Question

A box weighing 12,000 N is parked on a 36° slope. What will be the component of the weight parallel to the plane that balances friction *


Fg sin 36 degrees
Fg cos 36 degrees
Fg tan 36 degrees
Fg x N

Answer 1

Given :

A box weighing 12,000 N is parked on a 36° slope.

To Find :

What will be the component of the weight parallel to the plane that balances friction.

Solution :

The component of that will be parallel to the plane to balance friction is :

$F \ cos ( 90 - 36)^o\\\\F\ ( sin\ 36^o)$

Therefore, component of force to balance friction is F sin 36° .

Hence, this is the required solution.