Question

An object has three forces acting on it 1200N at 51.3 degrees and 1400N at 42.0 degrees,what is the magnitude and direction of the third force?

Answer 1

I will assume you want the equilibrant.

I will use vectors

R = 1200(cos51.3,sin51.3) + 1400(cos42,sin42)
= (750.291,936.516) + (1040.403, 936.828) carrying all decimals my calculator can hold
= (1790.694, 1873.299)

|R| = sqrt(1790.694^2 + 1873.299^2)
= 2591.493

tan(theta) = 1873.299/1790.694
theta = 46.29 degrees

so the resultant has direction 180 + 46.29 degrees= 226.29 degrees
and a magnitude of 2591.493 N

You might want to check my arithmetic by doing the question using the cosine law

Answer 2

thank s

Answer 3

I did not understand

Answer 4

three force acting on an object figure 1.32.which is in equilibrium .determine force A