A problem involving 'stress' and possible moment of force(?)

[JC2000]

Homework Statement

A light rod of length L is suspended from a support horizontally by means of two vertical wires P and Q of equal length as shown in the figure. Cross-section area of P is half that of Q and Young's modulus of P is double that of Q. A weight W is hung on the rod as shown. Find the value of x in terms of L so that the stress in P is the same as that in B.

Relevant Equations

$Y = \frac {F L}{\Delta L A}$
Stress = F/A

Given, $2A_P = A_Q$ (cross-sections) ... (1)
and, $Y_P = 2Y_Q$ ... (2)

We have $\frac {Y_P * x}{\Delta L} = \frac{Y_Q (L-x)}{\Delta L}$
Using (2) in the above expression we get $x = L/3$ whereas the correct answer is $x = 2L/3$

I feel my initial idea is flawed, and that I am overlooking something (?)
Thank you for you guidance.

 

[haruspex]

Problem Statement: A light rod of length L is suspended from a support horizontally by means of two vertical wires P and Q of equal length as shown in the figure. Cross-section area of P is half that of Q and Young's modulus of P is double that of Q. A weight W is hung on the rod as shown. Find the value of x in terms of L so that the stress in P is the same as that in B.
Relevant Equations: $Y = \frac {F L}{\Delta L A}$
Stress = F/A

View attachment 243586
Given, $2A_P = A_Q$ (cross-sections) ... (1)
and, $Y_P = 2Y_Q$ ... (2)

We have $\frac {Y_P * x}{\Delta L} = \frac{Y_Q (L-x)}{\Delta L}$
Using (2) in the above expression we get $x = L/3$ whereas the correct answer is $x = 2L/3$

I feel my initial idea is flawed, and that I am overlooking something (?)
Thank you for you guidance.

As you posted, stress is F/A. How is that affected by the modulus?
You seem to be making the strain the same.

 

[JC2000]

Stress = Strain * Y?
Thus if stress is the same then Strain*Y should also be the same?

If I compare Stress P to Stress Q, the relation between the areas is given, but the Force on P and Q would have to be calculated (or rather the tension) which I am not sure how to...

 

[haruspex]

Stress = Strain * Y?
Thus if stress is the same then Strain*Y should also be the same?

If I compare Stress P to Stress Q, the relation between the areas is given, but the Force on P and Q would have to be calculated (or rather the tension) which I am not sure how to...

You have assumed the change in length is the same, i.e. the strain is the same.
Forget about strain and modulus; just work with F, A and stress.

 

[JC2000]

In that case :

Let the force on the two wires be $T_P$ and $T_Q$, since the stress is the same on both wires
$\frac {T_P}{A_P} = \frac{T_Q}{A_Q}$ where $2A_P = A_Q$.

I realize that the forces on the wires will differ (hence the change in length differs) but I am not sure about how to calculate the forces (which is how the lengths get shoehorned into the equation?)...

I understand that the weight W would be distributed between the wires (assuming the rod has negligible mass).
If the weight was in the middle then W/2 would be the force on each wire, since it is x units from P and (L-x) from Q would this mean that the force on P be W/x and on Q by W/(L-x) ?

(Oof! Realised this does lead to the answer!)

I have some intuition as to why W is distributed as it is, but I am unable to put it explicitly. Would appreciate help in doing so.
Thank you!

 

[haruspex]

why W is distributed as it is

Take moments about the point where the weight is attached.

 

[JC2000]

So, If the moment of force has to be calculated for P about the point of attaching W it would be Wx? Likewise for Q it would be (L-x)W? How so?

In which case we have Moment of Force = Force * Distance but we need the force (= tension)...(?)

Really sorry for muddling up the basics here...

 

[haruspex]

the moment of force has to be calculated for P about the point of attaching W it would be Wx?

No. W has no torque about its point of attachment. What force acts on the rod where it is attached to P?

 

[JC2000]

No. W has no torque about its point of attachment. What force acts on the rod where it is attached to P?

The tension and gravity?

 

[haruspex]

The tension and gravity?

The rod is massless. There are three forces acting on the rod, thetwo tensions and W.
W has no moment about its point of attachment, so what is the torque balance equation about that point?

 

[JC2000]

Ah! I think I see now!
Since the rod does not rotate, the torques acting on it must be equal :
Thus, $T_P * x = T_Q * (L-x)$?

 

[haruspex]

Ah! I think I see now!
Since the rod does not rotate, the torques acting on it must be equal :
Thus, $T_P * x = T_Q * (L-x)$?

Right.

 

[JC2000]

Phew! But...
1. Say you had to calculate the tensions, is it possible? If not, what information would be required? How?

 

[haruspex]

Phew! But...
1. Say you had to calculate the tensions, is it possible? If not, what information would be required? How?

You know the sum and the ratio.

 

[JC2000]

You know the sum and the ratio.

I see the sum of the tensions being equal to W and the ratio being given by the torque relation.

Thank you for being so patient!

 

[haruspex]

I see the sum of the tensions being equal to W and the ratio being given by the torque relation.

Thank you for being so patient!

It was a deceptive question because they gave more info than you needed. I approve of that because it is more like real life. The first challenge is to work out what information is relevant.
(Or maybe there are more parts to this question?)

 

[JC2000]

It was a deceptive question because they gave more info than you needed. I approve of that because it is more like real life. The first challenge is to work out what information is relevant.
(Or maybe there are more parts to this question?)

I see! No, this is it. Or maybe they have picked up an auxiliary problem from somewhere and used it as the main problem...

Just realized, since we know the forces on the wires now, it would be possible to calculate how much each of them extend by as well...