Chapter 1: Q4E (page 124)

A man is dragging a trunk up the loading ramp of a mover’s truck. The ramp has a slope angle of $20 °$ , and the man pulls upward with a force $\vec{F}$ whose direction makes an angle of $30 °$ with the ramp (Fig. E4.4).
(a) How large a force $\vec{F}$ is necessary for the component $F_{x}$ parallel to the ramp to be 90.0 N?
(b) How large will the component $F_{y}$ perpendicular to the ramp be then?

Expert verified

(a)The magnitude of the force $\vec{F}$ 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 by step solution

The component of the force parallel to the ramp is

$F_{x} = 90 N$

The angle made by the force with the ramp is

$θ = 30 °$

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

$\vec{A} = A c o s θ \hat{i} + A s i n θ \hat{j} . . . . . (1)$

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

$\begin{array}{rcl} F_{x} & = & F \cos θ \\ F \cos 30 ° & = & 90 N \\ F & = & \frac{90}{\cos 30 °} N \\ = & 103.9 N \end{array}$

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

$\begin{array}{rcl} F_{y} & = & F \sin θ \\ = & 103.9 \times \sin 30 ° N \\ = & 51.95 N \end{array}$

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

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