Kinetic friction, tension, multiple masses

  • #1
Got a classic textbook physics problem here. Pretty sure I got it, but again, just wanting a "yay" or "nay" regarding if I'm doing it correctly.
And there's one little thing I'm not exactly sure of.

Homework Statement


http://img189.imageshack.us/img189/2345/picture3pft.png [Broken]
Two blocks, A and B, are placed as in Fig. 5 and con¬nected by ropes to block C. Blocks A and B weigh 20.0 N each, and the coefficient of kinetic friction between each block and the sur¬face is 0.40. Block C descends with constant velocity.
a) Draw two separate free-body diagrams showing the forces acting on A and on B.
b) Find the tension in the rope connecting blocks A and B.
c) What is the weight of block C?

Since the mass and radius of the pulleys are not given, we'll assume that they are ideal and have bearing on the equation.



Homework Equations


basic trig
F=ma


The Attempt at a Solution


For (a), all I had to do was doodle a bit, so let's not worry about that.

Since it's moving at a constant velocity, I know the net force has to be zero.
So, since mass A is 20 N, I just multiplied that by .4 for my force of friction.
Mu sub K A = 8 N.

For mass B, I did some fancy-pants vector smashing and wound up with a normal force of 15.9727N and a down-the-ramp force of 12.0363 N.
I multiply 15.9727 by .4 to get mu sub K B = 6.38908

So here's my first hiccup.
Friction is a two-way street (usually), so would the Fx of mass B actually be 12.0363-6.38908? (5.64722)
I'm not sure, but I don't think so.
Because ultimately mass C = 20+8+12.0363+6.38908 N, right?
Or would it be C = 20+8+5.64722+6.38908 N?

Such a small detail, but it's keeping me from finishing this assignment.

Either way, I think (but am not 100% certain) that the tension A-B will be equal to C-(12.0363+6.38908) or C-(5.64722+6.38908).

I'm right on top of the answer, but my brain has fused to the side of my skull, and I honestly can't figure out how to proceed.
 
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Answers and Replies

  • #2
PhanthomJay
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You're losing me there, friend. You said you drew Free Body Diagrams blocks A and B. Identify all the forces acting on each block, known and unknown. What are the equations associated with Newton's laws for each block?
 
  • #3
You're losing me there, friend. You said you drew Free Body Diagrams blocks A and B. Identify all the forces acting on each block, known and unknown. What are the equations associated with Newton's laws for each block?

I don't understand what you're trying to ask me.
 
  • #4
PhanthomJay
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I don't understand what you're trying to ask me.
Show us your chicken scratch doodle for the forces on A. You should then immediately be able to solve for T_AB. FBD's (Free Body Diagrams) that show all the forces acting on the object are essential; don't ignore them or brush over them quickly. Also, weight and mass are not the same.
 
  • #5
Tension isn't the part I'm having trouble with. Also, pretty well aware of the difference between weight and mass.
ninja edit: don't have a scanner. *shrug*
 
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  • #6
PhanthomJay
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Tension isn't the part I'm having trouble with.
ninja edit: don't have a scanner. *shrug*
a) Draw two separate free-body diagrams showing the forces acting on A and on B.
For (a), all I had to do was doodle a bit, so let's not worry about that.
I am very worried. If tension is not your problem, don't be afraid to tell us the value of T_AB (for starters) and we'll go from there.
 
  • #7
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Kinetic friction works in any direction, however, the vector always opposes the direction of motion (opposite to the velocity). This should resolve your question.
 
  • #8
PhanthomJay
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Also, pretty well aware of the difference between weight and mass.
since mass A is 20 N
Is it??
 
  • #9
I am very worried. If tension is not your problem, don't be afraid to tell us the value of T_AB (for starters) and we'll go from there.

Can't really know tension until I'm absolutely sure of Force C.

Here's a clip from the highlight reel.
Friction is a two-way street (usually), so would the Fx of mass B actually be 12.0363-6.38908? (5.64722)
I'm not sure, but I don't think so.
Because ultimately mass C = 20+8+12.0363+6.38908 N, right?
Or would it be C = 20+8+5.64722+6.38908 N?

This is the only thing keeping me from finishing the problem.
I'm not sure which calculation is the correct one.
 
  • #10
Is it??

Mass A is 20/9.81

2.03 kg.

Mass, lump, piece of matter, chunk, block.
I wasn't saying that the mass was 20N. I was identifying it as a mass.

Are you trying to help me or insult me?

SystemTheory said:
Kinetic friction works in any direction, however, the vector always opposes the direction of motion (opposite to the velocity). This should resolve your question.

Thanks. That's exactly what I was looking for.
 
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  • #11
PhanthomJay
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Mass A is 20/9.81

2.03 kg.

Mass, lump, piece of matter, chunk, block.
I wasn't saying that the mass was 20N. I was identifying it as a mass.

Are you trying to help me or insult me?
No insult intended, I just wanted to be sure you end up with the correct answer for part C. Now you are starting the problem backwards, so let me help you by first looking at Block A, which you are refusing to do. When you look at block A, identify the forces in the x direction, and use Newton 1. You have the friction force correct, 8 N. So, T_AB - 8 =0, per Newton 1, ot T_AB = 8 N . Now continue, looking at Block B forces, and ultimately, block C forces. Use FBD's.
 
  • #12
SystemTheory pointed me in the right direction.
Thanks for trying to help, Jay, but I was talking about tomatoes and you were talking about potatoes. No worries. No harm, no foul.

Let's just sweep this thread under the rug and pretend like it never happened.
 
  • #13
PhanthomJay
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SystemTheory pointed me in the right direction.
Thanks for trying to help, Jay, but I was talking about tomatoes and you were talking about potatoes. No worries. No harm, no foul.

Let's just sweep this thread under the rug and pretend like it never happened.
But your answer for the weight of C is wrong, no matter which one you choose! You are not drawing free body diagrams correctly.
 
  • #14
But your answer for the weight of C is wrong, no matter which one you choose! You are not drawing free body diagrams correctly.

Relax, man. It's just for extra credit. I'll get points for it whether I'm right or not just for the attempt and showing my work.
Besides, you can't possibly know if I'm drawing my free body diagrams correctly unless you've seen them.
Or you're psychic.






Edit: My answer ended up being 26.4 N for C. Tension I wound up with was 8 N between A and B.
 
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  • #15
PhanthomJay
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Relax, man. It's just for extra credit. I'll get points for it whether I'm right or not just for the attempt and showing my work.
Besides, you can't possibly know if I'm drawing my free body diagrams correctly unless you've seen them.
Or you're psychic.
Oh, sorry, I thought you were looking for help. Although your answer for the weight of C is incorrect, the answer doesn't mean that much, it's the method that is important. That's why I don't need to see your FBD's , because i know they are wrong, since your method is incorrect. Peace out!
:wink:
 
  • #16
PhanthomJay
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Torquescrew;2475:yuck:164 said:
Edit: My answer ended up being 26.4 N for C. Tension I wound up with was 8 N between A and B.
Looks Good!:smile:
 

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