Chapter 5.3, Problem 36PP

To determine

The value of angle $\theta$ for the given situation.

Answer to Problem 36PP

The value of $\theta$ is $26.6°$ .

Explanation of Solution

Given:

Consider the plane is inclined at an angle $\theta$ with the horizontal.

The mass of the suitcase is denoted by $m$ .

Consider the acceleration due to gravity of the Earth is $g=9.80{\text{m/s}}^2$ .

The component of the suitcase’s weight parallel to the plane is equal to half the component of suitcase’s weight perpendicular to the plane.

Formula used:

Consider a force $F$ is acting on an object at an angle $\theta$ with the horizontal.

The magnitude of horizontal component of the force $F$ is,

  $F_x=F\cos \theta$

The magnitude of vertical component of the force $F$ is,

  $F_y=F\sin \theta$

Calculation:

The free body diagram of the situation is shown below.

The parallel component of the suitcase’s weight is denoted by $F_{\text{H}}$ .

The perpendicular component of suitcase’s weight is denoted by $F_{\text{V}}$ .

The component of the suitcase’s weight parallel to the plane is,

$F_{\text{H}}=-mg\sin \theta$

The component of the suitcase’s weight perpendicular to the plane is,

$F_{\text{V}}=-mg\cos \theta$

The suitcase’s weight parallel component is one-half the perpendicular component.

  $\begin{array}{l}{F}_{\text{H}}=\frac{1}{2}{F}_{\text{V}}\\ -mg\sin \theta =\frac{1}{2}\left(-mg\cos \theta \right)\\ \left(\frac{\sin \theta }{\cos \theta }\right)=\frac{1}{2}\\ \tan \theta =\frac{1}{2}\\ \theta ={\tan }^{-1}\left(\frac{1}{2}\right)\\ \theta =26.6°\end{array}$

Conclusion:

Thus, the value of $\theta$ is $26.6°$ .