The Coefficients of Friction
The amount of force needed to overcome the force of static friction on an object, and the magnitude of the force of kinetic friction on an object, are both proportional to the normal force acting on the object in question. We can express this proportionality mathematically as follows:
where
is the
coefficient of kinetic friction,
is the
coefficient of static friction, and
N is the magnitude of the normal force. The coefficients of kinetic and static friction are constants of proportionality that vary from object to object.
Note that the equation for static friction is for the
maximum value of the static friction. This is because the force of static friction is never greater than the force pushing on an object. If a box has a mass of
10 kg and
=
0.5, then:
If you push this box with a force less than
49 newtons, the box will not move, and consequently the net force on the box must be zero. If an applied force
is less than
, then
= –
.
Three Reminders
Whenever you need to calculate a frictional force on SAT II Physics, you will be told the value of
, which will fall between 0 and 1. Three things are worth noting about frictional forces:
- The smaller µ is, the more slippery the surface. For instance, ice will have much lower coefficients of friction than Velcro. In cases where
, the force of friction is zero, which is the case on ideal frictionless surfaces.
- The coefficient of kinetic friction is smaller than the coefficient of static friction. That means it takes more force to start a stationary object moving than to keep it in motion. The reverse would be illogical: imagine if you could push on an object with a force greater than the maximum force of static friction but less than the force of kinetic friction. That would mean you could push it hard enough to get it to start moving, but as soon as it starts moving, the force of kinetic friction would push it backward.
- Frictional forces are directly proportional to the normal force. That’s why it’s harder to slide a heavy object along the floor than a light one. A light coin can slide several meters across a table because the kinetic friction, proportional to the normal force, is quite small.
Example
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A student pushes a box that weighs 15 N with a force of 10 N at a 60º angle to the perpendicular. The maximum coefficient of static friction between the box and the floor is 0.4. Does the box move? Note that sin 60º = 0.866 and cos 60º = 0.500. | |
In order to solve this problem, we have to determine whether the horizontal component of
is of greater magnitude than the maximum force of static friction.
We can break the
vector into horizontal and vertical components. The vertical component will push the box harder into the floor, increasing the normal force, while the horizontal component will push against the force of static friction. First, let’s calculate the vertical component of the force so that we can determine the normal force,
N, of the box:
