Chapter 9.5, Problem 33AYU

To determine

To find: The acute angle that a constant unit force vector makes with the positive $x\text{-axis}$.

Answer to Problem 33AYU

Solution:

${60}^{\circ }$

Explanation of Solution

Given:

Work done by the force in moving a particle from $\left(0,0\right)$ to $\left(4,0\right)$ equals 2.

Calculation:

Work $=F_x$

$x=$ displacement in force direction $=\sqrt{{(0-0)}^2+{(4-0)}^2}=4$ units in the positive $x$ direction.

$2=F\times 4$

$F=\frac{2}{4}=\frac{1}{2}N$ (the horizontal component of the force).

$F\left(horizontal\right)=F\times \text{ cos}\Phi$

$\text{cos}\Phi =\frac{F\left(horizontal\right)}{F}=\frac{\frac{1}{2}}{1}=\frac{1}{2}$

$\Phi ={\cos }^{-1}\left(\frac{1}{2}\right)={60}^{\circ }$