- #1
daphnelee-mh
- 66
- 4
- Homework Statement
- (attached below)
- Relevant Equations
- ΣFy=0
Please help to see whether I did correct or wrong, thank you.
'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 said: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.
You are not reading it correctly.daphnelee-mh said:'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?
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 said: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.
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 said: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
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 said: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.
You got Ty=mg. What is the relationship between Ty, T and theta?daphnelee-mh said: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.
Oh, I got it! Thank you very muchharuspex said:You got Ty=mg. What is the relationship between Ty, T and theta?
This is my answer and explanation for both casesChestermiller said:I think case b is more interesting. What do you get for case b?
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 said: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$$T=\frac{mg\sqrt{L^2+h^2}}{2h}=\frac{mg\sqrt{(\frac{L}{2})^2+(\frac{h}{2})^2}}{h}$$
Nodaphnelee-mh said: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?
Kindly adviseChestermiller said:No
daphnelee-mh said:I can know that T is proportional to L from your calculation
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.daphnelee-mh said:Kindly advise
Sorry for my misunderstanding, but I did correct my answer using the moment balance after your comment before. Thanks for you advise :)Chestermiller said: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.
A rigid body is an object that does not deform or change shape when subjected to external forces. In other words, the distances between points on the object remain constant, making it an idealized model for studying the equilibrium of objects.
Equilibrium of a rigid body refers to a state where the object is not moving or rotating, and the forces acting on it are balanced. This means that the net force and net torque on the object are both equal to zero.
To determine if a rigid body is in equilibrium, we must analyze the forces and torques acting on the object. If the sum of all forces acting on the object is zero, and the sum of all torques acting on the object is also zero, then the object is in equilibrium.
In static equilibrium, the object is not moving or rotating, and the forces and torques acting on it are balanced. In dynamic equilibrium, the object is moving or rotating at a constant velocity, and the forces and torques acting on it are also balanced.
The center of mass of a rigid body can be calculated by finding the weighted average of the positions of all the particles that make up the object. This can be done using the formula: xcm = (m1x1 + m2x2 + ... + mnxn) / (m1 + m2 + ... + mn), where m is the mass of each particle and x is its position.