Overcoming friction problems

1. Nov 8, 2007

crcowboyfan

1. The problem statement, all variables and given/known data

1. A 75kg box slides down a 25* ramp with an acceleration of 3.6 m/s^2
a. Find the coefficient of kinetic friction between the box and the ramp
b. What acceleration would a 175 kg box have on this ramp?

2. Relevant equations

Fg = (weight)(gravity)
??

3. The attempt at a solution

I started to try to find the force of gravity and the x and y components, but got confused on where to go from there

2. Nov 8, 2007

StringTH

Best way i found to start any questions with forces is a FBD/sketch, then bulid it up from what info your given.

2.
F=m*acceleration
Force cause by friction > Fs=(Coef Kf)*(force from ramp, as the block has mg on ramp, ramp pushes back ( paired forces).)
components of mg will help as well

From a FBD/sketch of the block we find N (force on block from ramp) to be mg*cos(25).
The Fs will act against the acceleration, ie jump out a plane the air resistance (Fs) wil push you up, opposite from gravity. So the Fs will act up the slope. Acting down the slope we have F=ma, the block moving at 3.6ms^-2, and we know the mass.

Forces(net): So going down the slope we have component of the weight mgsin25 and up (Coef Kf)*N. The block is moving down the slope so. ma=mgsin25-(Coef Kf)*N

b)
the force down the slope is ma=mgsin25-(Coed Kf)*N (as the friction is pulling up the slope) from components earlier.
now the acceleration down the slope, (cancelling m over eq)
a=gsin25- (Coef Kf)*N

Note: As it is presumably the same block with more mass, same surface, Coeffecient is the contact between the two surfaces, surfaces didn't change so remains the same. I always found myself if the mass changed wanting to change the coefficient.

Last edited: Nov 9, 2007
3. Nov 8, 2007

crcowboyfan

Ok so the force of gravity = (75 kg)(9.81m/s)= 735N

The Normal force = (735)(Cos25)= 666N

The force down the ramp= (75kg)(9.81)(Sin25)= 311N

Fnet= ?

How do I find the force of friction so that I can figure out the net force?

Last edited: Nov 8, 2007
4. Nov 8, 2007

crcowboyfan

Think I got it.

Fx= max
=(75)(3.6)=270

311-270=41

41/666= .061 = Coeff. friction

That agrees with the answer in the book

5. Nov 12, 2008

Fins_Wish

thank you for helping me out.
I posted a thread with this exact problem five minutes ago, seems like we were both having the same issues on the same problem in the same book.
weird, huh?
thank you so much for helping me.
i think i got it now. i was confused about the sine and cosine stuff. thanks for the helps!

-Fin

Last edited: Nov 12, 2008