How Do You Calculate Tension in Multiple Supporting Wires?

[PokeFan]

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

$$\Sigma$$M=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

 

[pongo38]

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?

 

[PokeFan]

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.

 

[pongo38]

Ah yes but did you check it by taking moments about some other axis? (always wise to do that)

 

[DeeNos]

Based on the given information, the first step in solving for the tension in each wire would be to determine the center of mass of the steel plate. This can be done by dividing the weight of the plate (87.8 lb) by the total area (18 in. x 18 in. = 324 in^2) to get a weight per unit area of 0.271 lb/in^2. The center of mass would then be located at (9 in., 9 in.) from the origin.

Next, we can use the principle of moments (\SigmaM=0) to solve for the tension in each wire. The moment of a force is equal to the force multiplied by the perpendicular distance from the axis of rotation. In this case, the axis of rotation is the center of mass of the plate.

For wire A, the moment of tension (TA) would be TA x 9 in. (perpendicular distance from the center of mass) and it would be clockwise. For wire B, the moment of tension (TB) would be TB x 9 in. (perpendicular distance from the center of mass) and it would also be clockwise. For wire C, the moment of tension (TC) would be TC x 18 in. (perpendicular distance from the center of mass) and it would be counterclockwise.

Setting up the equation \SigmaM=0, we can solve for the tension in each wire:

TA x 9 in. + TB x 9 in. - TC x 18 in. = 0

Since we know that A and B are equal, we can substitute TB with TA in the equation:

TA x 9 in. + TA x 9 in. - TC x 18 in. = 0

Simplifying, we get:

2TA x 9 in. - TC x 18 in. = 0

Solving for TA, we get:

TA = TC/2

Now, using the equation A + B + C = 87.8 lb, we can substitute TA and TB with TC/2 in the equation:

TC/2 + TC/2 + TC = 87.8 lb

Solving for TC, we get:

TC = 29.27 lb

Therefore, the tension in each wire is:

(a) TA = TC/2 = 14.635 lb
(b) TB = TC/2 = 14