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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
Pulley radius should not matter.
Posted By: AngusThe centre of gravity does not move because there is no net force.Oh... sounds like my housemates.
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)
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?
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) => 0Don'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.
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.
Posted By: Andrew PalfreymanPosted 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) => 0Don'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.
Posted By: Andrew PalfreymanPosted 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.
Posted By: joshsPosted By: Andrew PalfreymanPosted 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) => 0Don'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.