Friction Problem of a Crate Sliding Down a Ramp

  • #1
BBA Biochemistry

Homework Statement


A steel washer is suspended inside an empty shipping crate from a light string attached to the top of the crate. The crate slides down a long ramp that is inclined at an angle of 38 ∘ above the horizontal. The crate has mass 239 kg . You are sitting inside the crate (with a flashlight); your mass is 59 kg . As the crate is sliding down the ramp, you find the washer is at rest with respect to the crate when the string makes an angle of 71 ∘ with the top of the crate.

What is the coefficient of kinetic friction between the ramp and the crate?

Note: u=coefficient of friction

Homework Equations


Newton's Second Law: F=ma
Kinetic Friction: Fk=uN

The Attempt at a Solution



I have defined my xy-coordinate system with the x-axis along the ramp (negative going up the ramp) and the y-axis perpendicular to the ramp.

Forces on the crate: Fx= Mgsin38 - uMgcos38=ma. (Eq. 1)
There is no net force in the y-direction on the crate.

Gravity is acting in the positive x-direction (for how I defined the xy-coordinates), and friction acts opposite to the direction of motion, so it is negative. So Fx= (x-component of gravity)-(friction force). Friction force is equal to uN, and the Normal force is equal to the y-component of gravity (which is Mgcos38).

The washer is the trickier part to me. It seems that I need to use the washer to find the acceleration. Here are the equations I have for the forces on the washer.

Note: T=tension of rope on washer. Also, little m denotes mass of washer, whereas big M denotes mass of crate.

Forces on the washer: Fx= Tcos71+mgsin38. (Eq. 2)

Both gravity and the tension act on the string in the positive x-direction (I think).

Since m and T are not given, they somehow need to be eliminated, so I defined a second equation.

Forces in the y-direction on washer: Fy= Tsin71 - mgcos38=0. (Eq. 3)

Since the washer is not moving in the y-direction, the sum of the forces is 0.

I then solved Eq. 3 for T, getting:

T= [mgcos(38)] / [sin(71)]

I plugged this into Eq. 2, getting:

Fx= [mcos(38)cos(71)] / [sin(71)] - mgsin(38)=ma.

Since every term has m, they can all cancel out, leaving:

Fx= [cos(38)cos(71)] / [sin(71)] + gsin(38) = a = 6.3 m/s^2

Now that acceleration is known, there is only one unknown in Eq. 1, the friction coefficient.

Mgsin38-uMgcos38= Ma (This is just Eq. 1 restated)

Solving for u:

Since M is in every term, I can cancel out M: gsin38-ugcos38=a.

Rearranging the equation to isolate u: gsin38-a=ugcos38

Divide both sides by gcos38: tan38- [a] / [cos38] =u.

u= -0.35. I figure the negative is since it's just in the opposite direction.
 

Answers and Replies

  • #2
321
68
If you picture the situation, does the string hang in the downhill-half of the box, or the uphill-half of the box?
So should the x-component of tension be positive or negative (according to your coordinates)?

I am speaking in particular about this equation:
Forces on the washer: Fx= Tcos71+mgsin38. (Eq. 2)
 
  • #3
BBA Biochemistry
Turns out my significant digits weren't correct and that was the issue. The correct answer was 0.345. For anyone with a similar problem, I think my method was correct.
 
  • #4
321
68
But you said the x-acceleration is 6.3 m/s/s right? What if there is no friction, then the acceleration would be g*sin(38deg)≈6.0 m/s/s ....... So something is clearly not right if you calculated acceleration faster than the frictionless case!
 
  • #5
haruspex
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u= -0.35. I figure the negative is since it's just in the opposite direction
No, it's negative because you have the string in the upper half of the crate instead of in the lower half, as @Hiero pointed out.
 

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