# Finding kinetic force and net work

## Homework Statement

A 123-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 0.270. What should be the magnitude of P, so that the net work done by it and the kinetic frictional force is zero?

Fcos@=Fnet
F=ma
Fk = k*Fn

## The Attempt at a Solution

I drew a free body diagram.

horizontal force--->
|
| crate ^ normal force
v <---- frictional force (coefficient = .27)
weight

The normal force pushing on the crate is 1205.4 N. Multiplying this by the coefficient of kinetic friction gives me 325.46 N for the magnitude of Fk. This is in opposition to the crate's movement, so the horizontal force pushing back is also 325.46 (making the opposing force negative, really). Plugging this into Pcos@ gives a P value of 375.8 N. However, this is incorrect.

To make the net force zero (and therefore the net work zero), they should be opposites of each other, right? So where is the hole in my logic?

So my free body diagram didn't come out right.. the far left force (down) is the weight, and the force to the left is the frictional force. Sorry about that.. if confused, draw your own!

Astronuc
Staff Emeritus
P is pushing at an angle from below horizontal.

In order for no work to be performed, the frictional force would have to be zero, and the only way that will happen is if the net force of the crate on the floor is zero.

Find the vertical (upward) component of P which is equal to the weight of the crate.

The upward component of P is 1205.4, which would make P = 1205/cos30, which is 1391.41 N. Still not correct. Is that what you meant me to find?

PhanthomJay
Homework Helper
Gold Member
Do you have a diagram? Before you go too far with this problem, it needs to be clarified whether the pushing force is to the right and downward, or to the right and upward. When the problem says '30 degrees below the horizontal', that can be interperted either way...is it 30 degrees below the negative x axis, which gives it an upward component, or 30 degrees below the positive x axis, which gives it a downwrd component?. In any case, the appied force affects the normal force.

It's pushing downward, because that's how all of our problems are set up. The force is above the box and pointing 30 degrees below the horizontal.

PhanthomJay
Homework Helper
Gold Member
OK, so now you must calculate the normal force. What are all the forces in the y direction in your FBD? What are they in the x direction?

All the forces pushing y-direction: the vertical component of P, the weight of the block, and the normal force.
All x-direction F's: The horizontal component of P and the kinetic frictional force (does this go the x direction? I know static force does...)

PhanthomJay
Homework Helper
Gold Member
All the forces pushing y-direction: the vertical component of P, the weight of the block, and the normal force.
All x-direction F's: The horizontal component of P and the kinetic frictional force (does this go the x direction? I know static force does...)
Yes, correct on all counts. Now proceed as you have started, using Newton's law in each direction to solve for P and N.

so Fn = mg + Psin30
.27*Fn = Fk
Fk = -Pcos30

Fn = 1205.4 +.5P
.27*Fn = Fk
-Pcos30 = 325.458+.135P
-.866P = .135P +325.458
-1.001P = 325.458
P = 325.15 N

Does that look correct and/or is that the result you find? I couldn't decide whether Fk = -Pcos30 or Fk = Pcos30, but since they're inverses, I think I set that one up correctly.

PhanthomJay
Homework Helper
Gold Member
so Fn = mg + Psin30
.27*Fn = Fk
Fk = -Pcos30

Fn = 1205.4 +.5P
.27*Fn = Fk
-Pcos30 = 325.458+.135P
-.866P = .135P +325.458
-1.001P = 325.458
P = 325.15 N

Does that look correct and/or is that the result you find? I couldn't decide whether Fk = -Pcos30 or Fk = Pcos30, but since they're inverses, I think I set that one up correctly.
Almost, but you let the minus sign get the best of you. Since the horizonatl component of the pushing force must be exactly balanced by the friction force in order for no net work to be done, then using Newton 1, Pcos 30 - .27N = 0, or
Pcos30 = +.27N. This is the way you approached it in the y direction when you said that the normal force was equal to the weight plus Psin30....you didn't use a minus sign there for N, so you must be consistent throughout. Your answer is not correct.

So then .866P - .135P = 325.458
.731P = 325.458
P = 445.22 N

...Better? I didn't think about the total having to equal zero, but now I realize how obvious that is.

PhanthomJay