Answer:
[tex]F_r = 40.01N[/tex]
Explanation:
Given
Let the two forces be [tex]F_1\ and\ F_2[/tex]
[tex]F_1 = 80N[/tex]
[tex]F_2 = 70N[/tex]
[tex]\theta = 150\deg[/tex]
Required
Determine the resultant force [tex]F_r[/tex]
This will be solved using parallelogram law of resultant force.
The equation is:
[tex]F_r^2 = F_1^2 + F_2^2 + 2F_1F_2cos\theta[/tex]
Substitute values for [tex]F_1[/tex], [tex]F_2[/tex] and [tex]\theta[/tex]
[tex]F_r^2 = 80^2 + 70^2+ 2 * 70 * 80 * cos\ 150[/tex]
[tex]F_r^2 = 6400 + 4900 + 11200* cos\ 150[/tex]
[tex]F_r^2 = 6400 + 4900 + 11200* -0.8660[/tex]
[tex]F_r^2 = 6400 + 4900 - 9699.2[/tex]
[tex]F_r^2 = 1600.8[/tex]
Take the square root of both sides
[tex]F_r = \sqrt{1600.8[/tex]
[tex]F_r = 40.01N[/tex]
Hence, the resultant force is 40.01 N
