Finding tension to find work [SOLVED]

  • #1
JoeyBob
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Homework Statement:
See attached picture
Relevant Equations:
W = F * change in r
Confused on how to find tension in rope (this would be the force acting on mass 2 that is causing work). Fnet wouldnt be 0 since the speed isn't constant and no acceleration is given. I'm sure I'm suppose to use the masses in an equation somehow, but I'm not sure how.

It makes sense that the tension in the rope would be less than the force, but this doesn't help me find out how to actually calculate the tension.
 

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Answers and Replies

  • #2
Delta2
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Because the string doesn't stretch we can conclude that the accelerations (and the velocities and the displacements) of each mass are equal.

You should start by applying Newton's 2nd law in each mass. You ll get two equations with two unknowns: the common acceleration and the tension. You can solve the system of these two equations to calculate the two unknowns. Once you know the tension you can easily calculate the work done for the given displacement.
 
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  • #3
JoeyBob
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Because the string doesn't stretch we can conclude that the accelerations (and the velocities and the displacements) of each mass are equal.

You should start by applying Newton's 2nd law in each mass. You ll get two equations with two unknowns: the common acceleration and the tension. You can solve the system of these two equations to calculate the two unknowns. Once you know the tension you can easily calculate the work done for the given displacement.

Thanks, that got me to the right answer. One thing that doesn't make sense to me/confused me was that the question doesn't say there is a constant acceleration. In my head I imagined the masses moving from that force and then stopping at the end.

In hindsight the fact that it was a frictionless surface implies there was a constant acceleration I guess.
 
  • #4
Delta2
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The acceleration will be constant, IF all the forces at play are constant/time independent. We have no reason to assume that the Tension will be time varying, neither that the friction, if existed, would be time varying.
 
  • #5
Delta2
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Actually for this system if the external force of 48N was time varying (something like ##F_{ext}=48-3t## for example then the tension and the acceleration would be time varying. You could easily calculate both just as before by solving the system, except that now both acceleration and tension would be functions of time. So at the last step to calculate the work done you ll have to take the integral of tension $$W=\int T(t)dr=\int T(t) v(t) dt$$ where $$v(t)=\int a(t) dt$$ the velocity of the system (which also would be a function of time).
 
  • #6
Lnewqban
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... One thing that doesn't make sense to me/confused me was that the question doesn't say there is a constant acceleration. In my head I imagined the masses moving from that force and then stopping at the end.
The problem could have stated that the application of the force was constant, avoiding any assumption about a brief application of it.
For this case, you could imagine replacing that force with a pulley and a free-falling weight attached to a string.
For example (disregard numbers in diagram):

fig8.gif
 
  • #7
Delta2
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@Lnewqban i have my doubts if the two systems are equivalent. Maybe they are equivalent but with the external force being ##T_2## and not ##T_1## or the weight of the hanging mass.
 
  • #8
kuruman
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The problem could have stated that the application of the force was constant, avoiding any assumption about a brief application of it.
For this case, you could imagine replacing that force with a pulley and a free-falling weight attached to a string.
For example (disregard numbers in diagram):

View attachment 273477
One cannot disregard the numbers and call this diagram equivalent because the masses in the original problem are not equal. The ratio of the masses determines what fraction of the pulling force is the tension in the string between the masses.

A diagram, as shown on post #6, in which the mass on the left is ##m_2=3.4~\text{kg}##, the mass in the middle is ##m_1=3.8~\text{kg}## and the hanging mass is ##m_h=15.3~\text{kg}## would be exactly equivalent in the following sense: You can hide the pulley (assumed ideal) and hanging mass behind a screen and you will see exactly what you see in the original diagram, post #1, with a pulling force of 48 N.
 
  • #9
Lnewqban
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Yes, @Delta2 and @kuruman , both of your statements are absolutely correct.
The diagram is only one that I borrowed from a Google search.

Just wanted to show to the OP a general case in which a force was being continuously applied onto a system.
After reading his statement: "In my head I imagined the masses moving from that force and then stopping at the end", it seemed to me that he was confused about the duration of the action of the force in this problem.
 
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  • #10
JoeyBob
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Yes, @Delta2 and @kuruman , both of your statements are absolutely correct.
The diagram is only one that I borrowed from a Google search.

Just wanted to show to the OP a general case in which a force was being continuously applied onto a system.
After reading his statement: "In my head I imagined the masses moving from that force and then stopping at the end", it seemed to me that he was confused about the duration of the action of the force in this problem.

Yeah sometimes I fall into the trap of imagining physics scenarios as being real. Like frictionless surfaces aren't realistic and I sometimes forget to understand what a surface without friction implies.
 

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