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**1. Homework Statement**

A 30 kg neon sign is suspended by two cables, as shown. Three neighbourhood cats (5.0 kg each) find the sign a comfortable place. Calculate the tension in each cable when the cats are in the positions shown.

Ft1 = tension in left cable

Ft2 = tension in right cable

T=torque

**2. Homework Equations**

Xcm=(M1*x1+M2*x2+...)/Mtotal

ƩF(y) = 0

Ʃτ = 0

**3. The Attempt at a Solution**

Xcm = (5.0kg)(0.2m) + (30kg)(1m) + (5.0kg)(1.8m) + (5.0kg)(2.0m) / 45kg = 1.11 m from the left edge of the sign.

With left hand cable as reference point:

ΣT = Ft1*(0) + Ft2(1.6 m) - (45kg)(9.80 m/s^2)(1.11m-0.2m)=0

Ft2 = (45kg)(9.80)(0.91m) / 1.6 m = 250.82 N , 2.5 x10^2 N

∑Fy = Ft1 + Ft2 = mg

Ft1 = (45kg)(9.80m/s^2) - 250.82 N = 190.1 N 1.9 x10^2 N

I feel as though though I should somehow be taking into about the vertical distribution of the weight on the sign using Ycm but I'm not sure if I need to or how to go about doing that.

Thanks!