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Old Mar 7th 2018, 06:30 AM   #1
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Rotational motion problem

Hi everybody out there, hope you are all well,
I have come across a problem in the chapter I am working on in my Physics book and which has got me stumped for the moment.
I just can't see how the answer is to be gotten from the information we are given.
The problem is as follows :

Problem
A thin rod of length 'L' lies on the '+x-axis' with it's left end at the origin. A string pulls on the rod with a force F' directed towards a point 'P' a distance 'h' above the rod.

Where along the rod should you attach the string to get the greatest Torque
about the origin, if point 'P' is

(a) above the right end of the rod ?
(b) above the left end of the rod ?
(c) above the centre of the rod ?

Now I have worked out both parts (a) and (b) but I cannot solve part (c).
It is how to go about solving part (c) I am asking advice on.


My attempt - I noted that Torque = Force x lever arm ie T = F x l
and since point P is midway along the rod => l = L/2 .. so
Torque = F x (L/2)
but here is where I am stuck I do not know how estimate point of attachment of the string to maximize the torque. I know that the force is constant so the length is what changes the Torque up or down .. but how you go about estimating the value of 'x' (where along the rod and therefore the x-axis the string should be attached) I just don't know where to start with this one ?

The answer in the book is ..
x = (1/2) [1 + (2h/l)sqrd] , for l > 2h ; and x = l , for l < 2h

Can anyone help ?

Regards,
Jackthehat
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Old Mar 7th 2018, 09:31 AM   #2
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You can resolve the force F into horizontal and vertical components, and only the vertical component contributes to torque. So:

T = F h/sqrt (h^2+(x-L/2)^2)

Take the derivative dT/dx and set to zero, and solve for x.

The reason there is a different answer for h>L/2 is that the value for x using the above turns out to be greater than L, which can't be.
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Old Mar 8th 2018, 10:21 PM   #3
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Hi ChipB,
Thank you for that input you're explanation has helped me understand where I went wrong.
Regards,
jackthehat
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