Chapter 5.3, Problem 34PP

To determine

Parallel and perpendicular component of cup’s weight.

Answer to Problem 34PP

  $F_{\left|\right|}=1.116\text{N}$ and $F_{\perp }=4.17\text{N}$

Explanation of Solution

Given:

Angle of plane, $\theta =15°$

Mass of cup, $m=0.44\text{kg}$

Formula Used:

The components are given as:

  $F_{\perp }=F_g\cos \theta$ and $F_{\left|\right|}=F_g\sin \theta$

Where, $F_{\perp }$ component of weight perpendicular to inclined plane, $F_{\left|\right|}$ is component of weight parallel to inclined plane, $F_g$ is weight and $\theta$ is angle of inclination.

Calculation:

Consider the below diagram for the given condition,

  

Here,

Weight of book acting downward, $F_g$

Angle of inclined plane with horizontal, $\theta$

Angle of inclined plane with vertical, ${90}^0-\theta$

The component of weight parallel to inclined plane will make angle ${90}^o-\theta$ with the vertical due to corresponding angle.

So, the component of weight perpendicular to inclined plane will make angle $\theta$ with the vertical.

Now,

The component of weight perpendicular to inclined plane will be,

  $\begin{array}{c}{F}_{\perp }=F_g\cos \theta \\ =mg\cos \theta \\ =0.44\times 9.8\times \cos 15°\\ =4.17\text{N}\end{array}$

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

  $\begin{array}{c}{F}_{\left|\right|}=F_g\cos \left(90°-\theta \right)\\ =mg\sin \theta \\ =0.44\times 9.8\times \sin 15°\\ =1.116\text{N}\end{array}$

Conclusion:

Therefore, parallel and perpendicular components are $F_{\left|\right|}=1.116\text{N}$ and $F_{\perp }=4.17\text{N}$ .