- #1
Persimmon
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Homework Statement
A 30 kg neon sign is suspended by two cables, as shown. Three
neighborhood 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.
[URL=http://s1152.photobucket.com/user/rusalka4/media/cats_zpse2fe500d.png.html][PLAIN]http://i1152.photobucket.com/albums/p498/rusalka4/cats_zpse2fe500d.png[/URL][/PLAIN]
M = 30 kg
m = 5.0 kg
T1 = tension in right cable
T2 = tension in left cable
Homework Equations
ƩF(y) = 0
Ʃτ = 0
The Attempt at a Solution
ƩF(y) = 0 = T1 + T2 - g(3m + M)
T1 + T2 = 9.8(3*5 +30)
T1+T2 = 441
T1 = 441 - T2
For net torque, I chose the axis at the point where the third cat is sitting, the one that's hanging off the side. Mainly because I'm a bit confused as to what forces that cat is exerting on the beam and in what directions.
Ʃτ = 0 = 0.2*mg - 0.2*T1 + 1*Mg + 1.8*mg - 1.8*T2
0 = 9.8 - 0.2*T1 + 294 + 88.2 -1.8*T2
0.2*T1 + 1.8*T2 = 392
0.2(441 - T2) + 1.8*T2 = 392
88.2 + 1.6*T2 = 392
T2 = 189.875 ≈ 190 N
T2 = 441 - T2 = 251.125 ≈ 250
Is this correct?? That third cat is making me nervous, as is the fact that the height of the beam is given as 0.5m and I didn't use that anywhere. Did I correctly calculate the distance to the pivot point for each of the forces contributing to the net torque? I just took them as the perpendicular distance the pivot, ie the length away from the third cat.
Thanks in advance!