Static friction
Let us consider an example. Muffin the cat is trying to budge a wastepaper basket in order to see what is under it. Before she starts, the forces on the basket are those shown in Figure 61. She begins to push, but the basket does not budge. The force of friction has shown up and exactly balances her pushing force (Figure 62). When she pushes harder, the frictional force becomes larger, frustrating her effort (Figure 63).
Figure 61 
Figure 62 
Figure 63 
Figure 64 
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 a_{x} = 0 and a_{y} = 0.

Also,
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.
 If she does not pull on the rope, will he slide down into the river, which is infested with crocodiles?
 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 65). 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 F_{need}, that is, the force needed to keep him from sliding.
Figure 65
In Figure 66 we divide gravity into components.
Figure 66
The “vertical” equation becomes
N – mg cos 30 = (F_{net})_{y} = 0
because if Vincent does not move, then a_{y} = 0. Thus we have
N = mg cos 30 = (20 kg)(10 m/s^{2})(0.87)
= 170 N
The “horizontal” equation becomes
F_{need} – mg sin 30 = (F_{net})_{x} = 0
If Vincent is being held still by friction, or by friction and his sister’s rope, then we must have a_{x} = 0. Then we have
F_{need} = mg sin 30 = (20 kg)(10 m/s^{2})(1/2)
= 100 N
Equation (1) becomes
F_{s, max} = µN = 0.2(170 N) = 34 N
which is clearly insufficient. Beth must pull with a force
F_{Beth} = (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.