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Old Nov 9th 2017, 06:19 AM   #1
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Static Equilibrium

I'm having a lot of trouble getting my head around this problem-


A 3.6-meter long horizontal plank is held up by two supports. One support is at the left end, and the other is 0.80 m from the right end. The plank has uniform density and has a mass of 40 kg. How close can a 70 kg person stand to the unsupported end before causing the plank to rotate?

I've found a few explanations online but none of them make sense to me, could anyone break it down maybe with step by step explanations? I'd really appreciate it!
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Old Nov 9th 2017, 09:20 AM   #2
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I would start by setting up a "coordinate system". Take x= 0 at the left end, x= 3.6 at the right end. There is a support at the left end so indicate an upward force at x= 0 labeled "p". There is a support "0.80 m from the right end" so indicate an upward force at x= 3.6- 0.8= 2.8. The plank has uniform density and mass 40 kg so weight 40g Newtons (actually the "g" will cancel at the end so you could do everything in kg rather than Newtons) so indicate a downward force 40g at x= 1.8, the center of the plank. Finally, there is a downward force of 70g Newtons at an unknown position- call that y.

In order that this be in "equilibrium" the upward force must be equal downward force and clockwise torque must equal counter clockwise torque. The first condition gives p+ q= 70gy- 40g. Taking the torque around the left end, we have, for clockwise torque, 40g(1.8)+ 70gy and, for counter clockwise torque, 3.6 q. In order that the plank be in equilibrium, we must have 40g(1.8)+ 70gy= 72g+ 70gy= 3.5q. If the clockwise torque is greater than the counter clockwise, 72g+ 70gy= 3.5q, the plank will fall.
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Old Nov 9th 2017, 11:53 AM   #3
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Before immersing yourself in HallsofIvy's fully worked explanation,

Do you understand why the plank will tip up if you stand on the unsupported end?
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Old Nov 10th 2017, 05:36 AM   #4
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Originally Posted by studiot View Post
Before immersing yourself in HallsofIvy's fully worked explanation,

Do you understand why the plank will tip up if you stand on the unsupported end?
Because the right-hand support structure is the fulcrum?
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Old Nov 10th 2017, 06:07 AM   #5
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Perhaps a look at this link might give you a clue:

<Beam Balance>
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Old Nov 10th 2017, 06:18 AM   #6
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Originally Posted by HallsofIvy View Post
I would start by setting up a "coordinate system". Take x= 0 at the left end, x= 3.6 at the right end. There is a support at the left end so indicate an upward force at x= 0 labeled "p". There is a support "0.80 m from the right end" so indicate an upward force at x= 3.6- 0.8= 2.8. The plank has uniform density and mass 40 kg so weight 40g Newtons (actually the "g" will cancel at the end so you could do everything in kg rather than Newtons) so indicate a downward force 40g at x= 1.8, the center of the plank. Finally, there is a downward force of 70g Newtons at an unknown position- call that y.

In order that this be in "equilibrium" the upward force must be equal downward force and clockwise torque must equal counter clockwise torque. The first condition gives p+ q= 70gy- 40g. Taking the torque around the left end, we have, for clockwise torque, 40g(1.8)+ 70gy and, for counter clockwise torque, 3.6 q. In order that the plank be in equilibrium, we must have 40g(1.8)+ 70gy= 72g+ 70gy= 3.5q. If the clockwise torque is greater than the counter clockwise, 72g+ 70gy= 3.5q, the plank will fall.

Thanks! This makes a lot more sense, but I'm still getting the wrong answer...
I used a system of equations with two unknowns and plugged in 70y - 40= q and 72 + 70y = 3.5g.
I keep coming up with 2.4 m but the answer key says it's 0.23m ._.
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Old Nov 10th 2017, 06:47 AM   #7
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Originally Posted by TyrannosaurusWreck View Post
Because the right-hand support structure is the fulcrum?
Yes.

Are you happy now or do you need any more?
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Old Nov 13th 2017, 05:23 AM   #8
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Originally Posted by studiot View Post
Yes.

Are you happy now or do you need any more?
Well, I'm still having trouble getting the correct answer. The key says it's 0.23 but I keep arriving at 1.2. I used HallsOfIvy's explanation and a system of equations but it's still coming up with 1.2
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Old Nov 13th 2017, 06:31 AM   #9
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I'm afraid that HoI's explanation is a bit confused - he made the mistake of including forces (such as p and q) in the same equation as torque (70gy).

There's an easier way that doesn't require calculating the reaction forces. Realizing that the plank will tip when the reaction force of the left support is zero, you can write an equation for torque about the right support that includes just two terms: the man's weight times his distance from that support must equal the planks weight times the distance from the right support to the center of gravity of the plank:

70Kg x g x d = 40 Kg x g x 1 m

Solve for d, to give the distance from the right support to where the man is standing. His distance from the end of the plank is then 0.8-d.
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