Statis friction is the friction on a object that isn’t moving and up until the point at which it does move. Friction is a force caused by the resistance of a surface against which an object is trying to move so the magniude of the friction depends upon three things

1. The magnitude of the force in the direction parallel to the surface
2. The magnitude of the force acting on the object perpendicular to the surface (the normal force N)
3. The coefficient of friction acting between the object and the surface (this  is a measure of the “roughness” of the surface between 0 and 1 where 0 is a completely smooth surface and with coefficient 1 the object couldn’t be moved however large the force μ.

Firction acts parallel to the surface opposing motion and the normal is perpendicular to the surface

We know that if the object is in equlibrium then the resultant force in the direction parallel to the surface must equal zero and hence the firctional force must be equal to the resultant force acting on the object parallel to the plane. However, what we are interested in finding is the minimum force that must be applied in order that the object begins to move, that is the maximum value that the firction can take before it moves.

To find this maximum we simply multiply the normal force by the coefficient of friction. That is the firctional force R on a static object on a surface being acted upon by a normal force N and a force parallel to the surface F with coefficent of friction between the object and the surface μ is

F ≤ Nμ

and if the object is on the point of moving (that is the we applying the minimum force required to make the object move)

F = Nμ

Example

An object of mass 5kg is on a rough slope at 30 degrees to the horizontal. The acceleration due to gravity is g and the object is on the point of slipping down the slope. Find the coefficient of friction between the slope and the object.

We know from Newtons laws that the force of gravity acting on the object will have a equal force opposing it since the object is at rest. To use the formulea above we want to break this into two components parallel and perpendicular to the slope. The component perpendicular to the slope is the normal force N and the component parallel to the slope is the force R which is trying to move the object down the slope.

Firstly it is always useful to draw a diagram of what we are interested in.

The forces acting upon the object on a slope

Since we know that the object is on the point of slipping down the slope we can the use the eqaulity from the equations above

F = Nμ

So to find μ we need to first calulate F and N. Because the object is in equilibrium F = R .

So we can resolve the weight into directions parallel and perpendicular to the slope to get

N = Wcos(30) = (5g√3)/2

and

F = Wsin(30) = 5g/2

so we can re arrange to get