Question

A

Answer 1

Answer:

$F_r = 40.01N$

Explanation:

Given

Let the two forces be $F_1\ and\ F_2$

$F_1 = 80N$

$F_2 = 70N$

$\theta = 150\deg$

Required

Determine the resultant force $F_r$

This will be solved using parallelogram law of resultant force.

The equation is:

$F_r^2 = F_1^2 + F_2^2 + 2F_1F_2cos\theta$

Substitute values for $F_1$, $F_2$ and $\theta$

$F_r^2 = 80^2 + 70^2+ 2 * 70 * 80 * cos\ 150$

$F_r^2 = 6400 + 4900 + 11200* cos\ 150$

$F_r^2 = 6400 + 4900 + 11200* -0.8660$

$F_r^2 = 6400 + 4900 - 9699.2$

$F_r^2 = 1600.8$

Take the square root of both sides

$F_r = \sqrt{1600.8$

$F_r = 40.01N$

Hence, the resultant force is 40.01 N