Finding the coefficient of friction on a ramp

AI Thread Summary
To find the coefficient of friction on a ramp, the problem involves calculating forces acting on a trunk pushed up a 22-degree incline with 2.2 kJ of work over 3.1 m. The force exerted on the trunk is calculated as 710 N, equating the forces acting down the ramp (gravity and friction) to those pushing it up. The constant speed indicates that the net force is zero, leading to the equation 710 = mg(sin(22) + μcos(22)). However, the solution cannot be completed without knowing the mass of the trunk, which is necessary to isolate the friction coefficient. The expected answer is 0.6, but the lack of mass information leaves the problem unsolvable as presented.
Mastevvs
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Homework Statement



You do 2.2 kJ of work pushing a trunk at a constant speed 3.1 m along a ramp inclined upward at 22 degrees. What is the frictional coefficient between the trunk and the ramp?

Homework Equations



W = F*d
Ff = μmg * cos(θ)
Fg = mg * sin(θ)

The Attempt at a Solution



To find the force exerted on the trunk up the ramp, I divide 2200 J by 3.1 m to get 710 N.

Since the trunk is moving at a constant speed, the forces pulling the trunk down the ramp must equal the forces pushing it up the ramp.

So,

710 = Fg + Ff

710 = mg*sin(22) + μmg*cos(22)

710 = mg( sin(22) + μcos(22) )

...Aaaaand this is where I get stuck. I feel like there is some obvious way to either solve for or get rid of the mass that I'm just not seeing. My book gives 0.6 as the answer.

Any help would be greatly appreciated.
 
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Hey you told it moves up with constant speed 3.1m/s
then why are you dividing work with speed to get the force
 
Viru.universe said:
Hey you told it moves up with constant speed 3.1m/s
then why are you dividing work with speed to get the force
possibly:
1. that would not give force - work/speed has dimensions of force per unit time.
2. the time the work was done in was not given
3. the "3.1" figure is "3.1m" - not "3.1m/s". Spot the difference.

Mastevvs said:
I feel like there is some obvious way to either solve for or get rid of the mass that I'm just not seeing.
Welcome to PF;
Indeed: there is information provided in the problem that has not been used in the attempt at a solution. eg. it says that the movement was at constant speed - what does this tell you about the forces?

The way to handle this sort of problem is, usually, though conservation of energy.
Start by describing the energy changes that happen and recall how this is related to the work.
But in this case you have done enough groundwork already to get away without it.
 
Last edited:
Mastevvs said:
710 = mg( sin(22) + μcos(22) )

...Aaaaand this is where I get stuck. I feel like there is some obvious way to either solve for or get rid of the mass that I'm just not seeing. My book gives 0.6 as the answer.

Any help would be greatly appreciated.

Your work is correct and the problem can not be solved without the mass given.

ehild
 
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