Finding the Center of Mass of a Suspended Scaffold and Paint Cans System

AI Thread Summary
To find the center of mass of a suspended scaffold and paint cans system, the total mass of the scaffold (50 kg) and paint cans (80 kg) must be considered along with the tension in the right cable (830 N). The discussion highlights the importance of establishing a balance of torque around a chosen rotational axis, with the left side being suggested. Participants emphasize the need to determine the tension in the left cable to accurately locate the center of mass. The equation provided by the user can be simplified to find the horizontal distance from the right cable. The solution requires careful attention to the distances involved from both sides of the scaffold.
bravoman
Messages
9
Reaction score
0

Homework Statement


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 80.0 kg. The tension in the cable at the right is 830 N. How far horizontally from that cable is the center of mass of the system of paint cans?

12-32.gif


I can't seem to get the right answer, I believe I may be missing a force. I set the rotational axis to the left side

Homework Equations


Sigma F = 0

The Attempt at a Solution


I have tried to create a balance of torque equation:

(L/2)* MassScaffold * g + x*MassPaint*g - 830L = 0
 
Physics news on Phys.org
Hello Bravo, welcome to PF :)

Your relevant equation doesn't show up in your attempt at solution. What's the tension in the cable at the left ? When you know that, it's a piece of cake to establish where mg should be located (i.e. where the c.o.m. is) ! Then you still have to go to the c.o.m. of the paint cans only .
 
BvU said:
Your relevant equation doesn't show up in your attempt at solution.
I don't think you need it with the approach bravoman used. The attempt gives a linear equation which can be solved for x as everything else is known.
 
mfb is right and Bravo should continue on the path he (she?) started on. o:)
 
This should work. You didn't forget that x is the the distance from the left side, while the distance from the right side is needed for the answer?
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. 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. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top