Friction problem -- 2 blocks and a rope on an inclined plane

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Homework Statement


Capture.PNG

So, these 2 blocks are connected by a rope. Aluminuim block is 2kg and copper is 6kg. They are kept in motionless state via some kind of mechanism on steel surface. θ is 30 degrees. I was given dynamic and static friction coefficient (Aluminium μs=0,61 i μd=0,47, Copper μs=0,53 i μd=0,36.
Question is will these blocks move after mechanism that keeps them motionless turns off and I have to calculate force of tension in the rope. Friction of that little pulley is 0.

Solution for rope tension is 10N.

Homework Equations


Ff=N*μ
G=m*g

The Attempt at a Solution



I found G1 and G2:
G1=19.62N
G2=58.86N

I also found G2x and G2y:
G2x=sinθ*G2=29.43N
G2y=cosθ*G2=50.97N

After that I calculated force of static and dynamic friction:
Fsf1=G1s=11.97N
Fdf1=G1d=9.22N

Fsf2=G2ys=27N
Fdf2=G2yd=18.35N

Unfortunately, this is as far as I got because I don't understand and don't know what to do next since I never had any problems with static friction.
I hope someone can help me out on this one :)
 
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Answers and Replies

  • #2
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I think I may figured it out but i'm not 100% sure. So, if T=G2x-Fdf2 then T=11.08N. But it's not 10N like it says it should be. Also, should I use static friction or dynamic friction for calculating tension ?
 
  • #3
haruspex
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Ff=N*μ
Static and kinetic friction are different. The static coefficient only sets an upper bound on the frictional force:
|Fd|=μdN
|Fs|≤μsN
This distinction is important.
should I use static friction or dynamic friction for calculating tension ?
You should use static if the surfaces do not slide over each other and dynamic if they do.
So the first step is to determine whether the system can be static. Suppose it is static and work out the forces on that basis.
 
  • #4
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Static and kinetic friction are different. The static coefficient only sets an upper bound on the frictional force:
|Fd|=μdN
|Fs|≤μsN
This distinction is important.

You should use static if the surfaces do not slide over each other and dynamic if they do.
So the first step is to determine whether the system can be static. Suppose it is static and work out the forces on that basis.
Sorry, can you give me more info :). I have no idea were to start really.
 
  • #5
haruspex
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Sorry, can you give me more info :). I have no idea were to start really.
What is the max frictional force on the block on the ramp?
If it does not slip, what is the minimum tension?
Can the frictional force on the other block supply that tension?
 
  • #6
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What is the max frictional force on the block on the ramp?
If it does not slip, what is the minimum tension?
Can the frictional force on the other block supply that tension?
Well max friction should be Fsf2=G2ys=27N right ?
 
  • #7
haruspex
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Well max friction should be Fsf2=G2ys=27N right ?
Yes, so what is the minimum tension to prevent that block slipping?

By the way, it is rarely a good idea to plug in numbers so soon. Far better to keep everything symbolic as long as possible. There are many advantages, not least for others trying to follow your working.
In the present problem, for example, you might find that you did not keep enough significant digits to be able to resolve whether it is static, and you'll have to go all through it again.
 
  • #8
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Yes, so what is the minimum tension to prevent that block slipping?
T=G2x-Fsf2=29.43-27=2.43N
 
  • #9
haruspex
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T=G2x-Fsf2=29.43-27=2.43N
Ok. If that is the tension, will the left hand block slip?
 
  • #10
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Ok. If that is the tension, will the left hand block slip?
I dont think so...
Fsf1=G1s=11.97N
Friction is higher than tension force pulling left block so it shouldn't slip right ?
 
  • #11
haruspex
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I dont think so...
Fsf1=G1s=11.97N
Friction is higher than tension force pulling left block so it shouldn't slip right ?
Not exactly. As I pointed out, the static friction equation only gives you an upper bound for the frictional force. But you are right there is 11.97N available. So that's the first part answered.
The actual tension is problematic. We have only found the minimum tension. At maximum tension, either the left hand block is about to slip to the right or the right hand block is about to slip up the slope! The maximum tension for equilibrium is the lower of the two.
In practice, the actual tension depends on how the set-up was done. E.g. to get max tension we could start with a warm environment and cool it down. As the wire tries to shrink the tension increases until the left block slips right or the right block slips left. Or we could get this higher tension just by holding the wire a bit stretched while we place the blocks on the slope.
In short, there is no way to answer the question of tension as posted. The best you can do is to calculate a range.
 
  • #12
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Not exactly. As I pointed out, the static friction equation only gives you an upper bound for the frictional force. But you are right there is 11.97N available. So that's the first part answered.
The actual tension is problematic. We have only found the minimum tension. At maximum tension, either the left hand block is about to slip to the right or the right hand block is about to slip up the slope! The maximum tension for equilibrium is the lower of the two.
In practice, the actual tension depends on how the set-up was done. E.g. to get max tension we could start with a warm environment and cool it down. As the wire tries to shrink the tension increases until the left block slips right or the right block slips left. Or we could get this higher tension just by holding the wire a bit stretched while we place the blocks on the slope.
In short, there is no way to answer the question of tension as posted. The best you can do is to calculate a range.
Why does it say 10N for answer then?
 
  • #13
haruspex
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Why does it say 10N for answer then?
I have no idea.
 

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