A man pulls a box with a force of 70N along the positive Y-axis, while a boy pulls it with a force of 50N, making an angle of 80°counter clockwise with the positive Y-axis.

Question

(1)In what direction should the second boy apply a force of 60N so that the box will move along the positive Y-axis? Hint;two values are possible

Answer 1

well,
along the x axis: You want the net force in that direction to be zero.

70Sin0deg+50sin80+60sinTheta=0
sinTheta=-50sin80deg/60= -0.820673128
Theta=-(55.1522333 deg) or -(-55.1522333 deg+180deg) or -124.8deg

Answer 2

F1 = 70[90o] CCW from +x-axis
F2 = 50[170o] CCW from +x-axis.
F3 = 60[Ao] from +x-axis.

Sum of hor. components = 0:
70*Cos90 + 50*Cos170 + 60*Cos A = 0.
60*Cos A = 49.2.
A = 34.9 degrees.

Answer 3

50*sin280 = -60*sin A,
- 49.24 = -60*sin A,
sin A = 49.2/60,
A = 55.1o CW from +Y-axis = 34.9o CCW from +X-axis.