Physics 101
Homework Set 3 Solutions
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One Step Calculations:
Chapter 5:
Number 1:
a) if you add 100 km/hr North + 75 km/hr South you get 25 km/hr North.
One way to do this is to assign ‘+’ as North and ‘-‘ as South. Then the problem becomes v = +100 km/hr + (-75 km/hr) = +25 km/hr = 25 km/hr North
b) If both of the velocities are directed North, then
v = +100 km/hr + (+75 km/hr) = +175 km/hr = 175 km/hr North
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Exercises:
Chapter 4:
Exer 30: The only force acting on the coin is gravity. Gravity does not change. Therefore the force on the coin is its weight, regardless of whether it is halfway up, at its peak, or most of the way down.
Exer 32: There are two forces on you, gravity pulling down and the ground pushing up. When you are just standing there, those two forces sum to zero. This is because your velocity is not changing, therefore your acceleration is zero and therefore the net force on you is zero.
When you leap upwards, you are accelerating upwards. This means that the net force on you is upwards. Therefore, the ground is exerting an upward force on you that is larger than your weight. (Why does the force on you from the ground increase? It increases because you push harder on the ground when you are leaping upwards, therefore the ground pushes harder on you. But this is the subject of Chapter 5.)
Exer 50: When a car is moving in reverse, the driver applies the brakes. Therefore the car slows down. Therefore the acceleration is opposite in direction to the velocity. The velocity is backwards, therefore the acceleration is forwards.
Chapter 5:
Exer 4: a) action: hammer exerts a contact force on the nail :: reaction: nail exerts a contact force on the hammer
b) action: Earth exerts a gravitational force on the apple :: reaction: apple exerts a gravitational force on the Earth
c) action: action: helicopter blade exerts a force on the air :: reaction: air exerts a force on the helicopter blade
Exer 18: In order for a rope climber to move upwards, the net force on him must be upwards. There are two forces on the climber, the downward pull of gravity and the upward force of the rope on the climber. If the upward force of the rope on the climber is the action, then the reaction is the downward force of the climber on the rope. Therefore, the climber pulls down on the rope so that the rope will pull upwards on him. This is the same as the jumping up exercise from the previous chapter.
Exer 20: Assuming that the horse is amenable to reason and that the neighbors won’t laugh at me for arguing with a horse, I would say the following. The wagon will move if the net force on the wagon is nonzero. The horse exerts a force on the wagon. The reaction force from the wagon is exerted on the horse. This reaction force is irrelevant to whether the wagon moves.
Exer
22: The forces on the two wagons are equal and opposite by
Exer 36: The tightly stretched hammock is more likely to break. When you sit on the hammock, the hammock must counterbalance your weight (ie: provide an upward force equal to your weight). If the hammock sags a lot, then a lot of the force of the hammock will be vertical so that the horizontal and vertical components of the force will be comparable. If the hammock is tightly stretched, then the horizontal component of the force will be MUCH larger than the vertical component so that the total force will be much larger. We saw this in class where two guys exerted a LOT of force to hold up 2 kg with a weight of 20 N (or about 5 lbs) when the rope was tightly stretched.
Problems:
Chapter 4:
Prob 6: Two boxes have the same acceleration when a force F is applied to the first and 4F to the second. Since a = F/m, we know that
m1 = F / a
m2 = 4F / a = 4 (m1)
Therefore the second mass is four times larger than the first.
Chapter 5:
Prob 2: You push on the wall with a force of 40 N, therefore the wall pushes on you with a force of 40 N.
a = F/m = 40 N / 80 kg = 0.5 m/s2 (but only while you are still pushing on the wall)
Prob 4: To get the acceleration, we need the total force on the object. The total force F = sqrt((3.0 N)2 + (4.0 N)2) = 5.0 N. Therefore, a = F/m = 5.0 N / 2.0 kg = 2.5 m/s2
Estimation: Your car accelerates from 0 to 30 m/s as quickly as possible. We want to know the force exerted by the engine. To do this, we need to know the acceleration and the mass. To find the acceleration, we need to know the time it takes to accelerate. If the car is the sports car that my kids hope to get when they turn 16, it will take about 6 seconds. If it’s my car, then it will take about 10 seconds. (Any time in the range of 4 to 12 s is reasonable.) I’ll use 8 s. This means that the acceleration a = Δv / t = (30 m/s) / 8 s ≈ 4 m/s2. Now I need the mass. My small car has a mass of about 1 ton = 1000 kg. An SUV can have a mass of 2 or 3 (or even more) tons. Any mass between 1 and 5 tons is reasonable. I’ll use m = 2 tons = 2000 kg = 2*103 kg. Thus:
F = ma = 4 m/s2 * 2*103 kg = 8*103 N
Now we need to figure out how much force people can exert. I’ll assume that we can exert half our weight. My weight is 180 lb ≈ weight of 80 kg = 800 N. This means that I can exert a force of 90 lb = 400 N. Therefore it would take 20 people to exert that much force. Any answer between 5 and 100 will be acceptable.
Note: of course that I can only exert that force if I’m not moving too quickly!