# Another friction and inclined plane question.

## Homework Statement

you push a 325 newton trunk up a 20 degree inclined plane at a constant velocity by exerting a 211 newton force parallel to the planes surface.

a. what is the component of the trunks weight?
b. what is the sum of all the forces parallel to the planes surface?
c. what are the magnitude and direction of the friction force?
d. what is the coefficient of friction?

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## The Attempt at a Solution

a. sin 20 = Fx/325 --> Fx = 111.2 N?
b. not quite sure because i dont know if the Fx should be positive or negative.
c. 211 N for magnitude? same for Ff? due to equilibrium?
d. Ff= 111.2, Fn= 305.5, coefficient of friction= ?
--> equation --> Ff= coeff*Fn
--> 111.2= coeff*305.5
--> coeff= 0.36? i think. im not exactly sure what the 211 force should be labeled as (the stuff in red). i thought i was supposed to be the applied force, but i dont think thats right.

key:
Fn= normal force
Ff= frictional force
Fg= force due to gravity
Fx= x component
Fy= y component

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cristo
Staff Emeritus

## Homework Statement

you push a 325 newton trunk up a 20 degree inclined plane at a constant velocity by exerting a 211 newton force parallel to the planes surface.

a. what is the component of the trunks weight?
b. what is the sum of all the forces parallel to the planes surface?
c. what are the magnitude and direction of the friction force?
d. what is the coefficient of friction?

----

## The Attempt at a Solution

a. sin 20 = Fx/325 --> Fx = 111.2 N?
If the questions says "what is the component of the trunk's weight acting along the plane" then yes, this is correct. As it stands, the question is ambiguous.
b. not quite sure because i dont know if the Fx should be positive or negative.
The trunk is moving at constant speed.. what does that tell you about the net force acting on the trunk? FYI, Fx should have the opposite sign to the applied force, and the same sign as the frictional force, whatever you choose the sign conventions to be.
c. 211 N for magnitude? same for Ff? due to equilibrium?
No. The applied force is balanced by the friction and the component of the weight acting parallel to the plane.

BTW, thanks for setting out your problem clearly! The trunk is moving at constant speed.. what does that tell you about the net force acting on the trunk? FYI, Fx should have the opposite sign to the applied force, and the same sign as the frictional force, whatever you choose the sign conventions to be.

No. The applied force is balanced by the friction and the component of the weight acting parallel to the plane.

BTW, thanks for setting out your problem clearly! it tells you that the normal force and the Fy are equal to 325 N?

then, the applied force is negative and Ff and Fx are negative? so then, Fx and the applied are equal? or Fx and Ff are equal? or Fx and Ff are equal to the applied? lol.

..no problem. i figured it was a much better set up than before. thats only cuz the other one was my first post or whatever and i learned from that one. =)

oh yeah. and is d. right? i found the coefficient the same way i found the cofficient in the other problem about the skiing. except this time, they actually gave me real numbers i could work with.

cristo
Staff Emeritus
it tells you that the normal force and the Fy are equal to 325 N?
That's correct, but the fact that the trunk is moving at a constant velocity tells us that there is no net force acting on the trunk.
then, the applied force is negative and Ff and Fx are negative? so then, Fx and the applied are equal? or Fx and Ff are equal? or Fx and Ff are equal to the applied? lol.
Well, you've given every possible answer there, so one must be correct! Look at your diagram. I'll take positive as going up the ramp, so then the applied force is positive, and Fx and Ff are negative. So, Fa-Fx-Ff=0 => Fa=Fx+Ff.

..no problem. i figured it was a much better set up than before. thats only cuz the other one was my first post or whatever and i learned from that one. =)
[/quote]
Exactly! I don't think anyone minds a sloppy first post, but I think I speak for all homework helpers here; we appreciate when someone learns and posts a clear second question!

oh yeah. and is d. right? i found the coefficient the same way i found the cofficient in the other problem about the skiing. except this time, they actually gave me real numbers i could work with.

Well, you've given every possible answer there, so one must be correct! Look at your diagram. I'll take positive as going up the ramp, so then the applied force is positive, and Fx and Ff are negative. So, Fa-Fx-Ff=0 => Fa=Fx+Ff.

there was no point in b. that was the dumbest question possible. im so slow. its in equilibrium, so therefore, the sum of all the forces is 0, duh. wow.

lol. yet. but i will be soon =)

i dont quite understand what im doing for question c. 211 N?

Last edited:
cristo
Staff Emeritus
c says "what is the magnitude and direction of the frictional force?" You have this: Fa=Fx+Ff => Ff=Fa-Fx, where Fa is the applied force. You know Fa (211) and Fx (111.2), so plug into this equation. This gives the magnitude of the frictional force. Then its direction is, clearly, down the slope.

After you've done this, you can use Ff=(coefficient of friction)Fn to find the coefficient of friction.

oh okay. so i pretty much already found c from question b. alright. thank you.

cristo
Staff Emeritus