Q4E - University Physics with Modern Physics

Question

A man is dragging a trunk up the loading ramp of a mover’s truck. The ramp has a slope angle of  $\mathbf{20}\mathbf{°}$, and the man pulls upward with a force  $\overrightarrow{\mathbf{F}}$whose direction makes an angle of  $\mathbf{30}\mathbf{°}$ with the ramp (Fig. E4.4).

 (a) How large a force  $\overrightarrow{\mathit{F}}$ is necessary for the component  ${\mathit{F}}_{\mathbf{x}}$ parallel to the ramp to be 90.0 N?  

(b) How large will the component ${\mathit{F}}_{\mathbf{y}}$  perpendicular to the ramp be then?

Step-by-Step Solution

Answer

(a) The magnitude of the force$\overrightarrow{F}$   has to be $103.9 N$  for the component  $F_x$ parallel to the ramp to be 90.0 N.

 (b) The component  $F_y$ perpendicular to the ramp will be $51.95 N$

Step 1: Given data

The component of the force parallel to the ramp is

  $F_x=90 \text{N}$

The angle made by the force with the ramp is

  $\theta =30°$

Step 2: Vector components

The X  and  Y components of a vector $\overrightarrow{A}$  making an angle $\theta$ with the X axis can be summarized as

  $\overrightarrow{A}=Acos\theta \hat{i}+Asin\theta \hat{j} .....\left(1\right)$

Step 3: Components of the forces

From equation (1), the  x component of the force is

 $\begin{array}{rcl}{F}_x& =& F\cos \theta \\ F\cos 30°& =& 90 \text{N}\\ F& =& \frac{90}{\cos 30°} \text{N}\\ & =& 103.9 \text{N}\\ & & \end{array}$

From equation (1), the  y component of the force is

 $\begin{array}{rcl}{F}_y& =& F\sin \theta \\ & =& 103.9\times \sin 30° \text{N}\\ & =& 51.95 \text{N}\\ & & \end{array}$

The net magnitude of the force is  $103.9 \text{N}$ and the component perpendicular to the ramp is $51.95 \text{N}$ .