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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
F = (m_A + m_B)a = F ==> a = F/(m_A + m_B) = (20 N)/(10 kg + 5 kg) = 1.333 m/s²
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,
m_Aa = F - F_BA = F - m_Aa = 20 N - (10 kg)(1.333 m/s² ) = 6.67 N
The last system we consider is block B by itself. The only external, horizontal force here is , so
m_Ba = F_AB ==> F_AB = (5 kg)(1.333 m/s² ) = 6.67 N

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

EXAMPLE 4: Let's do another example of forces applied to a not so difficult situation so that we can contrast the method of doing problems both with and without friction. Suppose you are asked to tug a rope connected to a heavy box of mass m. The box slides along a frictionless floor but is low enough so that you have to hold the rope at an angle theta with respect to the horizontal (see the figure below).

You exert a tension T on the rope and the rope, in turn, exerts the same magnitude and direction tension T on the box. The free-body diagram is easy to draw...

We will show in the next section of the text that the tension can be considered as having both vertical and horizontal components: T_x = T cos(theta) and T_y = T sin(theta).
We can now ask the following question: If you maintain a constant tension in the rope while the box slides a known distance d along the floor, what is the speed of the box at d? Assume that the box was initially at rest and that T < mg

Solution: We can find the equations describing the change in motion now that the free-body diagram is known. Once the acceleration of the box is known, we can find its speed. For the vertical and horizontal equations we have

Note that since T < mg the box maintains contact with the floor (F_N > 0). Now that the horizontal acceleration is known, we can see that it is constant, hence we can use one of the equations for motion to find the speed after a distance d has been traveled:

So the velocity is the square root of 2a_xd. If we were to ask an additional question, what angle of inclination for the rope maximizes the speed we get for the same magnitude tension applied and the same distance traveled, we can see immediately that since a_x depends linearly on T cos(theta) and since cos(theta) is maximum for theta = 0, the optimal condition is to pull horizontally parallel to the floor. Is this also true if friction is involved? Let's see.

Friction

Now we can do a typical problem and add in friction. The force diagram for the block dragged across a floor by a string making an angle theta with respect to the horizontal that we considered before is now

The vertical change in motion is unaffected by the kinetic friction since the friction acts only in the horizontal direction. The magnitude of the kinetic frictional force is µ _kF_N = µ _k(mg - T sin(theta)) assuming that the vertical portion of the tension is less than the weight of the object! The horizontal acceleration is

We can calculate the velocity after a distance d has been traveled as before. Now, however, if we ask what angle theta maximizes the velocity, we can not be sure the answer is still 0. In fact, we can see that it is very likely not zero since the frictional force and the horizontal component of the tension both increase as theta goes to zero. As we increase theta, the horizontal component of tension goes down, but since the vertical component goes up, the normal force from the floor goes down and the frictional force also goes down. How do we solve this? First, note that the cosine of theta at zero is 1 and the sine of theta at zero is 0. For theta near zero, the cosine goes down more slowly than the sine increases as theta goes up. Hence, we expect that raising the angle will reduce the frictional force at a faster rate than the reduction of the horizontal component of the tension. To get the exact answer, we note that the velocity is largest when the acceleration is largest, so if we find the value of theta that maximizes the acceleration, we are done. The usual trick is to find where the first derivative or instantaneous rate of change of acceleration as a function of theta goes to zero. The rate going to zero describes an approach of the function value to a minimum or maximum. In our case,

describes a maximum since the acceleration increases as theta goes up from zero. We can prove this by asking Maple to graph the acceleration as a function of theta. In the plot below, I chose T = 50 N, m = 20 kg, and mu = 0.2.

What do the negative values of acceleration correspond to? These are not physical. When theta = 0.891 or so, the frictional force overtakes the horizontal component of the tension and the block stops. Since friction can not cause the object to move backwards, the block remains stationary for values of theta > 0.891. Our plot does show the acceleration going up as theta increases. The maximum occurs at about theta = 0.197. This angle minimizes the amount of pulling you have to do to get the largest acceleration.

What About Static Friction?

The static frictional force acts to keep objects from beginning motion against a surface. Its magnitude can be any value between 0 and a maximum value which is proportional to the normal force coming from contact between the object and the surface. We call the constant of proportionality for this maximum static frictional force mu_s, the coefficient of static friction.

The direction is such as to oppose the direction in which motion would occur if friction were not present.

Let's do an example. Consider the picture below:

Here, the floor exerts a static frictional pull against the block in order to oppose the motion that would otherwise occur due to the presence of F. The magnitude of f_k increases to value of the magnitude of F so that the net force in the horizontal direction is always zero. If F increases or decreases, then f_k increases or decreases to maintain f_k = F and zero acceleration in the horizontal direction. If continue to increase F, eventually it surpasses the value mu_smg and the box begins to move. Once this happens the box is subjected to a kinetic frictional force. In the case of most objects on most surfaces, µ _k < mu_s so once the object is in motion, the force needed to keep it in motion is smaller than the force needed to get it to start motion in the first place.

We can apply these ideas to a relatively straightforward case

Here, the question is the following: what is the smallest magnitude F can have and still maintain the block in a stationary position along the wall? The forces acting are shown in the free-body diagram above. The normal force depends on the external force F since the block does not fly into or away from the wall in the horizontal direction. Thus, F_N = F. And we know that f_s = mg if the block is not accelerating vertically. The maximum value of the static frictional force is f_s,max = µ _kF. Therefore, the condition we need is that the static frictional force just be able to maintain the block's vertical position against the pull of gravity. Therefore, we have F > = mg/µ _k The minimum value of F is the equality.

For a challenge, you can refer to an extra-credit problem which you can include with your homework.

Static and Kinetic Friction

You can develop the free-body diagrams by noting that, if we start with m₂, the simpler of the two blocks, there are only two points of contact, the rope and contact with block m₂. That means there are 4 forces acting on m₁. These are shown on the free-body diagram below. Note that the direction of kinetic friction is determined by the direction of motion of m₂ with respect to m₁. Since m₂ must move to the left when m₁ moves to the right, it's motion relative to m₁ is to the left, so the frictional force points to the right.

We expect that m₂ will not accelerate along the vertical position, so the equations governing its motion are:

Now we can look at the free-body diagram for m₂

Where we note that Newton's Third Law demands that if m₂ undergoes a frictional force because of its interaction with m₁, then m₁ must undergo a frictional force of the same magnitude, but oppositely directed. The equations of motion are:

We also note that if m₁ moves to the right by a certain distance, m₂ must move to the left by the same distance in the same time in order to keep the string from stretching. Therefore, we must impose a₁ = -a₂. Setting a₁ = a, we can solve our above expressions using Maple

We note that Maple gives the acceleration as the external force F - 2f_k divided by the sum of the masses. This makes intuitive sense because if we consider m₁ and m₂ to be a system, then the external forces are F acting on m₁ and f_k acting on both m₁ and m₂.