A man drags a 130 lb crate across a floor by pulling on a rope inclined at 20.0 degrees with the horizontal. If the coefficient of static friction is 0.550 , and the coefficient of kinetic friction is 0.350, find the acceleration (in feet/second^2) of the crate at the instant it breaks loose and starts sliding.

I converted lb to kg. Then I did m*g to get N. Then I did Fs=Ukfn and Fk=UkFn. Then a=fx-fk/m and got 6.43 by converting m/2^2 to ft/s^2.

but, that isn't the answer so i'm not sure what i'm doing wrong.

Question

A man drags a 130 lb crate across a floor by pulling on a rope inclined at 20.0 degrees with the horizontal. If the coefficient of static friction is 0.550 , and the coefficient of kinetic friction is 0.350, find the acceleration (in feet/second^2) of the crate at the instant it breaks loose and starts sliding.

I converted lb to kg. Then I did m*g to get N. Then I did Fs=Ukfn and Fk=UkFn. Then a=fx-fk/m and got 6.43 by converting m/2^2 to ft/s^2.

but, that isn't the answer so i'm not sure what i'm doing wrong.

shamim · Answer

Let the applied force is F

shamim · Answer

Horizontal component of F is
F*cos20
Right?

matt101 · Answer

130 lbs is about 60 kg, and multiplying by g (I'll round to 10 m/s^2 for easy numbers) gives you 600 N.

Notice the applied force is at an angle. For other questions we might need to worry about deconstructing the force into horizontal and vertical components, but here it actually doesn't matter. We're talking about friction, which by definition will be the horizontal force in this case. The force of kinetic friction will be less than that of static friction (you know that just by looking at the coefficients). The difference between the two represents the force (ma) that produces the acceleration once the box starts moving. So:

$$ma=f_{s}-f_{k}$$
This is the equation you have above. By plugging in the numbers, I get a = 2 m/s^2. Converting this back into ft/s^2, I got 6.56 ft/s...which is the same as what you got.

Looks like I'm getting the same answer as you for this. I'll think about it for a bit and see if I come up with anything...

matt101 · Answer

Oh got it!

matt101 · Answer

The normal force is NOT just mg. You need to account for the upward component of the applied force, which REDUCES mg by a value equal to Fsin20 (the crate is effectively "lighter" because part of its weight is being "cancelled" by the applied force). So instead of mg, the normal force is mg-Fsin20. So your updated equation is:

$$ma = f_{s}-f_{k}$$$$ma = \mu_{s}(mg-Fsin20)-\mu_{k}(mg-Fsin20)$$$$ma = (mg-Fsin20)(\mu_{s}-\mu_{k})$$
If you know the magnitude of the applied force, you can solve for a pretty easily. Do you have that number somewhere?

(Sorry, it's late and I can't think of a way to eliminate the F...)

anonymous · Answer

Thank you very much Matt101!!