(Check solution) Conceptual problem about the equilibrium of a rigid body

[daphnelee-mh]

Homework Statement

(attached below)

Relevant Equations

ΣFy=0

Please help to see whether I did correct or wrong, thank you.

 

[Chestermiller]

You omitted the force from the overhang pin connection at the wall. Also sine of angle is opposite over hypotenuse, not opposite over adjacent.

 

[haruspex]

As @Chestermiller mentions, you are missing the reaction force where the overhang meets the wall.
In a), moving the top of the tie down to D changes L.
In b), it's always a good idea to do a sanity check. Your algebra says the tension is minimised by moving the pin as close to the wall as possible. Does that seem reasonable? Does it seem like the overhang would be secure?
Hint: torque.

 

[daphnelee-mh]

As @Chestermiller mentions, you are missing the reaction force where the overhang meets the wall.
In a), moving the top of the tie down to D changes L.
In b), it's always a good idea to do a sanity check. Your algebra says the tension is minimised by moving the pin as close to the wall as possible. Does that seem reasonable? Does it seem like the overhang would be secure?
Hint: torque.

'The pin is connected to the building wall at A and to the center of the overhang B', isn't it a two force member?

 

[haruspex]

'The pin is connected to the building wall at A and to the center of the overhang B', isn't it a two force member?

You are not reading it correctly.
The rod ... is 'pin-connected' (i.e. connected by a pin, about which it could, in principle, rotate) to the wall. At the other end, it is pin-connected to the centre of the overhang.
The last sentence tells you the overhang is also pin-connected directly to the wall. This is the joint from which you are omitting the reaction force.

 

[daphnelee-mh]

You are not reading it correctly.
The rod ... is 'pin-connected' (i.e. connected by a pin, about which it could, in principle, rotate) to the wall. At the other end, it is pin-connected to the centre of the overhang.
The last sentence tells you the overhang is also pin-connected directly to the wall. This is the joint from which you are omitting the reaction force.

Thanks for your advise and corrected as below, can I assume that since the T is inversely proportional to theta, in both cases, theta decrease after moving, therefore T increase?

 

[haruspex]

Thanks for your advise and corrected as below, can I assume that since the T is inversely proportional to theta, in both cases, theta decrease after moving, therefore T increase?
View attachment 266331

Your algebra there is all good, but in order to answer the question you need to get T as a function of mg and theta. Having it as a function of Fx doesn't help because you do not know how changing theta will affect Fx.

 

[daphnelee-mh]

Your algebra there is all good, but in order to answer the question you need to get T as a function of mg and theta. Having it as a function of Fx doesn't help because you do not know how changing theta will affect Fx.

Ya, but I have no idea about it, I tried to total up the torque about A but ended with Fx = Tx which I got before.

 

[haruspex]

Ya, but I have no idea about it, I tried to total up the torque about A but ended with Fx = Tx which I got before.

You got T_y=mg. What is the relationship between T_y, T and theta?

 

[daphnelee-mh]

You got T_y=mg. What is the relationship between T_y, T and theta?

Oh, I got it! Thank you very much

 

[Chestermiller]

I think case b is more interesting. What do you get for case b?

 

[daphnelee-mh]

I think case b is more interesting. What do you get for case b?

This is my answer and explanation for both cases

 

[Chestermiller]

In case b, for the original attachment, if I take moments about the attachment point on the wall, I get $$mg\frac{L}{2}-T\frac{h}{\sqrt{(\frac{L}{2})^2+h^2}}\frac{L}{2}=0$$or$$T=\frac{mg\sqrt{(\frac{L}{2})^2+h^2}}{h}$$
For the case where the connection is moved to the end of the overhang, the similar moment balance is $$mg\frac{L}{2}-T\frac{h}{\sqrt{L^2+h^2}}L=0$$or$$T=\frac{mg\sqrt{L^2+h^2}}{2h}=\frac{mg\sqrt{(\frac{L}{2})^2+(\frac{h}{2})^2}}{h}$$

 

[daphnelee-mh]

In case b, for the original attachment, if I take moments about the attachment point on the wall, I get $$mg\frac{L}{2}-T\frac{h}{\sqrt{(\frac{L}{2})^2+h^2}}\frac{L}{2}=0$$or$$T=\frac{mg\sqrt{(\frac{L}{2})^2+h^2}}{h}$$
For the case where the connection is moved to the end of the overhang, the similar moment balance is $$mg\frac{L}{2}-T\frac{h}{\sqrt{L^2+h^2}}L=0$$or$$T=\frac{mg\sqrt{L^2+h^2}}{2h}=\frac{mg\sqrt{(\frac{L}{2})^2+(\frac{h}{2})^2}}{h}$$

I can know that T is proportional to L from your calculation, therefore the answer will be' increase' too when moved to outer position. Isn't it?

 

[Chestermiller]

I can know that T is proportional to L from your calculation, therefore the answer will be' increase' too when moved to outer position. Isn't it?

No

 

[daphnelee-mh]

No

Kindly advise

 

[haruspex]

I can know that T is proportional to L from your calculation

In @Chestermiller's algebra, L is the length of the overhang. This is the same for both cases, so how T varies as L varies is irrelevant. Similarly, h is the same in both. So just compare the two expressions and decide which is larger.

 

[Chestermiller]

Kindly advise

In case b, even though the angle is smaller, the moment arm for the tension is longer, and the longer moment arm wins out, so the tension is less.

You were advised to use a moment balance to solve this problem, but chose not to follow that advice. You were also advised that your force balance omitted the vertical force provided by the pin connection at the wall, but you neglected that advice also. Now you are wondering why you got the wrong answer.

 

[daphnelee-mh]

In case b, even though the angle is smaller, the moment arm for the tension is longer, and the longer moment arm wins out, so the tension is less.

You were advised to use a moment balance to solve this problem, but chose not to follow that advice. You were also advised that your force balance omitted the vertical force provided by the pin connection at the wall, but you neglected that advice also. Now you are wondering why you got the wrong answer.

Sorry for my misunderstanding, but I did correct my answer using the moment balance after your comment before. Thanks for you advise :)