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Can anybody check if I solved this correctly?(tension forces + pulley)

  • #1

Homework Statement


pulley_zps70f8d0bc.png

The pulley can rotate and has no friction.
Block [itex]m_2[/itex] is twice as big as [itex]m_1[/itex]
Find the tensions, [itex]T_1, T_2, T_3[/itex]

Homework Equations


Newton's second law


The Attempt at a Solution


Tension 1:
By creating a free body diagram only for the block, I got with Newton's second law:
[itex]T_1-mg=ma[/itex]
[itex]T_1=m(a+g)[/itex]

Tension 2:
[itex]T_2-2mg=ma[/itex]
[itex]T_2=m(a+2g)[/itex]

Tension 3:
[itex]T_3-T_1-T_2=ma[/itex]

Inserting above values for the other tensions:
[itex]T_3=ma+m(a+g)+m(a+2g)[/itex]
[itex]T_3=3m(a+g)[/itex]

Did I solve this correctly? What about the acceleration? Can I do anything about that?
 
Last edited:

Answers and Replies

  • #2
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  • #3
the pulley has no net force acting on it.
What exactly do you mean?
Are the 2 first tensions correct?
 
  • #4
So we have:

[itex]T_1-mg=ma[/itex]
[itex]T_1=m(a+g)[/itex]

[itex]2mg-T_2=ma[/itex]
[itex]T_2=2m(g-a)[/itex]

Since the tension is uniform on the string:
[itex]m(a+g)=2m(g-a)[/itex]
[itex]a=\frac{1}{3}g[/itex]

Substituting this into the first equation:
[itex]T_1-mg=\frac{1}{3}mg[/itex]
[itex]T=\frac{4}{3}mg[/itex]


The tension of the pulley since it has no net force acting:
[itex]T_3-T_1-T_2=0[/itex]

Since [itex]T_1=T_2=T[/itex]
[itex]T_3=2T[/itex]

Inserting the tensions:
[itex]T_3=2\frac{4}{3}mg[/itex]
 
Last edited:
  • #5
haruspex
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So we have:

[itex]T_1-mg=ma[/itex]
[itex]T_1=m(a+g)[/itex]

[itex]2mg-T_2=ma[/itex]
[itex]T_2=2m(g-a)[/itex]

Since the tension is uniform on the string:
[itex]m(a+g)=2m(g-a)[/itex]
[itex]a=\frac{1}{3}g[/itex]

Substituting this into the first equation:
[itex]T_1-mg=\frac{1}{3}mg[/itex]
[itex]T=\frac{4}{3}mg[/itex]


The tension of the pulley since it has no net force acting:
[itex]T_3-T_1-T_2=0[/itex]

Since [itex]T_1=T_2=T[/itex]
[itex]T_3=2T[/itex]

Inserting the tensions:
[itex]T_3=2\frac{4}{3}mg[/itex]
Assuming the pulley has no moment of inertia, that all looks correct. I wouldn't write the final answer as [itex]T_3=2\frac{4}{3}mg[/itex] though. It looks like you mean "two and four thirds".
 

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