- #1
itachipower
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wrong section
Last edited:
What Causes Incorrect Section Placement in Static Equilibrium and Torque?
- Thread starter itachipower
- Start date
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Homework Help Overview
The discussion revolves around a static equilibrium problem involving a horizontal scaffold supported by two cables, with additional weight from paint cans. Participants are exploring the calculation of the center of mass of the system and the associated torques.
Discussion Character
- Exploratory, Assumption checking, Mathematical reasoning
Approaches and Questions Raised
- The original poster attempts to calculate the torque using the left cable as the rotational axis, questioning the forces involved in their calculations. Some participants suggest reconsidering the torque value associated with the tension in the cable.
Discussion Status
Participants are actively engaging with the problem, with some offering insights about the equilibrium condition and the relationship between clockwise and anticlockwise moments. There is a mix of interpretations regarding the forces and torques involved.
Contextual Notes
There is mention of a specific expected answer, which may influence the discussion. The original poster expresses uncertainty about missing forces in their calculations.
- #2
itachipower
- 10
- 0
Homework Statement
I have been trying to solve this problem for a while now and I can't figure out what I am doing wrong...
Here is the problem:
In the figure below, a horizontal scaffold, of length 2.00 m and uniform mass 50.0 kg, is suspended from a building by two cables. The scaffold has dozens of paint cans stacked on it at various points. The total mass of the paint cans is 75.5 kg. The tension in the cable at the right is 780 N. How far horizontally from that cable is the center of mass of the system of paint cans?
Homework Equations
T = r x F
Sigma T = 0
The Attempt at a Solution
What I am doing is using the left cable as the rotational axis for Torque. So I get Torque due to the beam + torque due to the CoM of the paint cans = 780N
Therefore, (50)(9.8)(1m) + (75.5)(9.8)(x) = 780N
The answer is supposed to be .554 m. So I think I'm missing a force but I don't know what it is. Thanks for your help :)
- #3
adjacent
Gold Member
- 1,552
- 62
itachipower said:
Welcome to PF,Itachi (don't use sharingan on me) lol
The scaffold is in equillibrium.So,clockwise moment is equal to anticlockwise moment.You're 780N is wrong,it has to be the torque there.
780 x 2=1560Nm.
You are using the other cable as a reference point.So,you will get the distance from.Therefore,you have to subtract it from 2m.
- #4
adjacent
Gold Member
- 1,552
- 62
Welcome to PF,Itachi (don't use sharingan on me) lol
The scaffold is in equillibrium.So,clockwise moment is equal to anticlockwise moment.Your 780N is wrong,it has to be the torque there.
780 x 2=1560Nm.
You are using the other cable as a reference point.So,you will get the distance from.Therefore,you have to subtract it from 2m.
The scaffold is in equillibrium.So,clockwise moment is equal to anticlockwise moment.Your 780N is wrong,it has to be the torque there.
780 x 2=1560Nm.
You are using the other cable as a reference point.So,you will get the distance from.Therefore,you have to subtract it from 2m.
- #5
itachipower
- 10
- 0
Thank you!
- #6
adjacent
Gold Member
- 1,552
- 62
You are welcome
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