Recall: Newton's Laws of Motion

  1. If a body experiences no net external force, then its acceleration is zero: if SF=0 then a=0.
  2. SF=ma
  3. "For every action there is an equal and opposite reaction": F1 on 2 = -F2 on 1.
Demonstration: Office chairs.
Some Questions for discussion:
If a small car collides head-on with a massive truck, which vehicle experiences a greater inpact force? Which vehicle experiences a greater acceleration? What is wrong with the following argument? A horse attempts to pull a wagon from rest. If the horse pulls forward on the wagon, then Newton's third law holds that the wagon pulls back equally hard on the horse. Thus the forces cancel and nothing happens. Is it possible for an object to move if no net force acts on it?

Free-body diagram

When applying Newton's laws to solve a problem, the first thing that normally needs to be done is to identify the forces acting on the object(s). For instance, a crate being pulled by a rope experiences the tension in the rope (tension is a type of force), the pull of gravity on the center of mass of the crate, and the normal force of the floor against the crate. These forces act on the crate. The reaction forces to each of those listed act on other objects, so they are not listed. We can summarize these forces with a free-body diagram, a drawing of the crate and the forces acting on it. (diagram)

Solving Problems

When tackling problems involving something in equilibrium (sometimes called statics problems), it is best to use Newton's first law in component form (and we'll normally need only the x and y components, though an extension to z should be obvious): SFx = 0
SFy = 0
Example: P4-15
A block of mass m=2.0kg is held in equilibrium on an incline of angle h=60degrees by the horizontal force F, as in the drawing. (a) Determine the value of F. (b) Determine the normal force exerted by the incline on the block (ignore friction). 3 forces on the block: F, Fg, and n. Note that n is perpendicular to the surface of the incline! Let's make a table with a row for each force, and columns for the x and y components:
x component y component
F F0
Fg0-mg=-19.6N
n-n sin(60)n cos(60)
The sum of the x components is: F- n sin(60) = 0, so F=n sin(60). The sum of the y components is: -19.6N + n cos(60) = 0, so n=19.6N/cos(60) = 39N (the answer to part a). Putting this answer back into the equation from the x components, we find for the answer to part b, F=(39N)sin(60) = 34N. NOTE: For problems involving ropes, chains, etc., the tension in the rope is equal to the force it transmits to either end. If the object in a problem is not static, but is accelerating, then Newton's second law is needed. Again, it is convenient to work with the component equations: SFx = max
SFy = may
Example: P4-21
Assume that the three blocks in the figure move on a frictionless surface and that a 42N force acts as shown on the 3.0kg block. Determine (a) the acceleration given this system, (b) the tension in the cord connecting the 3.0kg and the 1.0kg blocks, and © the force exerted on the 2.0kg block by the 1.0kg block. In the y direction, the weight of the blocks is balanced by the normal force of the frictionless surface, such that SFy = 0. No motion occurs in the y direction. Let's turn our attention to the x direction. To determine the acceleration of the entire system (all 3 blocks), apply the second law, SF = ma, treating the 3 blocks as one object. Then m is the mass of the 3 blocks (plus rope), m=1+2+3kg=6.0kg. SF is the total exernal force on the blocks. The force of one block on another is an internal force, so it can be ignored for now. The only remaining force is the 42N applied force. The remaining quantity in the equation is the acceleration, a. Solving for a: a = SF/m = (42N)/(6.0kg) = 7.0m/s2. Now, regarding the 3 blocks as 2 separate groups, one being pulled right by the cord, and the other being pulled right by 42N and pulled left by the cord. The tension in the cord must be sufficient to supply the first pair with an acceleration of 7.0m/s2 (since the whole accelerates at this rate, then each part must also accelerate at this rate). The mass of the pair of blocks is 3.0kg, so the applied force from the tension in the cord must be F=ma=(3.0kg)(7.0m/s2) = 21N. Finally, to accelerate the 2.0kg block, the 1.0kg block must exert a force on it of F=ma=(2.0kg)(7.0m/s2)= 14N.

Friction

To finish this chapter on the laws of motion, we will discuss the force of friction. Friction arises when two surfaces slide past one another, like a book sliding across a table. Experiments reveal that friction has the following basic properties:
  1. The frictional force is independent of the contact are.
  2. The frictional force is proportional to the normal force on the object. (NOTE: this is not necessarily equal to the weight.) The constant of proportionality is called m.
  3. There is a static friction and kinetic (sliding) friction. Kinetic friction is constant, equals mkn, and is directed opposite the motion. Static friction is <msn, and directed to balance the net applied force.
Demonstration with weights and spring scale. Question: Describe a few examples in which the force of friction exerted on an object is in the direction of motion of the object.
Example: P4-41
Find the acceleration experienced by each of the two masses shown if the coefficient of kinetic friction between the 7.00kg mass and the plane is 0.250. Draw a free body diagram for both masses. Find the normal force for the 7kg mass, then, assuming it is moving up the plane, find the frictional force. This step is a bit subtle, since you will still get an answer if you choose the wrong direction for the frictional force, but afterward, a check of the answer will reveal that the force is pointing in the same direction as the motion.