Finding Force of friction without having coefficient or normal?

[DDRchick]

1. You slide a 326-N trunk up a 20.0° inclined plane with a constant velocity by exerting a force of 211 N parallel to the plane.

(a) What is the component of the trunk's weight parallel to the plane?

(b) What is the sum of your applied force, friction, and the parallel component of the trunk's weight?
2. Oh boy. Fp=sin(angle)-Fw
Fw=mg
Fn=cos(angle)x(Fw)
Ff=ux(Fn)
U=Ff/Fn

3. Alright, so I figured out part (a) by multiplying mass (326) and sin20 together, getting 111.5 N.
Now the second part is where I'm having trouble. I found Fn and Fp, as well as Fa.
Fn=cos20x(3194.8)
Fn=3002.1
Fp=sin20xFw(3194.8)
Fp=1092.7
Fa=211
u=Ff/3002.1

So how exactly do I find Ff and u?? :/

**(c) What is the size and direction of the friction force?
Size (magnitude)

Sorry to bother you guys again...I'm just at a loss of what to do x.x

Thanks in advance, and the u is the coefficient of friction.
:D

 

[heth]

1. You slide a 326-N trunk up a 20.0° inclined plane with a constant velocity by exerting a force of 211 N parallel to the plane.

"With a constant velocity" is question-code for something...

(Think Newton's Laws.)

 

[Rake-MC]

You need to look at the wording and units you're given in your question. When you're solving for part b), you have Fn = cos20*(3194.8). Where did you get 3194.8 from? It appears to me you multiplied 326 by gravity (9.81). However, you are told that the weight is 326 N . So it's already a unit of force (it has already taken gravity into consideration).

Therefore you only need to do cos(20)*(326) to find out the normal reaction force.

You then need to sum up all of the forces parallel to the plane, you are told the it is moving at a constant velocity. What does this tell you about the net force?

 

[DDRchick]

"With a constant velocity" is question-code for something...

(Think Newton's Laws.)

You need to look at the wording and units you're given in your question. When you're solving for part b), you have Fn = cos20*(3194.8). Where did you get 3194.8 from? It appears to me you multiplied 326 by gravity (9.81). However, you are told that the weight is 326 N . So it's already a unit of force (it has already taken gravity into consideration).

Therefore you only need to do cos(20)*(326) to find out the normal reaction force.

You then need to sum up all of the forces parallel to the plane, you are told the it is moving at a constant velocity. What does this tell you about the net force?

Oh okay so in since it's constant velocity...no acceleration?
And Fw is 326?
And therefore the net force is 211? I'm not sure if I'm right...

 

[heth]

Oh okay so in since it's constant velocity...no acceleration?

Yep.

And therefore the net force is 211? I'm not sure if I'm right...

If the net force was 211N, then the object would accelerate. But you already worked out that the object is moving at a constant velocity, so isn't accelerating...

 

[DDRchick]

oh. so it's 0?
Ohjeeze. So that makes everything 0...because it's at a constant velocity.

Ohmygoodness I didn't realize how simple that was lol. Thanks so much guys :D

 

[DDRchick]

(c) What is the size and direction of the friction force?
Size (magnitude)

Sorry to bother you guys again...:/ so I know that Ff=uxFn
Fn= cos20x(326)
Fn=316.3
U=?/306.3

Ugh help me :(

 

[heth]

Part b comes before part c for a reason. It's there to give you a clue about an easier way to do part c than what you're trying. Questions often work like this...

 

[DDRchick]

Well is Ff 0? If it is, it doesn't make any sense...:/

 

[heth]

Did you draw a sketch of the thing on the slope, and all the forces acting on it?

 

[DDRchick]

Yeah. I got the answer finally lol thanks so much for your help! :)