Kinetic Friction, Undefined Pushing Force, Find Normal Force.

[Saraballs]

Homework Statement

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?

Homework Equations

f_k = μ_kN
constant velocity indicates v₀=v_final and therefore a=0
no movement in the y direction, therefore ƩF_y=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)

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:
f_k = μ_kN
=(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:
P_y=29.9N*tan37
P_y=19.6N

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

 

[Redbelly98]

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.

 

[Saraballs]

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.

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.

 

[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 got to get thru 5 more months of it.

 

[Redbelly98]

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

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.

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!