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Finding acceleration of block

  • Thread starter Saitama
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  • #1
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Homework Statement


Three blocks A,B and C, whose masses are 9 kg, 9 Kg and 18 kg respectively as shown in the figure, are released from rest. The co-efficient of friction between the block A and the horizontal surface is 0.5 and The co-efficient of friction between the block B and the horizontal surface is 0.5. Find the acceleration of the block C just after the release. [Take g=10 m/s^2]. All strings and pulleys are ideal.


Homework Equations





The Attempt at a Solution


I can't figure out how the motion of block B will affect the motion of the other two blocks. I am unable to form any equation here.

Any help is appreciated. Thanks!
 

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Answers and Replies

  • #2
tms
644
17
Start by drawing a diagram showing all the forces on all the blocks. Use symbols, not numbers, in order to see what is going on (plugging in numbers early obscures the physics).

Block C is the key: without it, nothing will happen. When C is added, it will (maybe, depending on its weight and the other forces) pull B to the right.
 
  • #3
ehild
Homework Helper
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Homework Statement


Three blocks A,B and C, whose masses are 9 kg, 9 Kg and 18 kg respectively as shown in the figure, are released from rest. The co-efficient of friction between the block A and the horizontal surface is 0.5 and The co-efficient of friction between the block B and the horizontal surface is 0.5. Find the acceleration of the block C just after the release. [Take g=10 m/s^2]. All strings and pulleys are ideal.


Homework Equations





The Attempt at a Solution


I can't figure out how the motion of block B will affect the motion of the other two blocks. I am unable to form any equation here.

Any help is appreciated. Thanks!

Just take the accelerations of the blocks as aA, aB, and aC, and figure out how they are related. The length of the rope is constant.

ehild
 
  • #4
3,812
92
Start by drawing a diagram showing all the forces on all the blocks. Use symbols, not numbers, in order to see what is going on (plugging in numbers early obscures the physics).

Block C is the key: without it, nothing will happen. When C is added, it will (maybe, depending on its weight and the other forces) pull B to the right.
I have drawn the free body diagrams (see attachment) But still, I am not sure what to do next?
 

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  • #5
3,812
92
Just take the accelerations of the blocks as aA, aB, and aC, and figure out how they are related. The length of the rope is constant.

ehild
I have seen some examples in the past where this way of solving problems is used but I have never been able to use this way anywhere.
 
  • #6
tms
644
17
So far, so good. Now you see if you can figure out what some of the forces are in terms of the others, and you use F = ma.
 
  • #7
3,812
92
So far, so good. Now you see if you can figure out what some of the forces are in terms of the others, and you use F = ma.
2T-μmg=maB
T-μmg=maA
Mg-T=MaC

I have four variables and three equations. I am stuck here. :confused:
 
  • #8
ehild
Homework Helper
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How are the accelerations related?
Think: if A moves to the right by ΔxA and B moves to the right by ΔxB how much the horizontal part of the rope becomes shorter? And the vertical path longer?


ehild
 
  • #9
3,812
92
How are the accelerations related?
Think: if A moves to the right by ΔxA and B moves to the right by ΔxB how much the horizontal part of the rope becomes shorter? And the vertical path longer?


ehild
Does it become shorter by ΔxA+2ΔxB?
 
  • #10
ehild
Homework Helper
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Does it become shorter by ΔxA+2ΔxB?
Yes, for the horizontal part. And what is the downward displacement of C then?

ehild
 
  • #11
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Yes, for the horizontal part. And what is the downward displacement of C then?

ehild
Wouldn't that be again equal to ΔxA+2ΔxB? :confused:
 
  • #12
ehild
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Yes, it is. And what about accelerations? They are the second time derivatives of displacements.

ehild
 
  • #13
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Yes, it is. And what about accelerations? They are the second time derivatives of displacements.

ehild
aC=aA+2aB

Do I need to solve these four equations?
 
  • #14
ehild
Homework Helper
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Yes. Express all accelerations from the other three and substitute into aC=aA+2aB. You can use the numerical data it is easier.

ehild
 
  • #15
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Yes. Express all accelerations from the other three and substitute into aC=aA+2aB. You can use the numerical data it is easier.

ehild
Equation (i): T-45=9aA
Equation (ii): 2T-45=9aB
Equation (iii): 180-T=18aC

Multiplying (ii) with 2 and adding it with (i):
5T-135=9aC ...(iv)

Multiplying (iii) with 5 and adding it with (iv)
5(180-27)=99aC
Solving this equation doesn't give me the right answer. :(
 
  • #16
ehild
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Find all a-s and T. Show the results.

ehild
 
  • #17
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Find all a-s and T. Show the results.

ehild
I only need aC and my last equation doesn't give the right answer.
 
  • #18
ehild
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Find aA. If it is negative, the initial assumption that it moves was
not valid.

ehild
 
  • #19
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Find aA. If it is negative, the initial assumption that it moves was
not valid.

ehild
Yep, aA comes out to be negative. :)

What next? Assume that A moves left?
 
  • #20
ehild
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Such problems with ropes and friction, are tricky. You need to take into account that a rope can not push, and the friction can prevent some blocks from motion. You need to determine all accelerations and tensions to be sure that neither of them is negative.

ehild
 
  • #21
ehild
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Yep, aA comes out to be negative. :)

What next? Assume that A moves left?
No, A can not accelerate to the left, as the rope can not push. But it can be motionless. Assume that aA=0, and solve the equations.

ehild
 
  • #22
3,812
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You need to take into account that a rope can not push, and the friction can prevent some blocks from motion. You need to determine all accelerations and tensions to be sure that neither of them is negative.
Here are the values I got:
aA=(-5/11) m/s^2

aB=45/11 m/s^2

aC=85/11 m/s^2

T=450/11 m/s^2
 
  • #23
3,812
92
No, A can not accelerate to the left, as the rope can not push. But it can be motionless. Assume that aA=0, and solve the equations.

ehild
Assuming aA=0, T=45.
Substituting T in equation (iii),
180-45=18aC
This time too, I don't get the right answer. :(
 
  • #24
3,812
92
What is the "right answer"?
ehild
##\frac{70}{9} m/s^2##
And what I get is ##\frac{15}{2} m/s^2##.
 
  • #25
ehild
Homework Helper
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The problem is that you can not make the force of friction equal to mu mg, if A does not move. It is static friction then, what you do not know. Ignore the equation for aA. T is not equal to 45.

ehild
 

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