How Is the Net Work Zero with Both Push and Frictional Forces on a Crate?

[keemosabi]

Homework Statement

A 1.20 x 10² kg crate is being pushed across a horizontal floor by a force P that makes an angle of 30.0° below the horizontal. The coefficient of kinetic friction is 0.450. What should be the magnitude of P, so that the net work done by it and the kinetic frictional force is zero?

Homework Equations

W = FD

The Attempt at a Solution

I set the work done by friction 1200 x 9.8 x .450 x D equal to the work done by P which is P x cos 30 x D. I then canceled D out of both sides, and solved for P, and got 611.068 N. What did I do wrong?

 

[JoAuSc]

1.20 x 10^2 isn't 1200.

 

[LowlyPion]

Homework Statement

A 1.20 x 10² kg crate is being pushed across a horizontal floor by a force P that makes an angle of 30.0° below the horizontal. The coefficient of kinetic friction is 0.450. What should be the magnitude of P, so that the net work done by it and the kinetic frictional force is zero?

Homework Equations

W = FD

The Attempt at a Solution

I set the work done by friction 1200 x 9.8 x .450 x D equal to the work done by P which is P x cos 30 x D. I then canceled D out of both sides, and solved for P, and got 611.068 N. What did I do wrong?

There is not only the weight but also the Sin30 component of the downward force that goes into calculating the normal force that determines frictional resistance.

Cos30*P = u*m*g + u*Sin30*P is what you need to solve.

Oh and 1.2 x 10² is 120 as already noted by JoAuSc.

 

[keemosabi]

1.20 x 10^2 isn't 1200.

Oops, sorry about that. That was only a mistake in my typing; in my calculations I used 120.

 

[keemosabi]

There is not only the weight but also the Sin30 component of the downward force that goes into calculating the normal force that determines frictional resistance.

Cos30*P = u*m*g + u*Sin30*P is what you need to solve.

Oh and 1.2 x 10² is 120 as already noted by JoAuSc.

Ohhhh...I get it. Thank you so much for the help.

Edit: So if the 30 degree angle was pulling above the horizontal, I would subtract that from the normal force? In this case it's the opposite, but that's the general concept that I was missing, right?

 

[LowlyPion]

Ohhhh...I get it. Thank you so much for the help.

Edit: So if the 30 degree angle was pulling above the horizontal, I would subtract that from the normal force? In this case it's the opposite, but that's the general concept that I was missing, right?

Correct.

 

[keemosabi]

Correct.

Thank you for your help.

 

[keemosabi]

There is not only the weight but also the Sin30 component of the downward force that goes into calculating the normal force that determines frictional resistance.

Cos30*P = u*m*g + u*Sin30*P is what you need to solve.

Oh and 1.2 x 10² is 120 as already noted by JoAuSc.

Edit: Nevermind.