- #1
itachipower
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wrong section
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What Causes Incorrect Section Placement in Static Equilibrium and Torque?
- Thread starter itachipower
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SUMMARY
The discussion focuses on solving a static equilibrium problem involving a horizontal scaffold of length 2.00 m and mass 50.0 kg, suspended by two cables with a total paint can mass of 75.5 kg. The tension in the right cable is incorrectly stated as 780 N; the correct torque calculation should be 780 N multiplied by 2 m, resulting in 1560 Nm. The center of mass of the paint cans is determined to be 0.554 m from the right cable, requiring adjustments in the torque calculations to account for equilibrium conditions.
PREREQUISITES- Understanding of static equilibrium principles
- Knowledge of torque calculations and their application
- Familiarity with center of mass concepts
- Ability to apply Newton's laws in rotational dynamics
- Study the principles of static equilibrium in physics
- Learn about torque and its calculations in various scenarios
- Explore center of mass calculations for composite systems
- Investigate common mistakes in equilibrium problems and how to avoid them
Students studying physics, particularly those focusing on mechanics and static equilibrium, as well as educators seeking to clarify concepts related to torque and center of mass in practical applications.
- #2
itachipower
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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
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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
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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
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- 0
Thank you!
- #6
adjacent
Gold Member
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You are welcome
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