

3) Along the horizontal component of the tension force The vertical component of the tension balances the weight. The horizontal component of tension provides the centripetal force that points toward the center of the circle.

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:

3) 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, and it remains at rest. The truck accelerated away, essentially leaving the box behind!!

if a planet had twice the radius of the each what would its mass have to be in order to have the same acceleration of gravity

4 Me

Two blocks of masses 2m and m are in contact on a horizontal frictionless surface. If a force F is applied to mass 2m, what is the force on mass m ?

4) 1/3 F The force F leads to a specific acceleration of the entire system. In order for mass m to accelerate at the same rate, the force on it must be smaller! How small?? Let's see...

Antilock brakes keep the car wheels from locking and skidding during a sudden stop. Why does this help slow the car down?

4) µs > µk so static friction is better 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.

In which case does block m experience a larger acceleration? In (1) there is a 10 kg mass hanging from a rope and falling. In (2) a hand is providing a constant downward force of 98 N. Assume massless ropes.

5) case 2 In (2) the tension is 98 N due to the hand. In (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.

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?

5) the box does not move The static friction force has a maximum of µsN = 40 N. The tension in the rope is only 30 N. So the pulling force is not big enough to overcome friction.

A planet of mass m is a distance d from Earth. Another planet of mass 2m is a distance 2d from Earth. Which force vector best represents the direction of the total gravitation force on Earth?

The force of gravity on the Earth due to m is greater than the force due to 2m, which means that the force component pointing down in the figure is greater than the component pointing to the right. F2m = GME(2m) / (2d)2 = 1/2 GMm / d2 Fm = GME m / d2 = GMm / d2 2 (going a little to the right)

Two satellites A and B of the same mass are going around Earth in concentric orbits. The distance of satellite B from Earth's center is twice that of satellite A. What is the ratio of the centripetal force acting on B compared to that acting on A?

Using the Law of Gravitation: F=G x Mm/R2 we find that the ratio is .1/4

Two identical cement cylinders are attached to the opposite ends of a spring scale via very light ropes (the mass of which can be neglected) that run over frictionless pulleys as shown. When the same scale was suspended from the ceiling and one of the cylinders was hung from it, the scale indicated its weight is W newtons. What will the scale read in the configuration shown?

W newtons

Consider the three cases shown in the drawing in which the same force is applied to a box of mass M. In which case(s) will the magnitude of the normal force on the box be equal to (F sin θ + Mg)? case one has a right triangle w a cube on the hyp. a force is acting up on the cube. N is point up from cube Fg is point down from cube at angle case 2 has a cube sitting on a flat rectangle. there is a theta angle on the left of the cube, with the force F acting as hypotenuse, point OUT. N is point straight up from cube, Fg is pointing straight down from cube case 3 has the same cube and rectangle. everythign is the same but F is pointing IN

case 3 only

Consider two identical blocks, one resting on a flat surface, and the other resting on an incline. For which case is the normal force greater?

case A in Case A, we know that N = W. In Case B, due to the angle of the incline, N < W. In fact, we can see that N = W cos(θ).

A ping pong ball is shot into a circular tube that is lying flat (horizontal) on a tabletop. When the ping pong ball leaves the track, which path will it follow?

it will go straight Once the ball leaves the tube, there is no longer a force to keep it going in a circle. Therefore, it simply continues in a straight line, as Newton's First Law requires!

weight and tension will do what when in different directions

subtract

if tension and weight are both acting downward...

you must add them for the net force