Question

Two forces acting on a body are 500 N and 1000 N as shown in Fig. Determine the
third force F such that the resultant of all the three forces is 1000 N directed at 45° to x
axis​

Answer 1

Given:

Force 1 = 500 N at 30° from +ve x axis

Force 2 = 1000 N at 30° from +ve y axis

To find:

The  third force F such that the resultant of all the three forces is 1000 N directed at 45° to x axis.

Solution:

Evaluating all the components of the forces in x direction:

1000 sin 30° + 500 cos 30° + Fx = 1000 sin 45°

500 + 250√3 + Fx = 500√2

Fx = -225.9

Fx = 225.9 in the -ve x direction

Evaluating all the components of the forces in y direction:

1000 cos 30° + 500 sin 30° + Fy = 1000 sin 45°

500√3 + 250 + Fy = 500√2

Fy = -408.9

Fy = 408.9 in the -ve y axis

Total force = √(Fx² + Fy²)

= √(225.9² + 408.9²)

= 467.15 N and its angle α = tan α = 225.9/408.9= 0.55

Therefore the required force is of 467.15 N.