# Ridid body equilibrium, tension in hanging wires

1. Mar 26, 2008

### clope023

[SOLVED] ridid body equilibrium, tension in hanging wires

1. The problem statement, all variables and given/known data

A 60.0cm, uniform, 49.0N shelf is supported horizontally by two vertical wires attached to the sloping ceiling. A very small 25.0N tool is placed on the shelf midway between the points where the wires are attached to it.

a) find the tension in the left wire
b) find the tension in the right wire

2. Relevant equations

$$\Sigma$$Fx = 0 (condition for equilibrium)

$$\Sigma$$Fy = 0 (condition for equilibirum)

$$\Sigma$$$$\tau$$z = 0 (condition for equilibirum)

3. The attempt at a solution

$$\Sigma$$Fy = 0 = TL(.25m) + TR(.75m) - (25+49) = 0

$$\Sigma$$Fx = 0

tension in the left wire (torque measured from right wire)

$$\Sigma$$$$\tau$$z = 0 =TR(0) + TL(.6) + 25(.2) + 49(.3)

TL = 33N

tension in the right wire (torque measured from left wire)

$$\Sigma$$$$\tau$$z = 0 =TL(0) + TR(.2) + 25(.4) + 49(.3)

TR = 124N

checking with the FY equation I don't think I got the correct tensions, any help is greatly appreciated.

2. Mar 26, 2008

### Mapes

Check your units on the $\Sigma F_Y=0$ equation; you are mixing N-m and N. Why?

3. Mar 26, 2008

### Mapes

EDIT: Also, some of your distances look off. The moment around a point is the force multiplied by the lever arm (the perpendicular distance to the line of force).

4. Mar 26, 2008

### clope023

I see that now, I wasn't exactly sure what the vertical distances for the wire pertained to (or if they pertained to anything at all).

is it that the entire distance of the rod doesn't come to play, and it's only a distance of .4m (between the 2 wires) that's being examined?

5. Mar 26, 2008

### Mapes

It should help to draw a free-body diagram of the bar to make it clear what forces are applied.

6. Mar 26, 2008

### clope023

I'm aware of what forces are applied, the left tension, the right right tension, the weight of the tool and the rod; I'm pretty sure the two wires have normal forces with the ceiling (but wasn't sure if they apply here), but really it doesn't seem that the extra .20m to the right of the .75m long wire is relevant, to the problem.

7. Mar 26, 2008

### clope023

I got it.

I used TR(.4) + 49(.3) + 25(.2) = 0

TR = 49N

used TL(.4) + 49(.3) - 25(.2) = 0

TL = 25N

TR+TL = 74

74 = 74
.

8. Mar 26, 2008

### Mapes

Looks good. I assume you meant TL(.4) - 49(.1) - 25(.2) = 0 on the second equation.