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I'm in an algebra based physics class so I apologize if the way I approach the problem is different from what you might be used to. A buddy of mine is calc based physics always gets confused about how I set the problems up. Also, this is my first post! Thanks in advance for the help.

You are trying to slide a refrigerator down a ramp 10 degrees above the horizontal. The mass of the refrigerator is 192 kg and you need to exert a force of 500 N to make it just begin to move.

A.) what is the coefficient of static friction between the floor at the refrigerator?

B.) You find that after the refrigerator begins to move that you need only apply 350 N to keep it moving at a constant speed. What is the coefficient of kinetic friction?

C.) What force would you have to apply so that the refrigerator accelerated down the ramp at 0.5 m/s^2?

2. Relevant equations

Coefficient of static friction (μs) = fs/n where fs is static friction force

Normal force = y component of weight force

force = m*a

*for objects at rest or with no acceleration the sum of all forces must equal zero!

3. The attempt at a solution

A.) I drew a free body diagram and designated my axes with the x-axis parallel with the surface of the ramp and the y-axis is in the direction of the normal force and y-component of the weight force.

weight force (w) = 192 * 9.8 = 1881.6

y-component (wy) = cos(10)*1881.6 (which is also equal to the normal force)

x-component (wx) = sin(10)*1881.6

Since we have to find friction and the friction is in the x-direction we have three things to account for: wx, pushing force (fp), and fs. With that in mind I said:

wx + fp +fs = 0

and solved for fs. So:

sin(10)*1881.6 + 500 + fs = 0

This gives a friction force of about 826.7 so μs = 826.7/[cos(10)*1881.6]= 0.45. The thing is that my friend got a different answer on the test and he got it right. What am I doing wrong?

PART B:

I got this part right by taking the x-component of the weight force as sin(10)*1881.6 + 350 and dividing that by the normal force:

[sin(10)*1881.6 + 350]/[cos(10)*1881.6] = μk = 0.37

PART C:

I got this wrong on the midterm but I think I have a new solution:

fp + wx - fk = 192 kg * 0.5 m/s^2

fp + [sin(10)*1881.6] - 667 = 96

fp = 436 (Approximately!)

How does this last part look?

*edited to take out repetitive cut and pastes

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# Homework Help: Coefficient of static friction and kinetic friction problem!

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