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Drawing Free Body Diagrams for rope tension

  • Thread starter tascja
  • Start date
  • #1
87
0

Homework Statement


A man of mass 64.2 kg stands on a scaffold supported by a vertical rope at each end. The scaffold has a mass of 23.2 kg and is 3.2 m long. Suppose the man stands to the right from the middle of the scaffold that is a distance one sixth of the length of the scaffold. What is the tension in the right rope? What is the tension in the left rope?

The Attempt at a Solution


im always confused on how exactly to draw out the scenarios and start the question. For this particular problem i got this far:
1. straight bar for scaffold with an arrow pointing down, half way (the center of gravity)
2. a man closer to the right rope, with an arrow pointing down, (his weight F=mg)
3. the two ropes on either side are tension (arrows pointing up)

Now im not sure if there would be torque, because one rope has more force than the other, if so where would the pivot point be?? or if i should some how solve it be equating the tensions since the system is static??
Help would be greatly appreciated
 

Answers and Replies

  • #2
LowlyPion
Homework Helper
3,090
4
If it is in equilibrium (it is), then the ∑ F and the ∑ T are both 0.

So choose an end and balance out the torque.
 
  • #3
87
0
so the torque applied (right rope) = torque(scaffold) + torque(man)
Fr = Fr + Fr
Fr = (23.2)(9.81)(3.2/2) + (64.2)(9.81)(3.2*2/3)
F = 1707.7/3.2
F = 533.66 N
so the tension in the right rope is 534 N??
 
  • #4
LowlyPion
Homework Helper
3,090
4
I suppose you mean that

∑ F = 0 = Tl + TR - ∑ m*g

You should have the equation for the torque about end the man is farthest from.

And that should solve for the T of the rope he is closest to - the right Tension. And your answer looks OK.

Then you can figure the other tension directly from the first ∑ F.
 

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