How Do You Calculate Tension in Multiple Supporting Wires?

AI Thread Summary
To calculate the tension in three vertical wires supporting a uniform steel plate weighing 87.8 lb, the sum of the forces must equal the weight of the plate, leading to the equation A + B + C = 87.8 lb. Wires A and B are equidistant from the x-axis, indicating their tensions must be equal to prevent rotation. The solution involves applying the sum of moments about the x and y axes, as well as the sum of forces in the z direction. It's recommended to verify calculations by checking moments about different axes for accuracy. This approach ensures a comprehensive understanding of the tension distribution in the system.
PokeFan
Messages
2
Reaction score
0

Homework Statement



A uniform steel plate 18 in. square weighing 87.8 lb is suspended in the horizontal plane by the three vertical wires as shown. Calculate the tension in each wire ((a)TA , (b)TB, and (c)TC).

Homework Equations


\SigmaM=0


The Attempt at a Solution


both A and B are 9 in. from the x axis
C is on the x axis, so A and B have to be equal or it would rotate about x axis
A+B+C=87.8 or it would move in or out of the page
 

Attachments

  • 2tSgxQ.jpeg
    2tSgxQ.jpeg
    15.7 KB · Views: 1,454
Physics news on Phys.org
Good start. You have identified an axis about which moments must be equal. Actually ANY axis could be chosen, and the moments about that axis would have to balance. Why not choose an axis passing through two of the unknown tensions?
 
Thank you, I solved it using the Sum of the moments about the X and Y axis and the Sum of the forces about the Z.
 
Ah yes but did you check it by taking moments about some other axis? (always wise to do that)
 
Thread 'Have I solved this structural engineering equation correctly?'
Hi all, I have a structural engineering book from 1979. I am trying to follow it as best as I can. I have come to a formula that calculates the rotations in radians at the rigid joint that requires an iterative procedure. This equation comes in the form of: $$ x_i = \frac {Q_ih_i + Q_{i+1}h_{i+1}}{4K} + \frac {C}{K}x_{i-1} + \frac {C}{K}x_{i+1} $$ Where: ## Q ## is the horizontal storey shear ## h ## is the storey height ## K = (6G_i + C_i + C_{i+1}) ## ## G = \frac {I_g}{h} ## ## C...
Back
Top