5-28-97
Newton's laws of motion can be applied to an equilibrium situation, in which an object is at rest or is moving at constant velocity. They can also be applied to a non-equilibrium situation, in which there is an acceleration. For an equilibrium situation, all the forces are balanced, while in a non-equilibrium situation there is a net force.
At equilibrium there is no net force, and therefore no acceleration. An object in equilibrium is either stationary, or moving at a constant velocity. Let's do two examples of an equilibrium situation, one of a box at rest on an inclined plane (this is a technical physics phrase for a ramp), and another of the box sliding down the inclined plane at constant speed.
Example 1 - A box of mass 3.60 kg rests on an inclined plane which is at an angle of 15.0° with respect to the horizontal. A string is tied to the box, and it exerts a force of 5.00 N up the ramp. The coefficent of static friction is 0.40. What is the frictional force? Give the magnitude and direction.
The first step in solving this is to draw a picture, showing the box, inclined plane, and all the forces. We don't know which way the friction force goes, either up or down the incline, so let's just guess that it's down. (We could work out the component of the weight down the slope and compare it to the tension force to see, but it's fine to guess, too. If we guess wrong we'll get a minus sign at the end which will tell us that the force goes opposite to the way we guessed...the magnitude will be correct, though.)
It's a good idea to draw the free body diagram, also, which is similar to the first picture but without the inclined plane. Another difference is that the force of gravity has been split up into components. It is most convenient to break up forces into components parallel to the incline and perpendicular to the incline, because more of the forces are along these directions.
Applying Newton's second law in the y-direction gives:
The box is not accelerating in the y-direction, which is why the net force in that
direction is zero. Solving for the normal force gives:
This allows us to calculate the magnitude of the maximum possible static force of friction, which is:
Applying Newton's second law in the x-direction gives:
Again, there is no acceleration, giving a net force of zero. Solving for the static force of friction gives:
So, the magnitude of the force of friction is 4.13 N, well under the maximum possible value, which makes sense. The minus sign means that we guessed wrong for the direction on the diagram...the force is really up the slope, not down. This also makes sense, because the component of the weight down the slope is larger than the tension, so the tension needs a little help from friction to keep the box from sliding down the incline.
Example 2 - A box of mass 3.60 kg travels at constant velocity down an inclined plane which is at an angle of 42.0° with respect to the horizontal. A string tied to the box exerts a vertical force of 7.38 N. What is the kinetic coefficient of friction?
Again, start by drawing a picture. In this case, because the box is travelling down the ramp, we know the frictional force is kinetic, and that it is directed to oppose the motion, so it must be up the slope.
Next, draw a free-body diagram, splitting the forces into components along directions parallel and perpendicular to the inclined plane.
Again, apply Newton's second law to the forces in the y-direction.
Again, we can use this equation to solve for the normal force:
Applying Newton's second law in the x-direction gives:
The box is moving at a constant velocity, so that means the acceleration is zero. Solving for the kinetic force of friction gives:
The coefficient of kinetic friction can be found from the normal force and the frictional force:
Clearly this is a different box-inclined plane pair than in Example 1, because the coefficient of friction here is much larger. It's much harder to move the box on this inclined plane than on the inclined plane in the first example.
A non-equilibrium situation is one in which there is acceleration. Consider another example involving an inclined plane, only this time there will be two boxes involved.
Box 1, a wooden box, has a mass of 8.60 kg and a coefficient of kinetic friction with the inclined plane of 0.35. Box 2, a cardboard box, sits on top of box 1. It has a mass of 1.30 kg. The coefficient of kinetic friction between the two boxes is 0.45. The two boxes are linked by a rope which passes over a pulley at the top of the incline, as shown in the diagram. The inclined plane is at an angle of 38.0° with respect to the horizontal. What is the acceleration of each box?
The diagram for the situation looks like this:
The next step is to draw a free-body diagram of each box in turn. To draw these, it helps to think about which way the boxes will accelerate. The two boxes are tied together, and the heavier box will win...in other words, it will accelerate down the slope and the lighter one, on top, will be accelerated up towards the pulley. The pulley, by the way, simply changes the direction of the tension force. We're assuming that the pulley is massless and frictionless, so both boxes feel the same tension force. It's important to know the direction of the acceleration (or, if you don't know, to guess) and apply what you figured out (or your guess) consistently to both boxes.
The free-body diagram of box 1 is relatively complicated, with a total of 6 forces appearing. The free-body diagram for box 2 is a little easier to deal with, having 4 forces, so that's a good place to start.
For box 2 - start by summing the forces in the y-direction, where there is no acceleration:
This can be solved to give the normal force:
Now find the net force in the x direction, where there is an acceleration up the slope:
There are two unknowns in this equation, the tension T and ax, the acceleration. We can at least solve for T in terms of ax, like this:
Now move on to box 1. Going back to the free-body diagram and summing forces in the y-direction gives:
This equation can be solved to give the normal force associated with the interaction between the inclined plane and box 1:
Things are a little more complicated in the x-direction, but adding up the forces gives:
substituting in the expression we worked out for T, and what fB and fA are gives:
Moving the acceleration terms to the left side gives:
Solving for the acceleration gives:
