What Is the Acceleration of Block B in a Two-Mass Pulley System?

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SUMMARY

The acceleration of block B in a two-mass pulley system is definitively calculated to be g/7, while block A has an acceleration of g/7 as well. The relationship between the accelerations of the two blocks is established, with block B accelerating at twice the rate of block A. The tension in the strings is also analyzed, revealing that the tension in block B is half that of block A. This conclusion is reached through a series of equations derived from Newton's second law, specifically ma = F[net].

PREREQUISITES
  • Understanding of Newton's second law (ma = F[net])
  • Familiarity with pulley systems and tension in strings
  • Basic algebra for solving equations
  • Knowledge of acceleration relationships in mechanical systems
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  • Study the derivation of acceleration in pulley systems using Newton's laws
  • Learn about tension distribution in multi-mass pulley systems
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Students studying physics, particularly those focusing on mechanics, as well as educators and anyone interested in understanding the dynamics of pulley systems and acceleration relationships.

Neolight

Homework Statement


In the figure shown if the system is released from rest, acceleration of block B will be
(a). g/2
(b). g/3
(c). 2g/3
(d). g/7

Homework Equations


ma= F[net]
taking the direction of motion as positive
IMG_20170715_165758.jpg

The Attempt at a Solution



So my attempt was to first find tension in the string connected to Mass 3M from
3Ma= 3Mg-T
T= 3M(g-a)

now the tension in the second string connected to the M say T1 is half of T ( am i correct?)
so
T1= T/2= 3Mg/2 - 3Ma/2

now equation of motion for the mass M

Ma= T1-Mg
Ma= M(3g/2 -3a/2 -g)

a = 3g/2 - g - 3a/2
a + 3a/2 = 3g/2 -g
5a/2= g/2
∴ a= g/5

where did i go wrong , because i looked up the answer and it is g/7
please help...
[/B]
 
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Neolight said:

Homework Statement


In the figure shown if the system is released from rest, acceleration of block B will be
(a). g/2
(b). g/3
(c). 2g/3
(d). g/7

Homework Equations


ma= F[net]
taking the direction of motion as positive
View attachment 207210

The Attempt at a Solution



So my attempt was to first find tension in the string connected to Mass 3M from
3Ma= 3Mg-T
T= 3M(g-a)

now the tension in the second string connected to the M say T1 is half of T ( am i correct?)
so
T1= T/2= 3Mg/2 - 3Ma/2

now equation of motion for the mass M

Ma= T1-Mg
Ma= M(3g/2 -3a/2 -g)

a = 3g/2 - g - 3a/2
a + 3a/2 = 3g/2 -g
5a/2= g/2
∴ a= g/5

where did i go wrong , because i looked up the answer and it is g/7
please help... [/B]

Acceleration of block B is not 'a' i.e it is not equal to that of block A .

But they have a simple relationship .
 
Vibhor said:
Acceleration of block B is not 'a' i.e it is not equal to that of block A .

But they have a simple relationship .
can you please tell me the relationship between the two accelerations?
 
Neolight said:
can you please tell me the relationship between the two accelerations?

If the lower pulley goes down by a distance 'x' how much does block B go down ?
 
Vibhor said:
If the lower pulley goes down by a distance 'x' how much does block B go down ?
since the string isn't directly connected to the pulley so they won't travel the same distance, the string goes along the circumference of the pulley, this is as far as i can think sorry I'm really bad at this type of thinking
 
Neolight said:
since the string isn't directly connected to the pulley so they won't travel the same distance, the string goes along the circumference of the pulley, this is as far as i can think sorry I'm really bad at this type of thinking

No problem .

Suppose you hold block B i.e you do not allow it to move .Now let lower pulley go down by distance 'x' , how much string length of string whose one end is fixed at ground and other end connects block B , goes slack/loose ?

Hint: Carefully look at both the left and right parts of string going over lower pulley .
 
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Vibhor said:
No problem .

Suppose you hold block B i.e you do not allow it to move .Now let lower pulley go down by distance 'x' , how much string length of string whose one end is fixed at ground and other end connects block B , goes slack/loose ?

Hint: Carefully look at both the left and right parts of string going over lower pulley .
its half the circumference (πr) so if the pulley moves up say x distance the mass B will move πr +x ?
 
Neolight said:
its half the circumference (πr) so if the pulley moves up say x distance the mass B will move πr +x ?

No.

You do not need to worry about the part that lies on the pulley circumference . Why ?
Because before pulley moves down , πr length is over the pulley and after pulley moves down , same length πr is on the pulley circumference , so net change in the length of string as far as part that lies on the string is zero .

So forget about the part of string that lies on pulley circumference .

Just look at left and right parts of string .Pick a pen and paper and make a rough sketch . Without moving block B , move lower pulley down by a distance 'x' .

How much string length of left part gets loosened ?

How much string length of right part gets loosened ?

What is the total length of string that gets slack/loosened ?
 
Vibhor said:
No.

You do not need to worry about the part that lies on the pulley circumference . Why ?
Because before pulley moves down , πr length is over the pulley and after pulley moves down , same length πr is on the pulley circumference , so net change in the length of string as far as part that lies on the string is zero .

So forget about the part of string that lies on pulley circumference .

Just look at left and right parts of string .Pick a pen and paper and make a rough sketch . Without moving block B , move lower pulley down by a distance 'x' .

How much string length of left part gets loosened ?

How much string length of right part gets loosened ?

What is the total length of string that gets slack/loosened ?
hahah i get it now 2x

so acceleration of block B will be two times the accelaration of the pulley

i have done the calculation using this result and now the answer is g/7 ... thanks for your help , i really appreciate it
 
  • #10
Well done !
 
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  • #11
Neolight said:
i have done the calculation using this result and now the answer is g/7
Is this what you get for the acceleration of block B, or is it the acceleration of block A?
 
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  • #12
TSny said:
Is this what you get for the acceleration of block B, or is it the acceleration of block A?
It's for block B
 
  • #13
Neolight said:
It's for block B
Maybe I'm making a mistake, but I get that block A has the acceleration of g/7.
Do you agree with the following equations?

3Mg - TA = 3MaA

TB - Mg = MaB

TA = 2TB

aA = aB / 2
 
  • #14
TSny said:
Maybe I'm making a mistake, but I get that block A has the acceleration of g/7.
Do you agree with the following equations?

3Mg - TA = 3MaA

TB - Mg = MaB

TA = 2TB

aA = aB/2
Are you taking the acceleration of the block A and B as the same?
 
  • #15
Neolight said:
Are you taking the acceleration of the block A and B as the same?
No. See the last equation I wrote.
 
  • #16
TSny said:
No. See the last equation I wrote.
The acceleration of block B will be two times that of block A while tension of Block B will be half that of block A
 
  • #17
Neolight said:
The acceleration of block B will be two times that of block A while tension of Block B will be half that of block A
Yes, that agrees with the 3rd and 4th equations that I wrote.
 
  • #18
TSny said:
Yes, that agrees with the 3rd and 4th equations that I wrote.
I have done the math again ... And the result is

Acceleration for block A is g/7 and for B is 2g/7

Correct?

I made a mistake in a my previous math
 
  • #19
Neolight said:
Acceleration for block A is g/7 and for B is 2g/7
OK, good. That's what I get, too.
 
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