Question

Two forces of magnitudes 4.0 Newtons and 3.0 Newtons pull on a box. The forces make an angle of 40° with each other. What is the magnitude of a third force that must be applied to keep the box in equilibrium?

Answer 1

Answer:

6.6 N

Explanation:

Let's take the direction of the force of 4.0 N as positive x-direction. This means that the force of 3.0 N is at 40 degrees above it. So the components of the two forces along the x- and y-directions are:

$F_{1x} = 4.0 N\\F_{1y} = 0$

$F_{2x} = 3.0 N cos 40^{\circ}=2.3 N\\F_{2y} = 3.0 N sin 40^{\circ} = 1.9 N$

So the resultant has components

$F_x = F_{1x}+F_{2x}=4.0 N +2.3 N = 6.3 N\\F_y = F_{1y} + F_{2y} = 0 + 1.9 N = 1.9 N$

So the magnitude of the resultant is

$F=\sqrt{F_x^2 +F_y^2}=\sqrt{(6.3)^2+(1.9)^2}=6.6 N$

And in order for the body to be balanced, the third force must be equal and opposite (in direction) to this force: so, the magnitude of the third force must be 6.6 N.