Pulley with three masses what is T1?

[gap0063]

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

 

[Thaakisfox]

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 ;)

 

[CharliH]

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 (F_net=m_blocka_system) equation for each block. Then solve the equations for T₁ (or whichever tensions you need) and a_system.

 

[gap0063]

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??

 

[Thaakisfox]

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

 

[gap0063]

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...

 

[Thaakisfox]

Oh wait, that was a typo sorry, it should be:

m3g-T2= m3a

the previous equation didnt make any sense..;)

 

[gap0063]

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