Question

What is the magnitude of the forces when F=3, theta = 10 degrees. In kN?

у 4 kN F 15° 0 IT 30° х 6 kN

Answer 1

Solution:

у 4 kN 4* cos15 F*cos10 F 150 e F*sin 10 4* sin15 30° x 6 kN

Taking resoltion of horizontal forces,

F = 6+ (3 * sin 10°) - (4 * sin 15°)

F = 5.486kN

Taking resolution of vertical forces,

Fy = (3 * cos10°) + (4 * cos15°)

$F_{y}=6.818kN$

Therefore, Resultant force is,

$F=\sqrt{F_{x}^{2}+F_{y}^{2}}$

$F=\sqrt{5.486^{2}+6.818^{2}}$

$F=8.75kN$