Question

Two forces, F⃗ 1 and F⃗ 2, act at a point. F⃗ 1 has a magnitude of 9.60 N and is directed at an angle of 59.0 ∘ above the negative x axis in the second quadrant. F⃗ 2 has a magnitude of 6.60 N and is directed at an angle of 53.9 ∘ below the negative xaxis in the third quadrant.

A) What is the x component of the resultant force?

B) What is the y component of the resultant force?

C) What is the magnitude of the resultant force?

Answer 1

Given that,

F1 = 9.6 N ; at 1 = 59 Deg ;

F2 = 6.6 N and 2 = 53.9 Deg

(a)The x components of F1 and F2 will be:

F1 x = F1 cos1 = 9.6 x cos(59) = 4.94 N

F1y = F1 sin1 = 9.6 x sin(59) = - 8.23 N (above the negative x axis in the second quadrant)

F2x = F2 cos2 = 6.6 x cos(53.9) = 3.89 N

F2y = F2 sin2 = 6.6 x sin(53.9) = 5.33 N

We have to add the x and y components.

Fx = F1x + F2x = 4.94 N + 3.89 N = 8.83 N

Hence, X component of the resultant force = Fx = 8.86 N

(b)Fy = F1y + F2y = -8.23 N + 5.33 N = - 2.9 N

Hence, Y component of the resultant force = Fy = - 2.9 N

(c)Magnitude of the resultant force is given by:

F = sqrt [(Fx)² + (Fy)²] = [ (8.83)² + (2.9)² ] = 9.3 N

Hence, the magnitude of the resultant force = F = 9.3 N