Newton's Law and Friction (Ramp problem)

[Mad Season]

Homework Statement

A block is placed on a incline plane. A 21 N force is required to push the block up the incline with constant velocity. What is the weight of the block? Answer in units of N

Homework Equations

The force pushing the block=21 N with an angle of 13 degrees. The ramp has an angle of 21 degrees. Coefficient of friction=0.17. Ramp goes up from left to right.
fs = uN, F = ma, acceleration is 0.

Angle of force = phi, Angle of ramp = theta

The Attempt at a Solution

My attempt: (after drawing my free-body diagram)

F(y-component): N-mgcos(theta) = 0 --> N=mgcos(theta) --> Fsin(phi)-mgcos(theta) = 0
F(x-component): Fcos(phi)-mgsin(theta)-fs = 0 --> Fcos(phi)-mgsin(theta)-umgcos(theta) = 0

I multiplied the y-component with the negative u (or mu) to cancel out umgcos(theta)

Then: Fcos(phi)-uFsin(phi)-mgsin(theta) = 0, So I solved for mg and my answer was incorrect. What am I doing wrong? Any help appreciated, thanks.

 

[BvU]

What did you find when the solving was done?
I notice phi is different from the 21 degrees. Is phi wrt horizontal? Does its y component contribute to the friction force ?

 

[Mad Season]

Well the problem indicated that the force pushing the block up the ramp had an angle of 13 degrees. So denoting that phi, I used Fsin(phi) and Fcos(phi) respectively.

My answer in the end was about 54.85 N. Which was incorrect. No idea what I am doing wrong.

 

[BvU]

I take it the free body diagram decomposes mg along the slope and perpendicular to it.
You want to do the same with the 21 N force. What is the angle wrt the plane ?
The component perpendicular to the plane is to be added to mg $\cos\theta$ to determine the normal force.
That times $\mu$ is the friction force, along the slope, downwards. To be added to mg $\sin\theta$, also downwards. Call the sum F_down
No acceleration (i.e. constant speed) along the slope means the 21 N component along the slope is equal and opposite to this F_down

 

[HalfLostAndSearching]

I can confirm that your approach is correct. However, it seems like you made a small mistake in your calculations. When you multiplied the y-component with the coefficient of friction (u), you forgot to include it in the x-component as well. The correct equation should be: Fcos(phi)-uFsin(phi)-mgsin(theta)-umgcos(theta) = 0. This should give you the correct answer for the weight of the block. Keep up the good work!