Question

A box of mass 115 kg sits on an inclined surface with an angle of 59º. What
is the component of the weight of the box along the surface?

Answer 1

Answer:

966 N

Explanation:

The component of the weight along the direction perpendicular to the plane is:

$mg cos \theta$

where

m is the mass of the box, g the acceleration of gravity, $\theta$ the angle of the inclined plane.

Similarly, the component of the weight of the box along the surface is

$mg sin \theta$

acting down along the incline.

Here we have:

m = 115 kg

g = 9.8 m/s^2

$\theta=59^{\circ}$

Substitting in the last equation, we find the component of the weight of the box along the surface:

$mg sin \theta = (115)(9.8)(sin 59)=966 N$