Question

Given the scenario: A 45 kg. object is being pushed north with a force of 8N, and at the same time, it is been pushed east with a force of 6N. What will the magnitude and direction be of the resultant vector?

Answer 1

Answer:

10 N, $53.1^{\circ}$ north-east

Explanation:

The two forces applied on the object are perpendicular to each other, so we can find the magnitude of their resultant by using the Pythagorean theorem:

$R=\sqrt{(6 N)^2 + (8 N)^2}=10 N$

The direction of the force instead is given by

$\theta = tan^{-1} (\frac{F_n}{F_e})$

where F_n is the force pointing north, and F_e is the force pointing east, and the angle is measured with respect to the east direction. Substituting,

$\theta=tan^{-1} (\frac{8 N}{6 N})=tan^{-1} (1.333)=53.1^{\circ}$