Three forces with magnitudes of pounds, pounds, and pounds act on an object at angles , , and , respectively, with the positive x-axis. Find the magnitude and direction of the resultant force. Round answers to two decimal places.
A) pounds;
B)130.55 pounds;
C) pounds;
D) pounds;
E)130.55 pounds;

Question

Three forces with magnitudes of  pounds,  pounds, and  pounds act on an object at angles  ,  , and  , respectively, with the positive x-axis. Find the magnitude and direction of the resultant force. Round answers to two decimal places.
A) pounds; 
B)130.55 pounds; 
C) pounds; 
D) pounds; 
E)130.55 pounds;

Quizplus · Accepted Answer

Using the Law of Cosines, we can find the magnitude of the resultant force:
$R^2 = F_1^2 + F_2^2 + F_3^2 + 2F_1F_2\cos	heta_{12} + 2F_2F_3\cos	heta_{23} + 2F_1F_3\cos	heta_{13}$

$R^2 = (11 	ext{ pounds})^2 + (11 	ext{ pounds})^2 + (11 	ext{ pounds})^2 + 2(11 	ext{ pounds})(11 	ext{ pounds})\cos(60^\circ) + 2(11 	ext{ pounds})(11 	ext{ pounds})\cos(120^\circ) + 2(11 	ext{ pounds})(11 	ext{ pounds})\cos(180^\circ-40^\circ)$

$R^2 = 363	ext{ pounds}^2 + 363	ext{ pounds}^2 + 363	ext{ pounds}^2 + 2(11 	ext{ pounds})^2(0.5) + 2(11 	ext{ pounds})^2(-0.5) + 2(11 	ext{ pounds})^2(-\cos(40^\circ))$

$R^2 = 1305.52	ext{ pounds}^2$

$R= \sqrt{1305.52}	ext{ pounds} \approx 130.55 	ext{ pounds}$

Using the Law of Sines, we can find the direction of the resultant force:
$\frac{\sin	heta_{R}}{R} = \frac{\sin	heta_{13}}{F_3}$

$	heta_{R} = \sin^{-1}\left(\frac{F_3\sin	heta_{13}}{R}ight) = \sin^{-1}\left(\frac{(11 	ext{ pounds})\sin(40^\circ)}{130.55 	ext{ pounds}}ight) \approx 11.5^\circ$ measured counterclockwise from the positive x-axis.

Therefore, the magnitude and direction of the resultant force are approximately 130.55 pounds and 11.5 degrees counterclockwise from the positive x-axis, respectively. The best choice is B.

Three Forces with Magnitudes of Pounds, Pounds, and