Compute the distance x and tension of each cable

  • Thread starter Thread starter MrMechanic
  • Start date Start date
  • Tags Tags
    Cable Tension
AI Thread Summary
The discussion revolves around calculating the distance x and the tension in cables AC and BC supporting a thin steel plate weighing 82 lb/ft². Participants are attempting to determine the weight of the plate and its center of gravity, with one user calculating the total weight as 8856 lb. Key equations for solving the problem include the summation of forces in the x and y directions being zero, as well as the moments about a point. There is some confusion regarding the angles and the horizontal positioning of the plate, with suggestions to write equations based on the unknown angles and the center of mass. The problem requires clarification on the geometry of the setup to proceed effectively.
MrMechanic
Messages
46
Reaction score
0

Homework Statement


The thin steel plate, weighing 82lb/ft^2, is being lifted slowly by the cables AC and BC. Compute the distance x and find the corresponding tension in each of the cables.
[See attachments for Figure]

Homework Equations


Summation of X = 0
Summation of Y = 0
Moment=Rd

The Attempt at a Solution


So i tried to get the weight of the steel plate. here's the solution..
2(21) + 4(12) + 1/2(9)(4) = 108ft^2
now i multiply the given 82lb/ft^2 to cancel out the ft^2
i get 8856lb.. After that i don't know where to start next. Can you guys help me out?
I really think the problem is easy but i got lost on where to start...
 

Attachments

  • Untitled.png
    Untitled.png
    15.1 KB · Views: 688
Physics news on Phys.org
I've not done many of these but I imagine it will swing so that the centre of gravity is below the hook. So where is the centre of gravity?
 
MrMechanic said:

Homework Statement


The thin steel plate, weighing 82lb/ft^2, is being lifted slowly by the cables AC and BC. Compute the distance x and find the corresponding tension in each of the cables.
[See attachments for Figure]

Homework Equations


Summation of X = 0
Summation of Y = 0
Moment=Rd

What do you mean by
"Summation of X = 0
Summation of Y = 0
Moment=Rd"

The plate is moving with a small constant speed upward. What is the acceleration then? It is kept horizontal, it does not turn. What do these mean on the forces and torques (moments) acting on it?


MrMechanic said:

The Attempt at a Solution


So i tried to get the weight of the steel plate. here's the solution..
2(21) + 4(12) + 1/2(9)(4) = 108ft^2
now i multiply the given 82lb/ft^2 to cancel out the ft^2
i get 8856lb.. After that i don't know where to start next. Can you guys help me out?
I really think the problem is easy but i got lost on where to start...


ehild
 
so you mean forces along x-axis is = to 0 and also the torque.
And how would i find the two angles there.
Should I Sum all the torque along A?
 
The sum of the x components of the forces is 0, and also the sum of the y components of the forces.

Also the sum of all torques about any point is zero. You can choose A.

You need to write up three equations in terms of x and the the unknown angles.

AS CWatters wrote, x is related to the position of the CM.

ehild
 
The information doesn't seem to be complete. It looks like the top of the plate is horizontal, and it looks like AC and BC have the same length. But these cannot both be true, and neither is clearly given as true. At a guess, AB is horizontal. At the least, that's the easier version of the problem.
 
I've looked at it again and I think haruspex has a point. I think they are asking you to work out what value of x is needed for the top edge of the plate to stay horizontal (and not as I suggested above that it will rotate). This implies AC and BC are different lengths but the lengths aren't specified so that's ok.
 
Back
Top