Question

the magnitude and direction of two forces acting on an object are 120 pounds, s54°e, and 50 pounds, n56°e, respectively. find the magnitude, to the nearest hundredth of a pound, and the direction angle, to the nearest tenth of a degree, of the resultant force.
the magnitude is approximately □ pounds.
(do not round until the final answer. then round to the nearest hundredth as needed.)

Explanation:

Step1: Resolve forces into components

For 120 lb at 54°E:
$F_{1x} = 120\cos(54^\circ) \approx 120 \times 0.5878 = 70.536$
$F_{1y} = 120\sin(54^\circ) \approx 120 \times 0.8090 = 97.08$

For 50 lb at 56°E:
$F_{2x} = 50\cos(56^\circ) \approx 50 \times 0.5592 = 27.96$
$F_{2y} = 50\sin(56^\circ) \approx 50 \times 0.8290 = 41.45$

Step2: Sum components

$F_x = 70.536 + 27.96 = 98.496$
$F_y = 97.08 + 41.45 = 138.53$

Step3: Calculate magnitude

$F = \sqrt{F_x^2 + F_y^2} = \sqrt{98.496^2 + 138.53^2} \approx \sqrt{9691.46 + 19190.56} = \sqrt{28882.02} \approx 169.95$

Answer:

169.95