Question

A 15 kg box is pulled at an angle of 15* from the horizontal with a force of 55N In the coefficient of kinetic friction is 0.25. What is the normal force?

Answer 1

Net force along the axis, parallel to inclined plane is:

$F + m g \sin \alpha - F_f$, where $F = 55 N$ and $F_f = \mu N$ is static friction, $\alpha = 15^{\circ}$is the angle of inclination.

Assuming, we are pulling the box with no acceleration, $F + m g \sin \alpha - \mu N = 0$, from where $N = \frac{F + m g \sin \alpha}{\mu} \approx 372.3 N$.