Answer to Question #234777 in Mechanics | Relativity for adam

Question

Knowing that P = 120 lb and that the resultant of P and Q lies in the positive x-direction,

determine Q and the magnitude of the resultant.

Answer 1

From Force body diagram

$sinθ = \frac{3}{5} \\ cosθ = \frac{4}{5} \\ sinϕ = \frac{5}{13} \\ cosϕ = \frac{12}{13} \\ P= 120 \;lb$

Since there is resultant only in x-direction and no resultant force in y-direction.

$P_y=Q_y \\ Psinθ =Qcosϕ \\ 120 \times \frac{3}{5}= Q \times \frac{12}{13} \\ Q = 120 \times \frac{3}{5} \times \frac{13}{12} \\ Q = 78 \;lb$

Resultant

$R=Q_x+P_x \\ R= Qsinϕ + Pcosθ \\ R= (78 \times \frac{5}{13}) + (120 \times \frac{4}{5}) \\ = 30+96 = 126 \;lb$