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Old Nov 6th 2017, 08:37 AM   #1
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Finding a position vector of a pivot on a lever from its torque

Hi guys, I have a problem here which I'm really stumped on.

I have a rigid rod of 3.1m length, with two equal masses at either end. The pivot is placed so that the total torque is 25k N.m about the pivot, and I need to find it's location.

I've drawn a diagram, a basis, calculated that the Torque, 25k, will be given by A1F1-A2F2 where the forces are the weights of the masses and A1 and A2 are the position vectors in the i direction.

Any advice on how to approach it? I tried to define the A1 and A2 as x and 3.1-x but this isnt working as far as I can tell.

Any advice would be appreciated, cheers!
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Old Nov 6th 2017, 10:20 AM   #2
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Quote:
with two equal masses at either end.
I doubt the problem said exactly that


If there is a net moment acting on the rod then it is not in equilibrium.

So is it spinning?

If not, why not?

Vertical equilibrium can be satisfied if the assembly is supported at the pivot but this support force will not have any moment about the pivot.

What other information have you been provided with?
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Old Nov 6th 2017, 11:14 AM   #3
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Originally Posted by studiot View Post
I doubt the problem said exactly that

What other information have you been provided with?
The mass of the objects are 6kg each.

The other information says; The horizontal pole is supported by a pivot point somewhere along its length. Find the location of the pivot point if the total torque about the pivot is 25Nm in the K direction.

I'm just confused as to what to do, I've not got much knowledge beyond what I did last year and that was non-vector, where I simply defined unknown positions as a linear distance in terms of x.
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Old Nov 6th 2017, 01:21 PM   #4
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Well I was going to ask where the vectors came in as they only add difficulty and obscurity.

If each mass is 6kg then it has a weight of 6*9.81 = W Newtons

= 58.86N

Let the distance of one mass from the pivot be x and the other be y, with y> x, both in metres

Then x + y = 3.10

(given)

Since there is a net 25N-metre moment we must have

Wy - Wx = 25


So we have 2 equations in 2 unkowns, from which I make y = 1.76 metres and x = 1.34 metres.

By the way did you understand my first comment ?

Your lax wording has implied that there could be one or two 6kg masses at each end of the rod.

Coupled with the fact you omitted some of the detail from the question, first time round means you need to pay more attention to detail.

And in particular the old old advice

Read the frigging question properly.
That then carries over to writing things down properly.

Last edited by studiot; Nov 7th 2017 at 02:17 AM.
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Old Nov 6th 2017, 03:56 PM   #5
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My interpretation is that you have two masses of m kg at each end of a 3.1 m rod and a pivot point at distance x from the "left" end. The clockwise torque due to the mass at the right end is mg(3.1- x). The counterclockwise torque due to the mass at the mgx. Taking the counterclockwise torque to be negative, that is a net torque of mg(3.1- x)- mgx= mg(3.1- 2x) clockwise. If that is to be 25 Nm then mg(3.1- 2x)= 25 so 3.1- 2x= 25/mg, 2x= 3.1- 25/mg, x= 1.55- 12.5/mg meters.
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finding, lever, pivot, position, torque, vector



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