Discrepancy in force of friction in a problem

1. Mar 21, 2013

fredrogers3

1. The problem statement, all variables and given/known data
Here is the problem: A student pushes a 5.0 kg block at a constant speed 3 m across the ceiling of
her room using a force Fa
of magnitude 85 N and angle 75º. The coefficient of
kinetic friction between the block and the ceiling is 0.40.
Now, we must find the work done by the frictional force.

2. Relevant equations
My Newton Second Laws:
F, total,x direction= Facosθ-forcefriction=0
F, total, y direction=-Fn-mg+Fasinθ=0

3. The attempt at a solution
When I solve for the force of friction, my answer comes out different depending on how I solve it. For instance, if I use my x direction relations, I would get forcefriction=facosθ. Thus, forcefriction would be 85cos75= 21.9996
If I solve the y eqn for fn, I get a normal force of 33.1. This multiplied by the kinetic constant (.4) is not 21.9996. What is causing the differences here in the friction forces?

Last edited: Mar 21, 2013
2. Mar 21, 2013

rude man

You can't find the work done. You can only find the power dissipated by friction.

What is the net vertical force F pushing the block up against the ceiling?

Then how is power related to F, μk and the velocity?

3. Mar 21, 2013

haruspex

The problem appears to be overspecified, i.e. the given data are contradictory. I tried taking 75 degrees as the angle to the vertical, but it made it worse.

4. Mar 21, 2013

haruspex

It says the block is pushed 3 m at a constant speed, not at a constant speed of 3 m/s.

5. Mar 21, 2013

Staff: Mentor

Looks like the information in the problem statement is inconsistent.

6. Mar 21, 2013

fredrogers3

So are you saying that the force of friction could have different values based on how one solves the equations?

7. Mar 21, 2013

rude man

Oh yeh. Thanks.

8. Mar 21, 2013

rude man

I agree. Either the box doesn't touch the ceiling or there is not enough force left to push the box. Plus there is the overspecification problem on top of that ... ay ay ay!

9. Mar 21, 2013

rude man

If the force was properly defined then the coeff. of friction would ensue. You can't define the "push" (horizontal) force and also specify the mass along with the friction coeff.

10. Mar 21, 2013

haruspex

I'm saying the information provided is internally inconsistent. The answer you get will depend on which piece of information you ignore. E.g. if you ignore the given coefficient of friction then you can calculate it from the other data (and it won't be 0.4). Or you could ignore the magnitude of the force, or the angle of the force, or the mass of the block...
Ah - there is just one possibility. Maybe this isn't in the usual 9.81m/s2 gravity!

11. Mar 21, 2013

Aigik

Do you by any chance have professor Brown at NIU as your instructor? Working on this problem set too, and I'm stumped by the last problem (the one where it asks for the smallest angle she can push before the block falls), and stumbled upon this. That's definitely troubling that the problem is written badly (what a surprise), and can be interpreted multiple ways. When I solved for the frictional force, I solved for the normal force first (using the y equation), then plugged that into the equation for friction, and ended up getting a frictional force of around 13N. Have you figured out the last problem yet?

12. Mar 21, 2013

fredrogers3

I am in Professor Brown's class! I got the same frictional force that you got, but it did seem strange that solving for the frictional force in the other equation gave a different answer. Either way, I think for the last part, you have to assume that the the normal force must always be greater than zero in order to solve for theta.

13. Mar 21, 2013

SammyS

Staff Emeritus
Is the angle of 75° with respect to the [STRIKE]vertical[/STRIKE] horizontal ?

If so, I don't see a discrepancy, except that the horizontal component of force, Fa, is greater than the force of friction. If those are the only forces involved, then the block would accelerate and not move at a constant speed. If there is some other force, then the problem is fine.

I meant to say horizontal rather than vertical . DUH !

Last edited: Mar 21, 2013
14. Mar 21, 2013

Aigik

The angle of 75 degrees is with respect to the horizontal.

15. Mar 21, 2013

haruspex

It seems fair to assume that all relevant forces are mentioned, and it does say constant speed. Therefore I still see it as inconsistent.

16. Mar 21, 2013

rude man

Yes, the problem as stated is impossible.

OK, so of the 85N, sin(75)*85 goes to push up. That's 82N of vertical up-push. Subtract the weight of the box, you get 82 - 49 = 33 lbs net up-push. So horizontal force needed to set the box in motion is 33*0.4 = 13N but the 85N's horizontal component is 85*cos(75) = 22N. So either the angle had to be greater than 75 deg or the coeff of friction has to be greater.

17. Mar 22, 2013

SammyS

Staff Emeritus
Well, I suppose that makes sense. If one considers that there is some unmentioned force present, it would need to be horizontal in order to solve the problem -- not really a reasonable assumption.

If the problem had not stated the the speed was constant, it would have been fine.

18. Mar 22, 2013

haruspex

OTOH, as I posted, there is one way to make it consistent - don't assume the gravitational field is as at earth's surface. Maybe it's on the moon.