Hanging beam problem

[James10492]

Hi again, you guessed it, more mechanics:



A uniform beam AB, of mass M and length 2a, rests with its end A against a rough vertical wall. The beam is held in a horizontal position by a rope. One end of the beam is attached to the rod at a point C that lies .5a from B, and the other end is attached to the wall at a point D that lies vertically above A. The coefficient of friction between the wall and the beam is .4, and the angle between the rope and the rod is alpha.

Now the first part of the problem asks you to show, given the above information, and given that the beam is held in equilibrium, that the magnitude of the tension in the rope is equal to

[math]\frac{2Mg}{3\sin\alpha}[/math]
Taking moments about A,

[math]Mg*a=\frac{3}{2}a*\sin\alpha*T \\ T = \frac{Mg*a}{\frac{3}{2}a*\sin\alpha} \\ T = \frac{2Mg}{3\sin\alpha}[/math]
as required.

Now the next part is to find the horizontal and vertical components of the force exerted on the beam by the wall. Resolving forces horizontally,

[math]H = T\cos\alpha \\ H = \cos\alpha(\frac{2Mg}{3\sin\alpha}) \\ H = \frac{2Mg}{3\tan\alpha}[/math]
Now here is where I have a problem. Resolving forces vertically,

[math]Mg = T\sin\alpha +.4H+V \\ V = Mg - \frac{2}{3}Mg-.4(\frac{2Mg}{3\tan\alpha})[/math]
But, apparently, the solution is that V = Mg/3, as though you don't need to take into account the frictional force of the wall. Is this because V and F work in the same plane? (This doesn't sense to me, I am sure I have come across problems before where this is the case).

 

[topsquark]

Hi again, you guessed it, more mechanics:

View attachment 32683

A uniform beam AB, of mass M and length 2a, rests with its end A against a rough vertical wall. The beam is held in a horizontal position by a rope. One end of the beam is attached to the rod at a point C that lies .5a from B, and the other end is attached to the wall at a point D that lies vertically above A. The coefficient of friction between the wall and the beam is .4, and the angle between the rope and the rod is alpha.

Now the first part of the problem asks you to show, given the above information, and given that the beam is held in equilibrium, that the magnitude of the tension in the rope is equal to

[math]\frac{2Mg}{3\sin\alpha}[/math]
Taking moments about A,

[math]Mg*a=\frac{3}{2}a*\sin\alpha*T \\ T = \frac{Mg*a}{\frac{3}{2}a*\sin\alpha*T} \\ T = \frac{2Mg}{3\sin\alpha}[/math]
as required.

Now the next part is to find the horizontal and vertical components of the force exerted on the beam by the wall. Resolving forces horizontally,

[math]H = T\cos\alpha \\ H = \cos\alpha(\frac{2Mg}{3\sin\alpha}) \\ H = \frac{2Mg}{3\tan\alpha}[/math]
Now here is where I have a problem. Resolving forces vertically,

[math]Mg = T\sin\alpha +.4H+V \\ V = Mg - \frac{2}{3}Mg-.4(\frac{2Mg}{3\tan\alpha})[/math]
But, apparently, the solution is that V = Mg/3, as though you don't need to take into account the frictional force of the wall. Is this because V and F work in the same plane? (This doesn't sense to me, I am sure I have come across problems before where this is the case).

Your analysis looks good to me. Your given answer is wrong.

-Dan

 

[blamocur]

I am rather rusty on this things, but here is my understanding: the friction force does not belong in the equation. You first compute the vertical component without taking friction into account. If the result exceeds the friction force then there is no equilibrium.

To use a simpler example, say you have a body on a horizontal surface. The friction force is irrelevant unless you apply horizontal force to the body. As soon as you do you need to compare the applied force to the friction force to see if the body starts moving.

 

[Deleted member 4993]

Hi again, you guessed it, more mechanics:

View attachment 32683

A uniform beam AB, of mass M and length 2a, rests with its end A against a rough vertical wall. The beam is held in a horizontal position by a rope. One end of the beam is attached to the rod at a point C that lies .5a from B, and the other end is attached to the wall at a point D that lies vertically above A. The coefficient of friction between the wall and the beam is .4, and the angle between the rope and the rod is alpha.

Now the first part of the problem asks you to show, given the above information, and given that the beam is held in equilibrium, that the magnitude of the tension in the rope is equal to

[math]\frac{2Mg}{3\sin\alpha}[/math]
Taking moments about A,

[math]Mg*a=\frac{3}{2}a*\sin\alpha*T \\ T = \frac{Mg*a}{\frac{3}{2}a*\sin\alpha*T} \\ T = \frac{2Mg}{3\sin\alpha}[/math]
as required.

Now the next part is to find the horizontal and vertical components of the force exerted on the beam by the wall. Resolving forces horizontally,

[math]H = T\cos\alpha \\ H = \cos\alpha(\frac{2Mg}{3\sin\alpha}) \\ H = \frac{2Mg}{3\tan\alpha}[/math]
Now here is where I have a problem. Resolving forces vertically,

[math]Mg = T\sin\alpha +.4H+V \\ V = Mg - \frac{2}{3}Mg-.4(\frac{2Mg}{3\tan\alpha})[/math]
But, apparently, the solution is that V = Mg/3, as though you don't need to take into account the frictional force of the wall. Is this because V and F work in the same plane? (This doesn't sense to me, I am sure I have come across problems before where this is the case).

Did you draw a Free Body Diagram, showing all the forces acting on the bar and explicitly defining x & y axes? Please include dimensions.

You have included an image which is too dark for me discern any information. Please make sure that the picture you attach are legible.

 

[topsquark]

I am rather rusty on this things, but here is my understanding: the friction force does not belong in the equation. You first compute the vertical component without taking friction into account. If the result exceeds the friction force then there is no equilibrium.

To use a simpler example, say you have a body on a horizontal surface. The friction force is irrelevant unless you apply horizontal force to the body. As soon as you do you need to compare the applied force to the friction force to see if the body starts moving.

There has to be friction there otherwise the beam will slide on the wall, but you are correct. The friction force can only be calculated by knowing the net downward force on the beam. It isn't likely to be 0.4 H, which is the maximum before it moves.

Good catch!

-Dan

 

[The Highlander]

Did you draw a Free Body Diagram, showing all the forces acting on the bar and explicitly defining x & y axes? Please include dimensions.

You have included an image which is too dark for me discern any information. Please make sure that the picture you attach are legible.

(I see you've been busy. ?)
I was curious to see what the sketch was too (see below) but there's not very much to "read" in it other than what looks like it might be \(\displaystyle \frac{3}{2}\)\(\displaystyle a\) at the bottom.

​

 

[The Highlander]

Spot of advice @James10492.

When taking a pic of your work, try standing it upright instead of laying it down flat (it's very difficult to avoid the shadow of your phone/camera obscuring your work if it's directly above it).
If you're on a (Windows) PC, the built-in "Photos" App has very useful editing facilities (to Crop or adjust the Brightness, etc.) that can make your final pic much clearer. (See above.)
Most smartphones also have similar facilities available on them too but the most important bit is to get the best (clearest) picture you can when you take your first snap. ??