Pulley with three masses what is T1?

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In summary: Nshould T2 be bigger than T1?In summary, the tension in one string is the same. Hence if you just have two masses on either side of the pulley the tension of the string will be the same on both sides. However, if you add a third mass to one of the sides, then the tension in the string between the two masses on that side will be different than the tension in the string over the pulley connecting the other two masses.
  • #1
gap0063
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Homework Statement


A pulley is massless and frictionless. 3 kg, 2 kg (on the left), and 6 kg (on the right) masses are suspended
(a)What is the tension T1 in the string between the two blocks on the left-hand side of the pulley?
(b) What is the magnitude of the acceleration of the lower left-hand block?

Homework Equations


My prof gave an example T=m1g=m1a
m2g-t+m2a

The Attempt at a Solution


Can I solve this like if it was two masses?

My prof gave an example T=m1g=m1a
m2g-t+m2a
then we solved for T in equation one and plugged it into equation two... in that problem the Tension was the same on both sides but in the picture on the problem they have labeled them T1, T2 and T3 so I'm not sure
 
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  • #2
The tension in one string is the same. Hence if you just have two masses on either side of the pulley the tension of the string will be the same on both sides.

But this time on one of the sides there are two masses. The rope between these two masses has a different tension than the rope over the pulley connecting the other two.

So if T1 is the tension in the rope between the two masses on the same side then:

T1-m1g=m1a

if T2 is the tension in the other rope then:

T2-m2g-T1 = ...

and:

m3g-T2= ...

I will leave you to figure the ... out ;)
 
  • #3
You can solve it using the same general technique, but you need to calculate the tension for each string separately. You should be able to make a Newton II (Fnet=mblockasystem) equation for each block. Then solve the equations for T1 (or whichever tensions you need) and asystem.
 
  • #4
Thaakisfox said:
The tension in one string is the same. Hence if you just have two masses on either side of the pulley the tension of the string will be the same on both sides.

But this time on one of the sides there are two masses. The rope between these two masses has a different tension than the rope over the pulley connecting the other two.

So if T1 is the tension in the rope between the two masses on the same side then:

T1-m1g=m1a

if T2 is the tension in the other rope then:

T2-m2g-T1 = ...

and:

m3g-T2= ...

I will leave you to figure the ... out ;)

a=(m3-m2)/(m2+m3)*g
So I got T2= m2(g+a) => plug in a... and I got T2=29.4N

Then I used T2-m2g=m2a to get a=T2-m2g-m2
and plugged this a into T1-m1g=m1a and got T1=44.1N which is wrong...

Where did I go wrong??
 
  • #5
you calculated the acceleration wrong.
Dont leave out the T1 from the second equation ;)

Your equations should be:

T1-m1g=m1aT2-m2g-T1 = m2am3g-T2= m3T3
 
  • #6
Thaakisfox said:
you calculated the acceleration wrong.
Dont leave out the T1 from the second equation ;)

Your equations should be:

T1-m1g=m1a


T2-m2g-T1 = m2a


m3g-T2= m3T3

I used the last equation to solve for T2(which isn't T2=T3) and got T2=8.4 N

Then using the first equation I solved a= T1-m1g/m1 and plugged it into the second equation and got T2-m2g-T1=m2(T1-m1g/m1)

and got T1=5.04 N which is wrong...
 
  • #7
Oh wait, that was a typo sorry, it should be:

m3g-T2= m3a

the previous equation didnt make any sense..;)
 
  • #8
Thaakisfox said:
Oh wait, that was a typo sorry, it should be:

m3g-T2= m3a

the previous equation didnt make any sense..;)

Should T2 be bigger than T1?

I got T2=32.0727 N and T1= 19.2436
 

1. What is a pulley with three masses?

A pulley with three masses refers to a system where three masses are connected by a rope or cable and supported by a pulley. The pulley acts as a simple machine to change the direction of the applied force.

2. How is the tension (T1) calculated in a pulley with three masses?

The tension (T1) in a pulley with three masses is calculated by balancing the forces acting on the system. Using Newton's Second Law, T1 can be determined by equating the forces on each mass and solving for T1.

3. What factors affect the value of T1 in a pulley with three masses?

The value of T1 in a pulley with three masses is affected by the weight of each mass, the angle at which the rope is pulled, and the friction in the pulley system. Any change in these factors can alter the overall tension in the system.

4. How does the arrangement of the masses affect T1 in a pulley with three masses?

The arrangement of the masses in a pulley with three masses can impact the value of T1. If the masses are arranged in a way that creates an imbalance of forces, T1 will be different compared to when the masses are evenly distributed.

5. Can the value of T1 in a pulley with three masses be greater than the weight of the heaviest mass?

Yes, it is possible for the value of T1 in a pulley with three masses to be greater than the weight of the heaviest mass. This can occur if the angle of the rope is such that it creates a mechanical advantage, allowing a smaller force to be applied to lift a heavier mass.

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