Need help with friction question

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Homework Help Overview

The discussion revolves around a physics problem involving friction, inclined planes, and forces acting on a crate and a suspended washer. The crate, which has a specified mass, slides down a ramp at a given angle, while the washer remains at rest relative to the crate. Participants are tasked with determining the kinetic coefficient of friction between the crate and the ramp.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the inclusion of the observer's mass in the total mass of the crate, questioning whether the 172 kg includes the 67 kg of the person inside. There are discussions about the forces acting on the crate and the washer, including the effects of friction and the angles involved.

Discussion Status

Several participants have made attempts to analyze the forces at play, with some expressing uncertainty about how to incorporate friction into their calculations. There is an ongoing exploration of the relationship between the angles, forces, and the coefficient of friction, with no clear consensus yet reached.

Contextual Notes

Participants note that the problem does not clearly specify the direction in which the washer is hanging, which may affect the interpretation of the crate's acceleration. Additionally, the discussion includes references to the need for attempts to be made before receiving further guidance, in line with forum rules.

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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 34degrees above the horizontal. The crate has mass 172kg. You are sitting inside the crate (with a flashlight); your mass is 67 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 59degrees with the top of the crate.

What is the kinetic coefficient of the friction between the crate and the ramp?
 
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Is the mass of the observing person already included in the listed mass of the crate?
 
planck42 said:
Is the mass of the observing person already included in the listed mass of the crate?

nimboman said:
You are sitting inside the crate (with a flashlight); your mass is 67 kg. .

Does that answer your question about the observing person being listed in the mass of the crate?
 
alexratman said:
Does that answer your question about the observing person being listed in the mass of the crate?

Actually, it's still ambiguous to me. Does it mean that the crate is 172 kg with the person in there, or that the crate itself weighs 172 kg and then the person's 67 kg needs to be added in there for a grand total of 239 kg?
 
planck42 said:
Actually, it's still ambiguous to me. Does it mean that the crate is 172 kg with the person in there, or that the crate itself weighs 172 kg and then the person's 67 kg needs to be added in there for a grand total of 239 kg?
I believe the 172kg does not include the mass of the person.
 
But in any case, what's the question?
 
Sorry forgot to post the question. The question is what is the kinetic coefficient between the crate and the ramp
 
If the crate was moving at a constant velocity then the washer would hang straight down relative to the world outside the crate, or at 34 degrees to the roof of the crate. It's not, however, so the crate is accelerating in some way.

Note that the question doesn't make it clear in which direction the washer is hanging, but that doesn't matter...if it's one way then the crate is accelerating, if it's the other then it's decelerating, either way the coefficient of friction remains the same.

What have you got so far?
 
i don't really know how to start. need some enlightment!
 
  • #10
The forum rules say you have to make an attempt before clues can be posted...got any ideas at all? Doesn't matter if they're wrong, the whole point of science is that being wrong is just fine as long as you can work out why you're wrong.
 
  • #11
ok i suppose i can find the force acting parallel to the ramp when the crate is not moving.
which is (mass of crate + mass of me)*sin34*g. I'm not sure how to calculate the force when frictional force is taken into account. since friction is equivalent to normal force*friction coefficient, in this case, friction force=mgsin34*friction coefficient.
 
  • #12
oh wait, frictonal force will actually affect the normal force. so i can't use 34 as the angle
 
  • #13
Nope, right first time. The frictional force is there as well as the force parallel to the ramp. One counteracts the other. It's the difference between the two forces you need to look at. You've got a mass for the crate, you've got an acceleration for the crate (or, at least, you've got a pendulum...)...
 
  • #14
anyway, when the crate is moving with frictional force taken into account. the force parallel to the ramp is it mgcos59? so the frictional force is just mgcos34 - mgcos59?
but now i equate this frictional force to mgsin(theta)*friction coefficient. in this case i 34 for theta or 59?
 
  • #15
The Geoff said:
Nope, right first time. The frictional force is there as well as the force parallel to the ramp. One counteracts the other. It's the difference between the two forces you need to look at. You've got a mass for the crate, you've got an acceleration for the crate (or, at least, you've got a pendulum...)...

sorry if i phrased the question wrongly. the question is asking for the coefficient of the kinetic friction. so in this case we only take into account friction when the crate is moving
 
  • #16
My point was that with kinetic friction, there's a certain, special coefficient that will balance the acceleration due to gravity and leave the crate with a constant velocity down the slope.

But the washer is hanging at an angle, so there's an acceleration. We're not at the special, balanced coefficient.

How big the acceleration is (in terms of g), and therefore the coefficient of friction, is related to the angle the washer hangs at.
 
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  • #17
Soln:

Considering the Box:
Let M be combined mass, u be Fk coeff.
//Slope: Fx = Mgsin34 - uMgcos34 = Ma

Considering the Steel Washer:
Let m be mass of Steel Washer, T be tension in the string
Perpendicular to slope: Fy = Tsin59 - mgcos34 = 0
//Slope: Fx = Tcos59 + mgsin34 = ma
So combine the above 2 eqns: you get value of a.

Substitute a into first eqn and you'll get u. Hope this helps.
 
  • #18
hey thanks for ur help. i got it. ur from ntu aero?
 

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