Newton's Laws

Suppose a friend asks you for advice on an idea he or she wants to patent. The idea is for a method of lifting a person a few feet off the floor so that they can reach objects on tall shelves in a warehouse. Your friend has thought of three different methods that might work and wants your physics expertise to help pick out the best idea, i.e. the one that works with the least amount of effort.

Case 1: Case 2: Case 3:

The force diagram for the first case appears below

Now we do the force equations. For the box in which the person stands while pulling on the rope, we have (assume the positive direction is up)

T - FN - mbg = mba ==>
T = mba + FN + mbg

Since we are interested in the least amount of force the person would have to exert to pull themselves up, let's set the acceleration of the system to zero (it would initially have to be small and positive to get motion started, but after that, a = 0 is fine, i.e. the constant speed upward gets the job done). So, then the above equation simplifies to

T = FN + mbg

Before considering the forces on the person, note that the acceleration of the person must also be zero, that the normal force exerted by the person on the box must be equal in magnitude the normal force exerted by the box on the person by Newton's Third Law, and that the tension in the rope is assumed to be constant in magnitude throughout the rope, so

T + FN = mpg

Now let's add the equations for the box and the person together (this is just algebra at this point as all the physics of forces went into deriving the equations)

T + T + FN = mpg + FN + mbg ==>
2T = mpg + mbg ==>
T = ½ (mp + mb)g

So, not only is the bosun's chair lift possible, the person has to exert only half the weight of the box plus him or herself to do the lifting! You should go through this same analysis for the following problem.

The Weight

Our intuitive concept of weight is generally somewhat clouded by our lack of analytical experience with normal forces. What exactly is weight? While the tendency is to view it as the gravitational force, in fact, our experience of weight through our own senses is due to normal forces, not gravitational force.

Consider the following question:

A student of physics someday hopes to become an astronaut. She wishes to experience weightlessness to get accustomed to space travel. How do you achieve weightlessness while on the surface of the earth? Antigravity is definitely out for the foreseeable future. However, following Galileo's example of extrapolating from known phenomena, she explores the "funny" feeling you get in an elevator that stops suddenly after rising rapidly. Let's analyze it.

The forces on the student are shown in the image below

The normal force FM comes from the contact of the would-be astronaut with the scale, therefore, we must conclude that the scale does not measure mg, but FN. In normal cases these are the same since the normal force necessary to keep us from accelerating down into the scale is just enough to balance our weight. Here, however, the student is accelerating downward with acceleration a so the normal force must not supply force equal in the magnitude to mg. Instead, assuming the up direction is positive, we must have

FN - mg = -ma ==> FN = m(g - a)

Hence the scale measures a "weight" which is less than mg. If the elevator is accelerating in the upward direction.

we must have FN = m(g + a). Thus your apparent weight is greater than mg.

To achieve complete weightlessness the elevator has to accelerate downward at a rate a = g. This is free-fall and is not not usually allowed for standard elevator operating conditions.

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