Chapter 6, Problem 6.1P

A 150 Ib block rests on a horizontal floor. The coefficient of friction between the block and the floor is 0.30. A pull of 40 lb, acting upward at an angle of 30° to the horizontal, is applied to the block. Determine whether or not the block will slide.

To determine

The block will slide or not.

Answer to Problem 6.1P

The block will not slide.

Explanation of Solution

Given:

Weight of block $W=150lb$

Angle acting at pulling force on the block $\theta ={30}^{\circ }$

Pulling force on block $F=40lb$

Co-efficient of static friction $\mu =0.33$

Free body diagram.

First equating the vertical force to zero to find N.

   $\begin{array}{l}\sum Fy=0\\ W-N-F\sin \theta =0\\ N=150-(40\sin {30}^{\circ })\\ N=150-(40\times 0.5)\\ N=150-20\\ N=130lb\end{array}$

Therefore, the block is not going to move as the frictional force is more than the pulling force.

   $\begin{array}{l}F\cos \theta <f\\ F\cos \theta <\mu N\\ 40\cos {30}^{\circ }<0.3\times 130\\ 34.69<39\end{array}$

Conclusion:

Force on the block is not going to slide because the applied force on the block is not going to be enough to overcome the frictional force.