Question

A table of mass 20 kg is pushed at constant speed across a floor. If the coefficient of static friction is 0.9 and the coefficient of kinetic friction is 0.6, what horizontal force is necessary to keep the speed constant?g=10N/kg

Answer 1

The given problem can be exemplified in the following free-body diagram:

Now we sum the forces in the horizontal direction:

`F-F_f=ma_x`

Where:

`\begin{gathered} F=\text{ Horizontal force} \\ F_f=\text{ Friction force} \\ m=\text{ mass} \\ a=\text{ acceleration} \end{gathered}`

Since the movement is required to be at a constant speed the acceleration must be zero, therefore:

`F-F_f=0_{}`

Now we solve for the force "F":

`F=F_f`

Now we use the following definition of friction:

`F_f=\mu_kN`

Where:

`\begin{gathered} \mu_k=\text{ coefficient of kinetic friction} \\ N=\text{ normal force} \end{gathered}`

The kinetic friction coefficient is used due to the fact that we are considering the body as it is already in movement.

Replacing in the sum of horizontal forces:

`F=\mu_kN`

Now, to determine the value of the normal force we will add the forces in the vertical direction:

`N-mg=0`

The vertical forces add up to zero because there is no vertical movement. Now we solve for the normal force "F":

`N=mg`

Now we replace in the sum of horizontal forces:

`F=mg\mu_k`

Now we plug in the values:

Solving the operations we get:

`F=120N`

Therefore, a force of 120 Newtons is required.