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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

1 Answer

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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.

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