Static Friction:

A dockworker is trying to move a safe. It has a mass of 500 kg and he can supply 650 N of force. If the coefficient of static friction is 0.40, is that enough force to move it?

The worker must exceed the maximum static friction between the safe and the floor. The maximum static friction is found with this equation:

f_s = m_sFn

f_s = m_s(mg)

f_s = (0.40)(500 kg)(9.80 m/s²)

f_s = 1960 N

So, frustrated as the worker may be, he can’t move the safe.

Sliding Friction on a flat surface:

A dockworker slides a 10 kg crate across a greasy floor with an initial velocity of 10 m/s. The coefficient of kinetic friction between the crate and the floor is 0.08.

a) What is the value of kinetic Friction?

b) Draw the Free Body Diagram

c) What is the acceleration?

d) How far does the crate slide?

SOLUTIONS:

a) since f_k  =  m_kFn

we get:  f_k  = (0.08)(mass*gravity)

            f_k  =  (0.08)(10 kg * 9.8 m/s²)

            f_k  =  7.84 Newtons

b)

Label all the forces on the crate:

Notice that there is only ONE force acting horizontally on the object as it slides...friction. Kinetic friction. Velocity is NOT a force.

c)        net F_x     = ma

            -f_k     = ma

     -7.84 N    = (10 kg)a

               a    = - 0.784 m/s²

Notice that it’s negative.

d)  Use kinematics:

Dx = v_f²-v_i² / 2a

Dx = 0 - (10 m/s)² / 2(-0.784 m/s²)

Dx = + 63.8 m                 QED.....