FORCES AND MOTION: NEWTON'S LAW & ORDER

You need to know everything on this page very well. Work at it.

Newton's laws are simple in a way, but students frequently have trouble with them because the laws sometimes disagree with "common sense." For example, a truck smashes into a parked car. Do you know how the force of the car against the truck compares to the force of the truck against the car? How about in a boxing match when one lands a blow to the head of the other: there is a force on one guy's head and the force on the other's fist. How do they compare? Which is greater in each case?

One source of confusion with these things is that physicists use the word "exert" in an unusual way. To most people, a table cannot exert itself, so to talk of the force that the table exerts on a plate seems absurd. Exertion and hence exert should apply to things expending energy. There is physics-speak and regular English, it seems. This is why I like to avoid the word and simply say the force of the table on the plate ( or against the plate) and the force of the plate on the table.

(By the way, in the three examples of forces above, truck-car, fist-head and plate-table, all are equal and opposite pairs of forces. All boxing matches end in a tie if your only consideration is the force of one upon the other. If you find a force in nature that does not have an opposite, you will achieve all sorts of glory. It would revolutionize physics. These pairs are often called action forces and reaction forces, and for every action there is an equal and opposite reaction: Newton's 3rd law.)

Another thing that seems wrong is that when a thing is moving with constant velocity, the vector sum of all forces acting on it is zero (Newton's 1^st law). It might seem that there ought to be more force in the direction of motion than in the opposite direction. No, then it would speed up, and the acceleration could be found by Newton's 2^nd law, S F=ma. (Aristotle set back physics many centuries with his common sense notions that were completely wrong. Great philosopher, bad physicist.)

Another way of getting messed up with Newton's laws is by not identifying all forces and their directions. Here are the most common forces encountered in beginning physics courses:

1. Gravitational force (weight) = mg. Know your units for this.

2. The normal force is the force of a surface on something in a direction perpendicular to the surface. The "something" is in contact with the surface: resting on it, leaning on it.... (Normal means perpendicular in math-speak and physics-speak.) Normal force = whatever it takes to do the job. (If you stand on the floor and no other force besides gravity acts on you, then the normal force = mg upward in order to make the acceleration = 0. On the other hand, if you are accelerating upward in an elevator, the normal force needs to be greater than mg. (Why?) Incidentally, the way to measure normal force is with a scale. Scales in general do not measure weight, they measure the normal force acting on the object that is placed on the scale. For an object on an inclined plane, the normal force is perpendicular to the plane.

3. Strings pull, rods push or pull. Unless otherwise stated, strings in problems have negligible mass. For that reason, if two objects are attached to opposite ends of a string, the forces due to the string are equal in magnitude.

4. Friction. If an object is moving, it is tempting to say that friction always opposes the motion. Sorry, it aint that simple. If we simply have an object sliding on a stationary surface, it IS that simple, but you won't have trouble with those anyway. Say you have a brick sitting on top of another brick. Quickly pull the lower one in the +x direction out from under the upper. Figure out the directions of all horizontal forces on each brick. (E-mail me if you don't get it.) Sometimes it helps to imagine yourself being a brick, and think about the forces you would feel.

Here is another example- when a car is accelerating, the force on the car in the forward direction is the force of static friction between the tires and the road. Without that force, it would not accelerate.

Find out more about static friction here. If something is sliding, there is kinetic friction, and the friction force is found by f = m F_n, the coefficient of kinetic friction times the normal force. The coefficient m, often called m _k, is a property of the two surfaces involved. It real life it needs to be found experimentally; in books it tends to be given.

Here is a simple problem: Pull a wagon (mass m) on a horizontal surface with force F at angle q from horizontal. If the friction force has magnitude f, find the acceleration in terms of m, F, f, and q . Or if the coefficient of friction is known, find a. Here is a sketch and the solution.

Inclined plane problems: The most important thing is a sketch of the forces acting. The normal force on the object is perpendicular to the plane angling upward. The force of gravity is straight down, not perpendicular to the plane. Draw them carefully, and refer to a sketch in your text if you have any doubts about it. Now if the incline is just sitting there (not an incline in an accelerating truck, for example), and if a body is accelerating up or down the incline, the sum of the components of force perpendicular to the plane is zero, and the sum of components parallel to the plane is equal to ma. Be careful about the components of mg. If the incline is tilted angle q from horizontal, mg has a component mgsinq parallel to the plane and mgcosq perpendicular to the plane. Before reading further, what force is missing in the diagram? Let's say there is no friction.

Force not shown: the normal force on the mass above is equal to mgcosq in a direction up and to the left, perpendicular to the incline.

If the mass on the incline above has the normal force and mg, and no other forces acting (no friction), then
mgsinq = ma, so a = gsinq . This is a good approximation for a roller coaster when it is moving fairly slowly, but when it is moving fast, there is a lot of air drag. Then
mgsinq - f_drag = ma.

In principle you can use any coordinate system, but you will turn an easy problem into a hard one if you do not have the acceleration along an axis. Look at this to see how to prove that the acceleration of the frictionless mass on an incline is gsinq using the "wrong" coordinate system.

Circular Motion Problems: There is no new force here; you probably will have one or more of the forces referred to above: 1) mg, 2) normal force, 3) tension in a string, tension or compression in a rod, 4) friction. For circular motion there is acceleration toward the center equal to v²/r, and Newton's 2^nd law applies, so the procedure is to find components of all forces toward the center of curvature, (S F_c ) and set this equal to mv²/r. Want a derivation of a_c = v²/r?

Here are some examples:
For a car, bike or runner to turn a corner on a level surface without skidding, a static friction force toward the center is required.
For a banked turn on a road which is otherwise level, the normal force will have a component toward the center.
A pendulum will have tension toward the center of its circular arc (the pivot) and an mg component away from the center.
A conical pendulum goes in a horizontal circle, so tension has a horizontal component toward the center, and mg does not.
A roller coaster in a vertical plane has a normal force sometimes toward the center and sometimes away, and ditto for the component of mg.
A roller coaster doing a banked horizontal turn has a component of the normal toward the center. Want more detail on some of these examples?

DO LOTS OF NEWTON'S 2ND LAW PROBLEMS. THEY BUILD MENTAL MUSCLE. Don't worry about becoming too smart. Hard work pays off.

How to do 'em: Think about what is happening. What do you think is the direction of the acceleration? Many times you can answer this question immediately. Next, identify all forces acting on the object(s). The sum of all components perpendicular to the direction of acceleration is zero, and the sum of all components in the direction of acceleration is ma. Thus there are (at most) two equations (really 3, but most problems are two dimensional) to work with for each object. Try some easy problems first. E-mail me if you get stuck, and maybe I can give you a hint.

A big (10⁵ kg) jetliner is cruising along with a speed of 200 m/s at altitude 5000 m. What is the vector sum of all forces acting on it? Think about it. Do not click on the answer.

There was a horse that understood about the equal and opposite forces, so when he was hitched to the wagon, he said to himself, "If I exert a force F on the wagon, there will be a force -F on me, and these add up to zero, so the net force is zero and we cannot accelerate. So why bother? Hell no, I won't go!" The owner shot him of course, and had many a fine meal- all because a little knowledge is dangerous for a horse.

Your physics instructor will feel like shooting you if cannot reason better than the horse. The horse actually is correct about the F and -F, he just quit thinking a little too soon. You explain it. Do not read further until you have explained it to yourself.

I SAID DO NOT READ FURTHER...yeah, right. The horse did not consider all forces. The horse pushes backwards against the earth to get going and therefore receives the opposite, a forward force. It is that force of the earth on the horse that you use to calculate the acceleration. We consider all forces acting on the horse-cart system, and the force on the earth is not one of them, but the force of the earth on the horse is.

Here is a problem for you: Find the acceleration (in mph/s) of a 2 ton vehicle when a quarter ton net force acts on it, and do not convert units. (By the way, g is 22 mph/s.) How can you do it without converting units? Think in terms of proportions. Or don't bother, it is not in the mainstream. If you really want to know, click on this page and go near the bottom of it.

Here is the incline plane problem with m hanging on a string over a pulley and the other end attached to M. Three problems with M = 2.00 kg, m = 1.20 kg, q = 20.0^o:
1) There is no friction. Find the acceleration.
2) Find the minimum coefficient of static friction between M and incline to prevent acceleration.
3) The coefficient of kinetic friction is 0.200. Find the acceleration. (If static friction is holding it, you would need to get it started.)

Before you look at the solution, work hard to solve it. Identify all forces….

Go back to our main physics page by clicking on back. If you don't, your browser will byte you. Other main pages:

fluids, heat, electricity and magnetism
vibrations and waves
quantum
and for alphabetical listing of all kinds of junk, try my index.

e-mail me your ideas at fredrick.gram @ tri-c.edu (remove spaces)