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

10 N,
`53.1^(\circ)` north-east

Step-by-step 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=√((6 N)^2 + (8 N)^2)=10 N`

The direction of the force instead is given by

`\theta = tan^(-1) ((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) ((8 N)/(6 N))=tan^(-1) (1.333)=53.1^(\circ)`