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

966 N

Step-by-step 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`