HOMEWORK 5
SOLUTIONS

1. Al and George pushing the truck

(a) Here's the free-body diagrams for the car and the truck.



Both the truck and the Geo are accelerating. Therefore, the net horizontal force on each must be in the direction of the acceleration. If we ignore the air resistance and frictional effects of the non-driven tires not rolling perfectly, the forces must be ordered as follows:

f'R->G > NT->G = NG->T > fR->T > fR->G

That is, the friction force of the road on the Geo pushing it forward (and arising from the Geo's engine turning the tires so their friction is trying to push the road backwards) must be greater than the normal force the Geo feels pushing back on it from the truck (in order for it to accelerate forward). The normal force the truck exerts on the Geo and the Geo exerts on the truck must be equal by Newton's third law. For the friction pointing backwards, the friction felt by the truck must be less than the normal force pushing it forward. For the Geo, the friction force plus the push of the truck backwards should be less than the push of the road forwards (f').

(b) If the vehicles are accelerating uniformly for 5 minutes, start from an initial velocity of zero and end up at a velocity of 20 miles/hour, the average velocity is 10 miles/hour. So we can calculate the distance from



(c) Recall that when objects have an applied force but are not moving, the static friction force adjusts itself to counter the applied force (if it can), not to take its maximum value. Since the truck is not moving, the net force on it is 0 and the friction force between it and the road must be 1000 N backwards. Let's check to make sure this is possible. The normal force squeezing the truck's tires and the road together is in this case equal to the truck's weight, mg = (4000 kg)(10 N/kg) = 40,000 N. The maximum friction force is 0.1 times that or 4000 N. Clearly the car has to exert a bigger force.

2. Pushing a carriage

We can begin in a number of different ways. The most straightforward is to start with the velocity graph, since that is the one we have the most direct information about. It says at time t = 0 the carriage is moving with a constant velocity so on the velocity graph we draw a straight horizontal line. When she releases the carriage it slows and comes to a stop. We don't actually know how it does this, but if the only force acting on it is kinetic (moving) friction, that is likely to be a constant so, by N2, the acceleration should be constant. This would make the velocity a straight line angled downward on the velocity graph until it reaches 0 at time t2. This is the graph shown in the second picture below.

The position graph is easy to draw from this. When the velocity is constant, the position is changing at a constant rate. When the velocity is decreasing but positive, the position continues to increase, but curves over. This is shown in the first graph below.

The acceleration is easily inferred from the velocity graph. While the velocity is constant the acceleration is 0. While the velocity is decreasing at a uniform rate, the acceleration is constant negative. This is shown in the third graph below.

Since, by N2, the acceleration and the net force are proportional, the net force graph should look the same as the acceleration graph. The net force is made up of two parts: the friction, which is a constant negative force, and her push, which is positive. We easily draw the friction graph. We construct the graph of her push so that when it is added to the friction we get the net force graph, which we already know.

3. Pushing a box

Let's do these in a natural order.

The problem describes the force exerted by the worker. First it is a large force, then a smaller one, both positive. Finally, the force has to be negative. If we take the forces exerted in each region to be constant, we get that item (b), the force exerted by the worker could be graph E. Friction doesn't depend on speed so once the cart gets moving it should be constant and trying to keep the cart from moving, therefore negative in this example. Therefore, the force of friction, item (a) could be graph A.

The net force in this case is the sum of the horizontal force exerted by the worker and the friction. The vertical forces on the cart (gravity and normal force from the floor) cancel since the cart has no vertical acceleration. The sum of graphs A and C could be E, so this is what we choose for item (c).

Since a=F^net/m and m is constant, the graph for the acceleration of the cart and the net force it feels should have the same shape. Therefore, we choose for item (d) the graph E as well.

The last one to do is item (e), the velocity. We can get this by looking at the acceleration. Since it is experiencing constant acceleration in the first time segment, the velocity will increase linearly. In the second time segment the acceleration is 0 so the velocity should be constant. In the third time segment, the acceleration is constant negative so the velocity must drop linearly. Graph B does this.

Graph D represents the position, which was not asked for here.

4. The bridge at Mostar

(a) A falling object only feels the force of the earth pulling on it, it's weight. Therefore, its acceleration will be downward and its magnitude will be given by

a = F^net/m = W/m = mg/m = g

The key equations are



Using the last equation in the first , we get

Δy = -½ g (Δt)²

Putting in Δy = -23 m and solving for Δt gives



(b) After falling for a time Dt at an acceleration a, his change in speed would be

Δt = a Δt
vf = vi + a Δt
vf = 0 + (9.8 m/s²) (2.17 s) = (9.8) (2.17) m/s = 21.3 m/s

(c) It takes the diver 2.17 s to hit the water. The sound must then travel 23 m back up to the spectator. Sound travels at 340 m/s (gravity doesn't affect it), so it will take a travel time we can get from the equation



We know the speed of the sound and the distance, so we want to solve for the time:



The total time is therefore 2.17 s + 0.07 s = 2.24 s.

5. Riding an elevator

(a) Since the passenger is not accelerating, the two forces acting on him must be equal and opposite by N1. That is, the passenger's weight equals the upward (normal) force from the scale: Nscale->passenger = - Wearth->passenger. By N3, any pair of forces with their labels reversed -- two forces objects exert reciprocally on each other -- must be equal and opposite. Therefore Nscale->passenger = Npassenger -> scale. The sum of the two downward forces on the scale is also equal to the upward force on the scale, but that wasn't asked for.



(b) The only forces in the diagram that can change are the normal forces. We know that the weights don't change, no matter how an object moves (unless it moves far from the earth's surface). If the elevator is accelerating downward, it is trying to move away from the scale. So the normal force the elevator exerts on the scale will be reduced. Since the scale is trying to move away from the passenger, the normal forces between the two of them will also be reduced.

(c) When an object is accelerating, we can't apply N1 but N3 always applies no matter how the objects are moving. Therefore, now only the normal forces of the scale on the passenger and the passenger on the scale will still be equal.

(d) The scale reads whatever normal force it is exerting since that tells how much its spring is being squeezed. Therefore the value will be Nscale->passenger = Npassenger -> scale. The latter is the easiest to calculate from a = F^net/m. Since we know the motion of the passenger (assuming he stays on the scale), we should be able to get a. We also know m and one of the forces (the passenger's weight). There should therefore be only one unknown in our equation and we should be able to solve for it. Let's start by finding a.



Now we can figure out the normal force.