Question

An object acted on by three forces moves with constant velocity. One force acting on the object is in the positive xx direction and has a magnitude of 6.9 N; a second force has a magnitude of 4.5 N and points in the negative y direction. Find the direction and magnitude of the third force acting on the object.

Answer 1

Answer with Explanation:

We are given that

$F_1=6.9 N$

$F_2=4.5 N$

We have to find the direction and magnitude of the third force acting on the object.

Resultant force,F=$\sqrt{F^2_1+F^2_2}$

$F=\sqrt{(6.9)^2+(4.5)^2}=8.24 N$

The object moves with constant velocity .Therefore, net force on object is equal to zero

So,Third force,$F_3=F=8.24 N$

Direction,$\theta=tan^{-1}(\frac{F_2}{F_1})$

$\theta=tan^{-1}(\frac{4.5}{6.9})=33.02^{\circ}$

Angle lies in second quadrant because the direction of third force is opposite to the direction of the resultant force of F1 and F2.

Therefore,$\theta=\pi-\theta=180-33.02=146.98^{\circ}$