- #1
itachipower
- 10
- 0
wrong section
Last edited:
What Causes Incorrect Section Placement in Static Equilibrium and Torque?
- Thread starter itachipower
- Start date
AI Thread Summary
The discussion centers on a physics problem involving a scaffold in static equilibrium with paint cans stacked on it. The user is attempting to calculate the center of mass of the paint cans using torque equations but is struggling with the correct application of forces. A key point raised is that the user incorrectly identified the tension in the cable as a force rather than a torque, leading to confusion in their calculations. The correct approach involves recognizing that the clockwise and anticlockwise moments must balance, and the torque due to the tension should be calculated based on the full length of the scaffold. Ultimately, the user is encouraged to adjust their calculations to find the correct distance for the center of mass.
- #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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Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point?
Lets call the point which connects the string and rod as P.
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Let's declare that for the cylinder,
mass = M = 10 kg
Radius = R = 4 m
For the wall and the floor,
Friction coeff = ##\mu## = 0.5
For the hanging mass,
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First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on.
Force on the hanging mass
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There's also...
This problem is two parts. The first is to determine what effects are being provided by each of the elements - 1) the window panes; 2) the asphalt surface. My answer to that is
The second part of the problem is exactly why you get this affect.
And one more spoiler:
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