Question

If Mr. Mangan is pushing a 90 kg sled across the ground with a force of300 N and it is just about to move, what is the coefficient of staticfriction between the sled and the ground?

Answer 1

Given data:

* The mass of the sled is 90 kg.

* The force applied to move the sled is 300 N.

Solution:

The normal force acting on the sled is,

`F_N=mg`

where m is the mass of sled, g is the acceleration due to gravity,

Substituting the known values,

`\begin{gathered} F_N=90*9.8 \\ F_N=882\text{ N} \end{gathered}`

The force acting on the sled is the same as static friction force on the sled.

Thus,

`F=F_s`

where F is the applied force, and F_s is the static friction,

The static friction of the sled is,

`F_s=\mu_sF_N`

Thus, the coefficient of static friction is,

`\begin{gathered} F=\mu_sF_N \\ \mu_s=(F)/(F_N) \end{gathered}`

Substituting the known values,

`\begin{gathered} \mu_s=(300)/(882) \\ \mu_s=\text{0}.34 \end{gathered}`

Thus, the coefficient of static friction is 0.34.