What force P makes net work by friction and P equal zero?

[leezak]

A 121-kg crate is being pulled across a horizontal floor by a force P that makes an angle of 36.8° above the horizontal. The coefficient of kinetic friction is 0.212. What should be the magnitude of P, so that the net work done by it and the kinetic frictional force is zero?

the work of gravity and the normal force are going to be 0 because they are 90 degrees from the distance that the box is travelling. should the work of kinetic friction be 0 because it says "kinetic frictional force is zero"? that would mean that the net work is equal to the force of P * distance * cos 39.9 and we are looking for the force of P, but we don't know the distance... I'm confused can someone help? thanks

 

[Doc Al]

The problem doesn't say that "kinetic frictional force is zero"; it says that the net work done by the force P and the kinetic friction is zero. What does that tell you about the net horizontal force on the crate?

 

[leezak]

it means that the crate is in equilibrium, so would the net horizontal force be zero?

 

[Doc Al]

Right! Now use that fact to figure out what P must be.

 

[leezak]

okay... so i figured that the kinetic frictional force + the horizontal component must equal 0. so... -(.212)*(121)(9.8) + P*cos(36.8) = 0 and i found out that P = 313.95 N but that's the wrong answer so I'm not sure what i did wrong

 

[Doc Al]

Your calculation assumes that the normal force equals the weight of the crate. Not so. The vertical component of P reduces the normal force.

 

[leezak]

do i include the vertical component of force P in with the same equation making... the kinetic frictional force + the horizontal component of force P + the vertical component of force P = 0 ??

 

[Doc Al]

To calculate the friction, you need the normal force. That normal force depends on P. Set up force equations for vertical and horizontal components and solve for P.

 

[leezak]

ahh i got it thank you sooo much