Chapter 5, Problem 3STP

To determine

To Choose: The correct option.

Answer to Problem 3STP

Option B: 0.57

Explanation of Solution

Given:

The minimum force required to move a $50\text{ kg}$ crate is $280\text{ N}$ .

Formula used:

The maximum static friction force is given by,

  $F={\mu }_{s}N$

Where, ${\mu }_s$ is the maximum static friction coefficient and N is the normal force.

Calculation:

Since there is no motion of crate in the perpendicular direction of the surface, the normal force between crate and the surface will be balanced by the weight of the carte, that is

  $\begin{array}{c}N=W_{crate}\\ =m_{\text{crate}}g\end{array}$

Now, maximum static friction force acting on an object will be equal to the minimum force required to move that object.

Thus, it can be written as:

  $\begin{array}{c}{F}_{\min }={\left({\mu }_s\right)}_{\max }N\\ ={\left({\mu }_s\right)}_{\max }{m}_{cart}g\end{array}$

Where ${\left({\mu }_x\right)}_{\max }$ is the maximum static friction constant.

Thus, using given values in above equation:

  $\begin{array}{l}280={\left({\mu }_s\right)}_{\max }\left(50\right)\left(9.8\right)\\ {\left({\mu }_s\right)}_{\max }\approx 0.57\end{array}$

Conclusion:

Thus, the maximum static friction coefficient is 0.57.