Question

A set of forces are acting on an object. The first force, with a magnitude of 24.70 N, is directed at ─258o and

the second force, with a magnitude of 53.20 N, is directed at 37o north of east. The third force with a magnitude

of 166.50 N is directed at 270o, and the fourth force of 49.92 N is directed at 50o east of south. Solve for the

resultant force and its exact direction analytically.

Answer 1

Answer

Components of all force in x direction is written as

$F_x=-24.70sin12^°+53.20cos37^°+0+49.92sin50^°$

$F_x=$ -5. 14+42.49+38.24=75.59N

Now y component

$F_y=24.70cos12^°+53.20sin37^°-166.50-49.92cos50^°$

$F_y$ =24.16+32.02-166.5-32.09

=-142.41N

Now resultant force

$F=\sqrt{F_x^2+F_y^2}$

=$=\sqrt{(75.59)^2+(-142.41)^2}$

=161N

Direction

$\theta=\tan^{-1}(\frac{F_y}{F_x})$

$=\tan^{-1}(\frac{-142.41}{75.59})$

$=-62.11^°$