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**fusilli_jerry89****Member**- Registered: 2006-06-23
- Posts: 86

Sorry no toher forum is wroking rightnow so i am forced to post here.

a 2.0 kg brick rides on a 20kg sled and is dragged upa22 degree slope by a rope witha tension of 1000 N. Theropeis parallel totheslope. The friction coefficient(both static and kinetic) between the brick and sled is 0.30 and between the sled and the ground is 0.15. Predict each mass's acceleration. I got 0.8 m/s/s for thebrick which i'm sure is right, and for the sled sled I got 40 m/s but the teacher says it is 45 m/s/s.

Teacher's method:

1000-20(9.8)sin22-0.15(22)(9.8)sin22-5.4=20a

a=45.5m/s/s

My method:

1000-81-30-5.4=22a

a=40m/s/s

*Last edited by fusilli_jerry89 (2006-10-18 12:47:25)*

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**fgarb****Member**- Registered: 2006-03-03
- Posts: 89

This is a pretty complicated problem, so don't feel at all bad for struggling with it

First point: whenever you write down a force equation, make sure you understand what it is the forces are being applied on. Be sure to draw a free body diagram for this object before you start and then transfer the force arrows into your mathematical formulas.

I'm not going to draw a picture, but I'll list the forces that I think are acting on the sled:

-An upwards normal force from the incline

-A downwards normal force from the brick

-The tension force from the rope

-Gravitational force

-Friction from the incline

-Friction from the brick (the sled is not moving at the same speed as the brick, so this is an issue)

Once you have the arrows drawn, you can write down the math expressions for each force - breaking them up by components. Again, I really think it's a bad idea to plug in numbers when you do this. There are *many* advantages to working with symbols all the way up to the very end, and only plug them in for your final answer. It'll save you a lot of headaches when you end up with a wrong answer and you want to work backwards to find your mistake.

As I was saying, when you write down your equations you need to keep in mind what the forces are acting on. In this case they're all acting on the sled - not the sled+brick together. So when you use F=ma, the mass is only the mass of the sled. It looks as though you also have another mistake here - I think the -30 in your expression is wrong, but since you didn't leave it in symbolic form I can't tell how you got here. I think you did everything else fine though.

Here's what I get for an answer:

[align=center]

Mu-sub-s is the coefficient of friction between the incline and the sled, and Mu-sub-b is between the sled and the brick here. Most of our expressions match up when you plug the numbers in, but I apparently disagree with the official answer in that I have cos instead of sin for the mu_s term instead of sin. So let me take this opportunity to show you how cool leaving things as symbols is: since I left this in symbolic notation I think I can easily prove I am right. Behold the power .

What if the incline weren't at 22 degrees but much steeper - almost 90 degrees instead? As you make the angle very close to 90 degrees, you should logically expect the normal forces (and as a result, the friction forces as well) to vanish - the objects no longer push against each other. Logically, all you should have left is gravity and the tension force, so there should be no mu's left.

Here's the test. Plug in theta = 90 degrees. My answer:

[align=center]

[/align]This makes sense. Gravity is pulling down and the rope is pulling up. But that's it. Teacher's answer (I think I subbed the symbols back in correctly):

[align=center]

[/align]And somehow friction from the incline is still resisting the motion. This does not make sense to me.

So I think this is a mistake in your teacher's answer, though it is possible I'm not thinking about this the right way (it's been a while since I did these kinds of assignments). But either way, I hope this can help your thinking about how to approach these problems, which is what is really more important.

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