This week we focus on forces. Re-read chapters 4, 5 and 6 in your textbook.

 

Summary.

A force is easily understood as any push or pull. When Isaac Newton studied dynamics and expressed his famous first, second and third laws he was thinking of a net force as two things:

First, the net force is the sum of all forces acting on a body.

Second, the net force on a body can be determined by watching the body change its momentum (by accelerating). The magnitude of a net force is given by the rate of change of the momentum of the body against which the net force is directed.

Chapter 4. Newton's First Law. Often called the law of inertia, the First Law simply says that if the net force on an object is zero, its state of motion will not change (neither magnitude nor the direction of its momentum would change -- it could either remain at rest or in motion at constant velocity).

Chapter 5. Newton's Second Law. Often referred to by: F = ma. This statement accurately predicts that an object's acceleration should be directly proportional to the net force that acts against it and inversely proportional to the mass of the object.

-- In a direct proportion, as one quantity increases, the second quantity must increase in lock-step as the product of the first with a third quantity called the proportionality constant. The standard form is y = kx where k is the constant and y is a dependent variable that changes each time the independent variable x changes. k is the quotient: y/x

-- In an inverse proportion, as one quantity increases, the second quantity must decrease so that the product of the two quantities does not change. The standard form of an inverse proportion is therefore: k = x•y.

-- Once again, a better way to characterize "net force" is as the rate of change of momentum. If we account for all forces in our determination of net force, then we have: F = ∆p/t = m∆v/t.

Chapter 6. Newton's Third Law. Often called Action - Reaction. Force occur in pairs. When object A exerts a force on object B, then object B must exert a second force back on A of equal magnitude but opposite direction. Important: the only forces that change the momentum of an object are those forces that act on it never those exerted by it.

 

Some simple problems:

1. A 2.0 toy car slows from 4 m/s to 1.5 m/s over 5 seconds. At what rate does its momentum change? How much net force acts against its motion? What is its rate of acceleration?

∆V = (1.5 - 4) = -2.5 m/s ∆p = m∆v = 2•-2.5 = -5 kg•m/s and ∆p/t = -5/5 = -1 kg•m/s/s for the rate of change of momentum

-1 N of net force.

The rate of acceleration was -2.5 m/s÷5 s = 0.5 m/s/s.

2. Two hockey players face each other. They are angry. Assume the ice to be frictionless. Hockey player A has a mass of 100 kg. Hockey player B has a mass of 120 kg. Hockey player A exerts a 500 N force against the chest of hockey player B towards the north lasting for 0.5 seconds.

a. What happens (to each player) if hockey player B is taken by surprise and does not react to the push?

Then the 500 N force will act equally on each. 500 N toward the north on B and 500 N toward the south on A. Each will experience a change in momentum equal to F•t or 500 N • 0.5 s = 250 kg m/s

b. At what rate and in what direction is each moving after the interaction?

∆p = m∆v so ∆v = ∆p/m. For A: -250 kg•m/s ÷ 100 kg = -2.5 m/s or 2.5 m/s south for player B 250 kg•m/s ÷120 kg = +2.08 m/s or 2.08 m/s north.

c. What happens if hocky player B reacts by grabbing the uniform of player A (assume the uniform remains on and intact)?

If one player holds onto the other, the forces between them will be internal to a single system. They can flail about with as much wild insanity as they want, but until they turn loose of each other, neither will acquire an acceleration away from the other (or from the very spot where the fight starts).

3. A horse pulls a cart. The horse has a mass of 375 kg. The cart has a mass of 200 kg. The horse pushes backward on the ground with a force of 1200 N. The ground exerts a friction force against the motion of the cart of 50 N.

a. At what rate does the momentum of the horse-cart system change?

Net force is the same as the rate of change of momentum. This horse-cart system has two forces: 1200 N from the ground forward and 50 N from the ground backward. That's a net force of 1150 N. The rate of change of momentum must be 1150 kg•m/s.

b. What is the rate of acceleration for the horse-cart system?

The mass is 200 + 375 = 575 kg. Acceleration is F/m so 1150 kg•m/s ÷ 575 kg = 2 m/s/s

c. What is the magnitude of the force pair between the horse and the cart?

Horse: F = ma and m = 375 kg and a = 2 m/s/s so m•a = 375 •2 = 750 N. The horse gets 1200 N forward from the ground, so the cart must pull back with 450 N to account for the net force forward.

Cart: F = ma and m = 200 kg and a = 2 m/s/s so m•a = 200 • 2 = 400 N. The cart gets 50 N backward from the ground so requires a forward pull of 450 N to balance that plus provide the impetus for acceleration.

Note that horse's pull on the cart and the cart's pull on the horse are part of a single "force pair" or "interaction". The third law requires that those two forces be of equal magnitude but opposite direction. If it had turned out differently we would know that there was a mistake.