Tension and Equilibrium: Hanging sign

[cassie123]

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

Homework Equations

Xcm=(M1*x1+M2*x2+...)/Mtotal
ƩF(y) = 0
Ʃτ = 0

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!

 

[Qwertywerty]

You should find Ycm too .

τ= r × F = r F sin(θ) where θ is angle between r and F .
Com is not at the top surface of the sign and so the torque of mg is is not horizontal distance multiplied by mg .

 

[haruspex]

I feel as though though I should somehow be taking into about the vertical distribution of the weight on the sign using Ycm

No need. In fact, there's no benefit in finding Xcm either. Just sum the torques from the different weights.

the torque of mg is is not horizontal distance multiplied by mg .

Yes it is. The horizontal distance is the perpendicular distance from the reference axis to the line of action of the force.

 

[Qwertywerty]

Haruspex , with regard to your two points , obviously calculating resultant of all initially is not required .

However , to your second , no it is not . The force of gravity acts at a point below the point from which torque is balanced . So only one component of M(resultant)g produces a torque .

 

[haruspex]

Haruspex , with regard to your two points , obviously calculating resultant of all initially is not required .

However , to your second , no it is not . The force of gravity acts at a point below the point from which torque is balanced . So only one component of M(resultant)g produces a torque .

The point of action is not crucial. What matters for torque is the line of action.
See https://www.physicsforums.com/insights/frequently-made-errors-mechanics-moments/

 

[Qwertywerty]

Ok I think I just made a mistake .

τ = r⊥ × F
or r × F ⊥ .

You were saying r⊥×F and I r × F⊥ .Please excuse my mistake . I wasn't thinking clearly .

 

[cassie123]

No need. In fact, there's no benefit in finding Xcm either. Just sum the torques from the different weights..

Thanks for your reply! Does this mean that where I used (1.11m-0.2m) as the distance in the torque equation for the sign, I should have just used (1.0m-0.2m)?

And Qwertywerty thanks for adding your thoughts as well!

 

[cassie123]

.

Thanks for your reply! Does this mean that where I used (1.11m-0.2m) as the distance in the torque equation for the sign, I should have just used (1.0m-0.2m)?.

Nevermind, I gave this way a shot and then the tensions come out as equal. Which wouldn't make sense.

 

[haruspex]

Thanks for your reply! Does this mean that where I used (1.11m-0.2m) as the distance in the torque equation for the sign, I should have just used (1.0m-0.2m)?

I mean, for each force, find its torque about the reference axis (vertical force times horizontal displacement) and add these up. Be careful with signs. Equate the result to zero.