Chapter 5, Problem 85A

To determine

Component of weight parallel to inclined plane.

Answer to Problem 85A

  $F_{\left|\right|}=123.32\text{N}$

Explanation of Solution

Given:

Weight of box, $F_g=215\text{N}$

Angle of inclined plane with horizontal, $\theta =35°$

Formula Used:

Component of weight parallel to inclined plane is $F_g\sin \theta$ .

Calculation:

From the given condition, the figure obtained is shown below

  

As, angle of inclined plane with horizontal, $\theta =35°$

So, angle of inclined plane with vertical, $90°-\theta$

Therefore, the component of weight parallel to inclined plane will also make angle $90°-\theta$ with the vertical.

So, the component of weight parallel to inclined plane will be,

  $\begin{array}{c}{F}_{\left|\right|}=F_g\cos \left(90°-\theta \right)\\ =F_g\sin \theta \\ F_{\left|\right|}=215\times \sin 35°\\ =123.32\text{N}\end{array}$

Conclusion:

Component of weight parallel to inclined plane is $F_{\left|\right|}=123.32\text{N}$ .