Which forces am I missing in the FBDs?

  • #1

Homework Statement


Two blocks with pulleys attached lie on a frictionless surface and are connected by a massless rope strung over the pulleys as shown at right. A force of magnitude F is applied to the second block. Draw a FBD for a system consisting of Block 1 and its pulley, and also a system consisting of Block 2 and its pulley. (See the screenshot attached.)

Homework Equations


F=ma?
I don't think there are any specific equations needed...

The Attempt at a Solution


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Obviously, there are normal forces by the table on both systems as well as weight, and from each pulley we have two tension forces emanating. The system with block 2 also has an additional force of magnitude F towards the right. However, apparently there are still more forces I am missing, but I can't seem to find them...
 

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Answers and Replies

  • #2
Orodruin
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Unfortunately, your description of what you actually did is too vague for me to understand what you did and what you got. Please provide the equations you obtain from your FBDs with appropriate descriptions of what each term represents.
 
  • #3
haruspex
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Look at the ends of the rope.
 
  • #4
Chandra Prayaga
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What is your attempt at drawing the FBDs?
 
  • #5
@Chandra Prayaga — please see the screenshot attached.

@Orodruin

since neither system is accelerating vertically, we have
N_(table,1) - W_(Earth,1) = 0 &
N_(table,2) - W_(Earth,2) = 0
where 1 denotes the system that has box 1 (System 1) and 2 denotes the system that has box 2 (System 2).

Horizontally, for System 1, making the right positive, the two tension forces make possible the system's acceleration (from what I have currently):
2T_(rope,1) = m_(system 1)*a_1

For System 2, horizontally,
F - 2T_(rope,2) = m_(system 2)*a_2

@haruspex — I'm sorry, but it's still not obvious to me what I'm missing... Could you elaborate?
 

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  • #6
haruspex
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it's still not obvious to me what I'm missing... Could you elaborate?
What is the rope attached to at its ends?
 
  • #7
Orodruin
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Horizontally, for System 1, making the right positive, the two tension forces make possible the system's acceleration (from what I have currently):
2T_(rope,1) = m_(system 1)*a_1

For System 2, horizontally,
F - 2T_(rope,2) = m_(system 2)*a_2
As haruspex points out. You are missing some forces. A rope will also exert a force on anything it is attached to. You cannot just ignore those forces.
 
  • #8
@haruspex, @Orodruin

The rope is attached to box 1 on one end and the table on the other end. But I don't think the table is exerting any direct forces on System 1...
So would I get something like this (attached)?
 

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  • Screen Shot 2018-08-12 at 8.26.44 PM.png
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  • #9
haruspex
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@haruspex, @Orodruin

The rope is attached to box 1 on one end and the table on the other end. But I don't think the table is exerting any direct forces on System 1...
So would I get something like this (attached)?
That looks right.
 
  • #10
Alright, thank you for all the help, everyone!
 

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