Question

Five forces act on an object.
(1) 59 N at 90°
(2) 40 N at 0°
(3) 81 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

Answer on #37758 – Physics – Mechanics

Five forces act on an object.

(1) 59 N at 90°

(2) 40 N at 0°

(3) 81 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?

Solution

We can split the forces into their two components, $x$ and $y$.

$$\text{For } x: F(x) = -F \cdot \cos(\theta)$$ $$\text{For } y: F(y) = -F \cdot \sin(\theta)$$

The sum of all $x$-components and $y$-components must be zero for equilibrium.

$$\text{For } x: 0 + (-40) + 0 + 40 + (-50) \cdot \frac{1}{2} + F_x = 0$$ $$F_x = 25\,N$$ $$\text{For } y: -59 + 0 + 81 + 0 + (-50) \cdot \frac{\sqrt{3}}{2} + F_y = 0$$ $$F_y = 21.3\,N$$

The total force $F$:

$$F = \sqrt{F_x^2 + F_y^2} = \sqrt{(25\,N)^2 + (21.3\,N)^2} = 32.8\,N$$

The angle between the X-axis and the force is:

$$\theta = \arctan\left(\frac{21.3\,N}{25\,N}\right) = 40.4{}^\circ$$

**Answer**: magnitude of the force: 32.8N; angle between the X-axis and the force: 40.4°