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1.
I'm a bit stuck on a problem here. Maybe the 'trap can unblock me?

A uniform beam is initially horizontal and suspended from two lines, one at each of its ends. Each line is looped over a pulley and has a weight attached. There is also a weight hung directly from the midpoint of the beam. These 3 weights are much larger than the weight of the beam itself, so we can disregard the beam weight. From left to right, then, the forces due to the weights are +1, -(2 + d), +(1 + 2*d), where d is a lot less than 1.

How will the beam move, when released from the initial horizontal position?
2.
Surely that is from a Beer and Johnston problem set.
3.
Nope entirely self-made for a nefarious purpose I will share, if I can get it to play ball.
4.
Pulleys have same radius, of course?
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CommentAuthoralsetalokin
• CommentTimeMar 4th 2014 edited

If d is a lot less than one, the forces reduce to +1, -2, and +1, so the beam will just rise up until it hits the pulleys, remaining horizontal the whole time.

5.
I'm sure I remember those problems. It must be in there somewhere, but I can't get to my hard copy at the moment.

http://www.mhhe.com/engcs/engmech/beerjohnston/wmdemo/Content/Problem_Statements.html
6.
Posted By: alsetalokinIf d is a lot less than one, the forces reduce to +1, -2, and +1, so the beam will just rise up until it hits the pulleys, remaining horizontal the whole time.

If d = 0, it will neither rise nor fall
+1, -2, and +1, = 0
and there is no turning couple anywhere.

You probably know that the "couple" about an axis (equivalent to linear force in a linear system) is the force applied times the distance at which it is applied. So by inspection, the resultant couples are (< 0 is anticlockwise):

But there are no fixed axes. That's why I am having trouble figuring out the motion.
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CommentAuthoralsetalokin
• CommentTimeMar 4th 2014 edited

Yes that's right, I was thinking you would be giving it a little push so it would rise up. If d is positive it will sink. The "fixed axis" you are seeking is the centre of mass of the system, and that will depend on how you have the hanging central weight hung. Compute your torques and moments about that axis. But since the system is symmetrical, there will be no torques left over to do anything. You can set the angle from horizontal of the beam, then give it a nudge in either direction up or down thru the CofM and it will coast up or down , keeping that same angle. Won't it?
7.
Forget the pulleys and ropes. Just put your end masses on the ends of the beam, and pretend that they have negative mass, or reverse the sign of "g" for them.

With the suspension, unless the ropes are very long, there will be complications as the pulls won't be vertical after the beam departs from horizontal, if it does.
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CommentAuthorAngus
• CommentTimeMar 4th 2014

The centre of gravity does not move because there is no net force. Find the rotation by picking any point and figuring torque and moment of inertia about it.
8.
Posted By: AngusThe centre of gravity does not move because there is no net force.
Oh... sounds like my housemates.
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CommentAuthorAngus
• CommentTimeMar 4th 2014

That can be very peaceful (no force) or very stressful (no net force).
9.
Posted By: AngusThe centre of gravity does not move because there is no net force. Find the rotation by picking any point and figuring torque and moment of inertia about it.
There is indeed a net force for nonzero d, and its value is d = 1 - (2 + d) + (1 + 2*d)

As for the rotation, it is clearly anticlockwise, because it has that sign about any point on the beam one chooses. I picked 3 points and already showed that for each of them.
• CommentAuthorjoshs
• CommentTimeMar 4th 2014

Posted By: Andrew PalfreymanI'm a bit stuck on a problem here. Maybe the 'trap can unblock me?

A uniform beam is initially horizontal and suspended from two lines, one at each of its ends. Each line is looped over a pulley and has a weight attached. There is also a weight hung directly from the midpoint of the beam. These 3 weights are much larger than the weight of the beam itself, so we can disregard the beam weight. From left to right, then, the forces due to the weights are +1, -(2 + d), +(1 + 2*d), where d is a lot less than 1.

How will the beam move, when released from the initial horizontal position?

Solve it by working out the tensions. You can get tripped up by sin(x) here. As the beam rises towards the pulleys, the beam weight ends up with huge gain against the side weights as 1/sin(x) => 0.
10.
Posted By: joshsSolve it by working out the tensions. You can get tripped up by sin(x) here. As the beam rises towards the pulleys, the beam weight ends up with huge gain against the side weights as 1/sin(x) => 0
Don't follow that sin(x) thing. The forces always act vertically, whatever the beam orientation. As for calculating the tensions, I already know their initial values - because I specified them.
11.
Posted By: alsetalokinIf d is positive it will sink.
No, it will rise
Posted By: alsetalokinThe "fixed axis" you are seeking is the centre of mass of the system, and that will depend on how you have the hanging central weight hung.
Like I said, a weight is attached to the beam centre. It can be directly stuck there.
Posted By: alsetalokinBut since the system is symmetrical, there will be no torques left over to do anything.
You wrote this right below where I computed the torques. Hmm.
• CommentAuthorjoshs
• CommentTimeMar 4th 2014

Posted By: Andrew Palfreyman
Posted By: joshsSolve it by working out the tensions. You can get tripped up by sin(x) here. As the beam rises towards the pulleys, the beam weight ends up with huge gain against the side weights as 1/sin(x) => 0
Don't follow that sin(x) thing. The forces always act vertically, whatever the beam orientation. As for calculating the tensions, I already know their initial values - because I specified them.

Well maybe I don't have the right picture from your description. But if you picture a high tension power line, it should be obvious to you that when you suspend a mass under tension that it sags. Next think of why that is. If the cable goes almost straight down, then the up component of the tension vector is nearly 1.0. But if the cable is nearly level, then the up component of the tension vector is nearly zero.
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CommentAuthorAngus
• CommentTimeMar 4th 2014 edited

Posted By: Andrew Palfreyman
Posted By: AngusThe centre of gravity does not move because there is no net force. Find the rotation by picking any point and figuring torque and moment of inertia about it.
There is indeed a net force for nonzero d, and its value is d = 1 - (2 + d) + (1 + 2*d)

As for the rotation, it is clearly anticlockwise, because it has that sign about any point on the beam one chooses. I picked 3 points and already showed that for each of them.

Ahhh...correct. My blunder. I also missed your intention that the bar was massless.

The masses of the weights, which are the only thing with a mass in the problem, I assume, are 1/g,(2+d)/g and (1+2d)/g. The net force is d, and is up. Therefore the centre of mass accelerates upward at a rate of d/(g(4+3d)). You can figure out where the centre of mass is by considering the masses all to be attached to the bar at the point where the string attaches. The centre of mass is at the solution to x/g+(L/2-x)(2+d)/g+(L-x)(1+2d)/g=0 where L is the length of the bar and x is the distance of the centre of mass from the left end.

The effective moment of inertia about the centre of mass can be figured out in the same way, by considering the weights attached to the bar. Calculate the torque as well and figure the angular acceleration.

This only works for the initial motion. As joshs said everything changes as soon as it actually moves and you have to figure in the angles of pull. You can follow the same reasoning and do it if you really want to know that badly.

It also only works if you are pulling on the strings. As soon as any net force on a string becomes compressive, you violate the Fundamental Principle of Engineering.
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CommentAuthoralsetalokin
• CommentTimeMar 4th 2014 edited

You guys crack me up. First you say that d is very very much smaller than 1, which normally means you can ignore it by setting it to zero. It is obvious that the specified initial condition is symmetrical wrt torques around the _real_ "axis of rotation" which is at the centroid. Presumably the strings and pulleys are spaced such that the strings can remain vertical: with the bar horizontal, that is. Even if the strings from the pulleys are straight down at the start, if the bar rotates then the ends are closer together in the x direction and the strings are no longer vertical. There will be a restoring force caused by the horizontal component of the tension that tends to keep the bar horizontal especially as it gets closer to the pulley suspension. (Assuming the pulleys now are the same size and their axes are at the same y.)
I don't see how any unbalanced torques could arise unless there is asymmetric friction in the pulleys or your initial "nudge" either up or down isn't vertical and directly through the centroid. Once an unbalanced torque does happen then the bar may swing from horizontal but as it goes up. Why should it rotate anticlockwise instead of clockwise if everything is symmetrical about the center of mass? Did I misread the initial condition?

I'm confurrsed by the sign convention, I guess. Did I read wrong? I thought d was the beam's mass. If d is zero the thing is in balance vertically. If d is greater than zero the thing is heavier than the two weights on the ends of the strings, so the beam will descend. Won't it? But if d is very very much smaller than 1, then you can neglect it unless you are going to Mars or somewhere on this beam.

Posted By: joshs
Posted By: Andrew Palfreyman
Posted By: joshsSolve it by working out the tensions. You can get tripped up by sin(x) here. As the beam rises towards the pulleys, the beam weight ends up with huge gain against the side weights as 1/sin(x) => 0
Don't follow that sin(x) thing. The forces always act vertically, whatever the beam orientation. As for calculating the tensions, I already know their initial values - because I specified them.

Well maybe I don't have the right picture from your description. But if you picture a high tension power line, it should be obvious to you that when you suspend a mass under tension that it sags. Next think of why that is. If the cable goes almost straight down, then the up component of the tension vector is nearly 1.0. But if the cable is nearly level, then the up component of the tension vector is nearly zero.

And in the real world it is impossible to pull that cable so that it is completely horizontal rather than a catenary arch, because to do so the tension in the cable must rise without limit.

Disclaimer: I am not an engineer nor do I play one on TV. And it is problems like these that I must thank for that.
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CommentAuthorAngus
• CommentTimeMar 4th 2014 edited

I agree with all you say(ETA - Oh except about d being the bar's mass). I presume the problem is to figure out the INITIAL accelerations, as my post actually says somewhere in the mists of trying to take this problem seriously.

There +is+ a torque simply because there is a couple. That is how the bar winds up oscillating around some cock-a-mamie angle sometime within the first three seconds. Just after one end gets jammed in a pulley.