Friction

• Example 1:
  

  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 lecture that the tension can be considered as having both vertical and horizontal components: T_x = T cos(q) and T_y = T sin(q).
  We can now ask the follow 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

  y motion:	FN + T sin(q) - mg = 0 ==> FN = mg - T sin(q)
  x motion:	T cos(q) = max==> ax = T cos(q)/ m

  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:
  

  v² = v₀² + 2a_xd

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(q) and since cos(q) is maximm for q = 0, the optimal condition is to pull horizontally parallel to the floor. Is this also true if friction is involved? Let's see.

How Does Friction Change Things?

First, we need to know some things about the frictional force. To begin with, let's start with case of an object moving across a floor which has friction. Experimental observation tells us that the frictional force, at low velocities, is constant as long as the object is moving across the surface and is not moving too rapidly. The direction of the frictional force is opposite to the direction of motion and its magnitude is proportional to the normal force which maintains contact between the object and the surface, in this case the floor. We call this type of friction kinetic friction. Intuitively, it makes some sense that the kinetic friction should depend on how strongly the object is being pressed into the surface it's sliding on. The fact that the frictional force is independent of the surface area in contact just reflects an intuitive notion that, if the area is large, the average force per unit area smushing the object and surface together is small, but the area is large. If the area of contact is small, then the average force per unit area smushing the surfaces together is large. These effects of force per unit area and amount of area in contact largely offset each other so that the friction depends just on the total amount of force maintaining the contact. The constant of proportionality is usually referred to as the coefficient of kinetic friction and is denoted by the greek letter, m.

Now we can return to our previous problem and add in friction. The force diagram now becomes

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 friction is m_kF_N = m_kmg - T sin(q) since we previously found the relationship between the normal force, mg, and T. 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 q 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 q goes to zero. As we increase q, 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 q at zero is 1 and the sine of q at zero is 0. For q near zero, the cosine goes down more slowly than the sine increases as q 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 q 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 q 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 q goes up from zero. We can prove this by asking Maple to graph the acceleration as a function of q. In the plot below, I chose T = 50 N, m = 20 kg, and m = 0.2.

What do the negative values of acceleration correspond to? These are not physical. When q = 0.891 or 51° 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 q > 0.891. Our plot does show the acceleration going up as q increases. The maximum occurs at about q = 0.197 or about 11° . This angle minimizes the amount of acceleration you get for a given tension in the rope.

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 m_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.

• Example 2:
  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 m_smg and the box begins to move. Once this happens the box is subjected to a kinetic frictional force. Since m_k < m_s, 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 = m_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/ m_k The minimum value of F is the equality.

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.

• Example 3:
  Suppose a penny sits on a turntable (pre-CD device for playing music; mostly used by cave dwellers) which spins at 45 rpm (revolutions per minute) while sitting on a horizontal surface. The penny is 10 cm from the center of the turntable. What is the minimum coefficient of friction that keeps the penny on the turntable?

  Solution:
  Draw a picture.

  

  The forces acting on the penny are shown in the figure above. Note that the only force that can act in the horizontal direction is friction. We want static friction since we assume the penny does not move. To make this happen for the minimum coefficient of static friction, we want the static friction to satisfy the centripetal condition by assuming its maximum value. So

  fs, max	= mv2/r
  mk, min FN	= mv2/r
  mk, minmg	= mv2/r ==>
  mk, min	= v2/(g*r)
  	= ((45 revs./min.)(1 min./60 s) (2*p*r/rev.))2/(g*r)
  	= ((¾ )2 (4 p2) (0.1 m)s-2)/(9.8 m/s2)
  mk, min	= 023