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Static friction

Let us consider an example. Muffin the cat is trying to budge a waste-paper basket in order to see what is under it. Before she starts, the forces on the basket are those shown in Figure 6-1. She begins to push, but the basket does not budge. The force of friction has shown up and exactly balances her pushing force (Figure 6-2). When she pushes harder, the frictional force becomes larger, frustrating her effort (Figure 6-3).


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Figure 6-1


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Figure 6-2

There is a maximum for friction, and that is shown in Figure 6-4 as a dashed vector, labeled Fmax. If she can manage to push harder than the theoretical maximum, then the basket will move. At that time we no longer have a problem in static friction.

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Figure 6-3

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Figure 6-4


There are two principles here:
If there are no slipping surfaces, then the static friction, which acts parallel to the surface, has whatever magnitude it needs in order to maintain nonslipping surfaces. This generally involves solving the whole force equation with ax = 0 and ay = 0.


If the calculated force of friction is greater than the theoretical maximum, then static friction is not relevant, and the problem needs to be reconsidered with kinetic friction. The maximum is given by
where  the coefficient of static friction, depends only on the materials involved. It has no units and is generally less than 1.

Beth (45 kg) has tied a rope around her brother’s waist. Vincent (20 kg) is on the slippery slope of a river bank making an angle 30˚ with the horizontal. The coefficient of static friction between him and the bank is 0.2.

  1. If she does not pull on the rope, will he slide down into the river, which is infested with crocodiles?
  2. If so, what is the smallest force she must exert parallel to the bank in order to keep him from slipping?

First, we need to DRAW A DIAGRAM (Figure 6-5). First, there is the force of gravity. The bank is touching him, so there is a normal force. There is also friction and possibly the rope. At any rate, these act along the bank, so we label them Fneed, that is, the force needed to keep him from sliding.


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Figure 6-5


In Figure 6-6 we divide gravity into components.


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Figure 6-6


The “vertical” equation becomes

N – mg cos 30 = (Fnet)y = 0

because if Vincent does not move, then ay = 0. Thus we have

N = mg cos 30 = (20 kg)(10 m/s2)(0.87)

= 170 N


The “horizontal” equation becomes

Fneed – mg sin 30 = (Fnet)x = 0


If Vincent is being held still by friction, or by friction and his sister’s rope, then we must have ax = 0. Then we have

Fneed = mg sin 30 = (20 kg)(10 m/s2)(1/2)

= 100 N


Equation (1) becomes

Fs, max = µN = 0.2(170 N) = 34 N


which is clearly insufficient. Beth must pull with a force

FBeth = (100 – 34)N = 66 N


If Beth pulls with a force 66 N, then static friction provides 34 N, enough to keep Vincent from slipping down. If she wants to pull Vincent up the slope, then she must pull hard enough to exceed the static friction maximum in the other direction. That is, friction would pull down the slope, and she would have to pull up the slope with a force (100 + 34)N = 134 N.



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