Question

Five forces act on an object.
(1) 62 N at 90°
(2) 40 N at 0°
(3) 85 N at 270°
(4) 40 N at 180°
(5) 50 N at 60°
What are the magnitude and direction of a sixth force that would produce equilibrium?

Answer 1

Question

Horizontal Forces:

$$F _ {h r e s} = F _ {2} - F _ {4} + F _ {5} \cdot \cos 6 0 {}^ {\circ} = 4 0 - 4 0 + 5 0 \cdot \frac {1}{2} = 2 5 N.$$

Vertical Forces:

$$F _ {v r e s} = F _ {1} - F _ {3} + F _ {5} \cdot \sin 6 0 {}^ {\circ} = 6 2 - 8 5 + 5 0 \cdot \frac {\sqrt {3}}{2} \approx 2 0. 3 N.$$

Resultant Force:

$$F _ {r e s} = \sqrt {F _ {v r e s} ^ {2} + F _ {h r e s} ^ {2}} = \sqrt {2 0 . 3 ^ {2} + 2 5 ^ {2}} \approx 3 2. 2 N. \text{ The direction:}$$ $$\cos \alpha = \frac {2 5 \mathrm {N}}{3 2 . 2 \mathrm {N}} \Rightarrow \alpha = \arccos \left(\frac {2 5 \mathrm {N}}{3 2 . 2 \mathrm {N}}\right) \approx 3 9 {}^ {\circ}.$$

So, we can say that sixth force should be $32.2\mathrm{N}$ in magnitude and $180{}^{\circ} + 39{}^{\circ} = 219{}^{\circ}$ in direction.

Answer: magnitude: 32.2 N; direction: $219{}^{\circ}$