Kinetic Friction, Undefined Pushing Force, Find Normal Force.

1. Dec 30, 2011

Saraballs

1. The problem statement, all variables and given/known data
A crate is pushed at constant velocity across a rough, horizontal surface by a push force P. The crate has a weight of 73N. Coefficient of kinetic friction between the crate and the floor is 0.41. Vector force P points at an angle of 37 degrees below the horizontal. What is the magnitude of the normal force?

2. Relevant equations
fk = μkN
constant velocity indicates v0=vfinal and therefore a=0
no movement in the y direction, therefore ƩFy=0

These are random things that I have found online but I'm not sure if they're true:
Fcosθ=f (The horizontal component of the push force is equivalent to to friction)

3. The attempt at a solution
The problem is, I don't have the magnitude of the push force. I don't know where to go from here, but I have tried many things.

To find the kinetic friction:
fk = μkN
=(0.41)(73N)
=29.9N

Using Fcosθ=f, I assume that 29.9N is the horizontal component of the push force. I then find the vertical component of the push force:
Py=29.9N*tan37
Py=19.6N

I then add this to the weight to find the Normal Force:
73+19.6=92.6N

The Normal Force is not 92.6N. This problem is from an exam I failed last semester, and I have the correct answer now, but I can't get to it.
Thank you in advance, for anyone who can help me out.

Last edited: Dec 30, 2011
2. Dec 30, 2011

Redbelly98

Staff Emeritus
Welcome to Physics Forums.

You'll need to draw a force (free-body) diagram for the crate, and then look at x- and y-components of the net force.

3. Dec 30, 2011

Saraballs

Thank you.
I have drawn a free body diagram.
There is a Push force which is exerting some negative vertical force, some positive horizontal force (this is at a 37 degree angle).
There is a frictional force pointing in the negative x direction.
There is mg (73N as specified in the question). The Normal force is what is unknown. I can upload a picture of this free body diagram if this helps more.

4. Dec 30, 2011

Saraballs

Alright, after spending about 8 hours trying to figure this question out, googling, flipping thru my textbook, asking all my friends, I've finally got it (with some help from the online version of my textbook-- I had to piece together several different problems to get this).

Recognizing that there is a=0,

ƩF_x = Pcosθ-f_k = 0
Therefore, Pcosθ-(f_k)(μ_k)(N)=0

ƩF_y = Psinθ+N-mg = 0
Therefore, N=mg-Psinθ

** Recognize at this point that N (normal force) is not mg. So we are missing the magnitude of the force P, and N (which will be the final answer).
-- Start by finding P.
Substitute (mg-Psinθ) for (N) in the first formula [Pcosθ-(f_k)(μ_k)(N)=0].
-- From here you should be able to isolate and solve for P. This is the magnitude of the Push force.
-- Next, using the angle, you can extract the vertical component of the Push force (P).
-- You should be able to solve for the normal force from this.

// Side note, signing up and asking a question here was honestly my last resort. It was just ironic that I happened to give in 1 hour before I solved this problem. I will definitely be back for more help during the next semester of Physics, though. Just gotta get thru 5 more months of it.

5. Dec 30, 2011

Redbelly98

Staff Emeritus
Good job, but note that it should be -P sinθ for the y-forces, since the crate is being pushed with a downward angle. That will change things, but your method is sound.

Good job recognizing that constant velocity → a=0

In general, listing the forces as you have done, along with having drawn the figure for yourself, will usually be good enough.

Good luck with the rest of your course!