Force Components, Friction, etc. Check my work?

[Summer95]

Could someone credible please check my work on this one?

Homework Statement

This problem is "pulling a sled up a slope"
All relevant data is in the picture. I wanted to find F_P (The force one has to pull to start the sled moving up with slope).

mass = 26.0kg
μ_s = 0.096
slope of hill = 12 degrees
F_P = 23 degrees (from horizontal)

I did all my work in the rotated reference frame where the x direction is in line with the slope of the hill.

Homework Equations

I fully believe I got the right answer which is why I want this to be double and triple checked to make sure I really do understand everything here. I have my midterm on Friday so now is the time to learn!

The Attempt at a Solution

The solution I got was F_P = 76.9N

Thank you so much in advance! <3‿<3

 

[rude man]

I got a slightly higher number. To wit, 78.4N.

I notice a term μ_sF_psin(11) in your work which I did not have.

 

[Summer95]

I got a slightly higher number. To wit, 78.4N.

I notice a term μ_sF_psin(11) in your work which I did not have.

Thanks for replying :) I see exactly what you did differently so I'll explain and can you tell me where/if I wrong in the following?

To find the F_friction it is dependent on F_N which ordinarily would be equal to the y component of F_g. However, there is also a y component to F_P (an applied force in the y direction) which reduces F_N by F_PY. In other words because you are pulling up on the object the normal force is reduced. If this were not the case there would be a net force in the y direction and it would accelerate off the slope. That is where μ_sF_psin(11) in my work comes from. It is part of 0.096[mgcos(12)-F_Psin(11)] where F_Psin(11) is the y component of the applied force F_P.

 

[rude man]

Well, guess what, I agree! Except one thing bothers me: the force F_psin(11) is not applied at the c.g. of the sled, but at one edge. So the tendency of that vertical component of F_p is to tilt the sled upwards rather than lift it evenly off the ramp. So the part of the surface contact the sled makes with the ramp that is closest to the rope will get "lighter" but the part on the low end will hardly be affected. I guess you can say it averages out. Certainly I was remiss in neglecting that vertical component. So congrats! I'd say the apprentice beat the master except I'm no master!

 

[Summer95]

Thanks! That is an interesting point. Would it balance out perfectly or would that factor into a more accurate answer?

 

[rude man]

I think it would balance out and your analysis was spot-on.