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Force Revisited

Application of the laws of motion requires us to understand how to write down equations which accurately describe the situation at hand. The use of mathematics as a ``second language'' is never more useful than it is when ``word'' problems must be solved. Now that you have had an introduction to Newton's laws and the mathematics behind them, its necessary to practice the art of translating words to equations. In some cases this is straightforward (after some practice), but, in other cases, it can be devilishly difficult. Part of the reason for this is that written language depends heavily on context to carry meaning (spoken language has an even greater dependence on context). Equations, however, must be literal in their description. Information that is automatically assumed when communicating in language must be explicitly included in equations. Assumptions, hunches, trivia, boundary conditions, etc. must all be correctly described by the equation you write down in translation of the problem or your solution will usually be meaningless. This is the frustrating part of learning how to apply quantitative laws and is the reason that doing physical science problems is as much an art as it is a science. Even though the laws of nature can be rigorously applied, knowing what to leave out and what MUST be included is your goal in learning physics and chemistry.

We will begin force problems by using examples that don't ``need'' to solve the differential equation implicit in Newton's Second Law. For these problems, we typically are interested only in the acceleration, which is usually constant, or the force itself. These problems are the easiest to do mathematically, but represent only a small subset of the interesting problems that can be done.

Before starting with example problems, we need to be reminded of the standard set of assumptions that usually go into our problem-solving.

1.

There are three ``tools'' you are usually required to use over and over in freshman physics. These are the pulley (sometimes we refer to a block and tackle if two or more pulleys are connected by rope), the inclined plane, and the string or rope. Unless you are explicitly told otherwise, you need to assume that pulleys and strings are massless and without friction. Inclined planes are assumed to be infinitely massive (in other words fixed in place) and perfectly rigid. Strings do NOT stretch and inclined plane surfaces do not deform under any circumstances.

2.

All surfaces are frictionless and there is no air resistance unless you are explicitly told otherwise. Sometimes you are not directly told that a surface has friction, but it is indicated by saying that the surface has a ``coefficient of friction''. We will probably not have time to consider friction for the interdisciplinary course, so this will likely not be a consideration.

3.

Under the assumptions listed in item 1, tension is constant in magnitude in any single, i.e. unbroken, piece of string. Since a string goes smoothly around a pulley, the tension in any string or rope around a pulley is constant throughout the string.

4.

Forces are measured in units of Newtons (abbreviated N). One Newton is 1 kg times 1 m/s² .

Now that we have given sufficient warning, let's jump right into applications by considering some classic problems. Problems on more interesting subject matter appears in the homework exercises. Remember that when you are asked to specify a force, you need to give the magnitude, the direction, and what the force acts on to be complete.

SOLUTION: Even a smooth lift with zero acceleration will win the bet, so the minimum force needed on the lineman is a net force upward which is equal to his weight. As soon as the woman begins to pull, the rope gets taut and the tension in strands 1, 2, 3, and 4 shown in the figure above start to increase. As long as the sum of tensions , , , and is less than Mg, the weight of the lineman with mass M, then the lineman stays on the ground. He begins to ascend as soon as the sum of the tensions acting on him exceed his weight by even the tiniest fraction. Thus, the condition for lifting him is

We note that all four tensions are acting on the lineman and that, since the rope is continuous, the tensions in all four strands are the same. Since the woman weighs (54.5 kg) x (9.8 m/s² ) = 534.1 N, she can easily lift the lineman just by hanging on the rope.

SOLUTION: Leaving everything in equations for the moment, we first determine all the forces acting on each block. These are shown in figure b. above. For this problem, we note that all the action of interest happens in the horizontal direction, thus we do not need to consider gravity at all since we assume that the surface the blocks sit on is level. Therefore, we can consider the problem to consist of three systems. In the first system, we consider the two blocks as a unit. In this case, the internal forces between them are irrelevant. The only external horizontal force is 1.3 m/s² . Therefore, by Newton's Second Law

So the acceleration of the system, and therefore of both blocks, is 1.3 m/s² . We now consider just block A by itself as a system. This system has two horizontal forces, and . Thus,

The last system we consider is block B by itself. The only external, horizontal force here is , so

Newton rests easy once again.

SOLUTION: We begin, as always, by drawing the forces on the two blocks. These are shown in the figure below.

Again, the horizontal forces are all that count for block 2 since we do not expect any motion in the vertical direction for that block. For block 1, there are no horizontal forces so there is only motion in the vertical direction. We must be careful to follow our sign conventions in setting up the force equations from Newton's Second Law. For block 1, the vertical motion is controlled by:

For the horizontal motion of block 2, we have

We notice now that we have three unknowns, , , and T, but only two equations. The third constraint is that if m_1 moves down by a certain distance, h, then m_2 must move the same distance h to the right if the string between blocks 1 and 2 cannot stretch. Therefore, any motion of m_1 must be accompanied by the same amount of motion by m_2 in the same time. So, we are constrained to have . Notice that we need the minus sign since points down in the negative y direction in our figure above, while points to the right along the positive x direction. We simplify life by writing to eliminate subscripts. Now our equations are

This is trivial to solve for the two unknowns, T and a, but since we have Maple handy and we're typing our solutions anyway, let the computer do the dirty work with the following Maple command:

solve({-m1*a = T - m1*g, m2*a = T}, {a,T});

Maple promptly returns

We can easily substitute in values for m_1, m_2, and g to get a = 1.4 m/s² and T = 84 N. Therefore, block 1 moves -1.4 m/s² downward while block 2 moves with acceleration 1.4 m/s² to the right. The tension is the string is 84 N.

Note that Maple gives us the explicit power to explore the behavior of our solution beyond the singular values included in the problem. This is a good way to check on how sensible our solution is. It should not violate our physical intuition (unless it gives us some insight we did not have before!). Suppose, for example, that you are asked to evaluate what happens as the mass m_1 is increased beyond the value of m_2 by some large amount or what happens if m_1 is much less than m_1. Physical intuition tells us what to expect: if m_1 is much larger than m_2, then the acceleration should be close to its free-fall value of 9.8 m/s² . Case IV, i.e. it doesn't matter much that a teeny weight m_2 is attached via the string to m_1, the mass of m_1 dominates. On the other hand, if m_1 is teeny compared to m_2, we expect the acceleration to be very small since the weight of m_1 is providing all the force to move both m_1 and m_2. What about the tension? What should happen to it in these two extremes? Your physical intuition may not be up to predicting it just yet. That's where Maple offers us the ability to build intuition by exploration. For example, we can make use of Maple's limit command to find the behavior of our solution for various limits.

# Begin by typing in our solution or asking Maple to
# solve our simultaneous equations for us.
a := m1*g/(m1 + m2);    T := m_1*m_2*g/(m_1 + m_2);
#
# Now just have Maple calculate the extremes for us
limit(a, m1=infinity);
                         g
limit(a, m1=0);
                         0