Rotational Motion and Torque problem

[jackthehat]

Homework Statement

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 the 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

2. Homework Equations
torque (T) = l x F = l sin (angle) F , where l = lever-arm [/B]

The Attempt at a Solution

I was able to get correct solutions for parts (a) and (b) but I cannot work out how to get a correct solution for part (c).

The book gives the following as the correct solution for part (c) ..
x = (l/2) (1 + [2h/l]), for l > 2h and x = l , for l > 2h

I used the above equations substituting (l/2) for the lever arm and tried to get a general equation for the attachment point on the bar using Pythagoras theorem (as this would form a right-angled triangle where point 'P' to the bar is the hypotenuse) but I cannot get anything near the correct solution. In addition I am unsure as to why we have two distinct solutions for this problem.
Can anyone help ?[/B]

 

[Isaac0427]

I used the above equations substituting (l/2) for the lever arm and tried to get a general equation for the attachment point on the bar using Pythagoras theorem (as this would form a right-angled triangle where point 'P' to the bar is the hypotenuse) but I cannot get anything near the correct solution. In addition I am unsure as to why we have two distinct solutions for this problem.

In this problem you are attempting to maximize the lever arm, that is r*sin(θ). Find an expression for sin(θ) in terms of r, l, and h. Then plug that in for sin(θ) and find the r that maximizes the expression.

 

[haruspex]

I cannot get anything near the correct solution.

We cannot tell where you are going wrong if you do not post your attempt (as is anyway required by forum rules).

 

[jackthehat]

We cannot tell where you are going wrong if you do not post your attempt (as is anyway required by forum rules).

Hi Haruspex,
I did summarise my attempt at the problem in my questio, but I will repeat what I did below ...

torque (T) = l x F = l sin (angle) F , where l = lever-arm

I used the above equation substituting 'l/2' for the lever arm and tried to get a general equation for the attachment point on the bar using Pythagoras theorem (as this would form a right-angled triangle from point 'P' to attachment point on x-axis, and point 'p' perpendicular to x-axis, where point 'P' to the attachment point is the hypotenuse) but I cannot get anything near the correct solution.
I also have no idea why the book shows 2 solution for this problem .. one for l>2h and another for l<2h ?

 

[haruspex]

I did summarise my attempt

I know. Repeating it does not help. There is no way I can tell from that what equations you wrote down or how you manipulated them.

 

[jackthehat]

Hi again Haruspex,
Sorry about being vague about my attempt .. I will give a more detailed explanation of what I did below ..

I first used the equation .. Torque = F L (where F is the force applied and L is the distance from the position where force is applied and the axis of rotation)
In this case .. L = l / 2 (where l = 1/2 length of the rod)
If we then construct a right-angled triangle connecting point P vertically to the bar (on x-axis) (call this point A and so line 'PA' will be the 'opposite' side of the triangle .. and PA = h (height of P above axis)... the 'adjacent' side of the triangle is the part of the x-axis from point A on x-axis to the attachment point (point B) somewhere on the x-axis ... and so the 'hypotenuse' will be the line joining point P to the attachment point of the string on the bar .. that is the line 'PB'.
From this we see that .. PA / L = Tan theta => h / (l/2) = Tan theta
Now we also know that .. Torque = F sin theta
So [F (h / (l/2)] = F sin Theta => (h / (l/2)) = sin Theta => Tan Theta = sin Theta

This is where i got totally lost I didn't know where to go from here and I couldn't see how this helps me get the answer listed in the book or how and why there are two solutions for this problem. Can you help point me in the right direction ?
Jackthehat

 

[haruspex]

Torque = F sin theta

I assume you meant FL sin(θ), but that is not right.
You have defined θ as angle POA. Did you mean to define it as angle PBA?

 

[jackthehat]

Hi Haruspex,
Yes you are right I did mean FL sin theta. This was just a careless error on my part, and yes I did mean the angle PBA. On my first attempt at tackling this problem I thought the correct angle was POA, but when thinking about it in subsequent attempts, I realized that it's the angle at the point of attachment on the bar, which can vary and not at the origin which is fixed. I was copying my old working from my notebook and copied the wrong piece of working (from my first attempt) in error. Hope that clears things up.
Jackthehat

 

[haruspex]

Hi Haruspex,
Yes you are right I did mean FL sin theta. This was just a careless error on my part, and yes I did mean the angle PBA. On my first attempt at tackling this problem I thought the correct angle was POA, but when thinking about it in subsequent attempts, I realized that it's the angle at the point of attachment on the bar, which can vary and not at the origin which is fixed. I was copying my old working from my notebook and copied the wrong piece of working (from my first attempt) in error. Hope that clears things up.
Jackthehat

Ok, now FL sin(θ) makes more sense, but in fact it is wrong.
The easiest way to solve this is to resolve the force F into vertical and horizontal components. One of them has torque FL sin(θ), but the other also exerts a torque about O.

 

[haruspex]

I should also explain why FL sin(θ) is wrong.
In vectors, the torque is $\vec r\times\vec F$, where $\vec r$ is the position vector of a point in the line of action of $\vec F$. Converting to scalars gives $|\vec r|.|\vec F|\sin(\theta)$, where θ is the angle between the vectors (measured from $\vec r$ to $\vec F$ anticlockwise).
Point A is not in the line of action of F, but point B is, so you should choose OB as $\vec r$ and the torque is Fr sin(θ).

 

[jackthehat]

Hi Haruspex,
Thank you for having the patience to stick with me on this one, and for making the effort to explain why 'FL sin theta' is wrong. However I am sorry to say that in taking into account your explanations above and giving the matter much thought I am no nearer understanding how we get either ..
a) two solutions for this problem or b) how from the equation ... 'Fr sin theta' to the solutions given in the book, which are ...
x = (l/2) (1 + [2h/l]), for l > 2h and x = l , for l > 2h
I am afraid, if anything, I am even more confused than I was when I first posted this. Sorry.
Jackthehat

 

[haruspex]

Hi Haruspex,
Thank you for having the patience to stick with me on this one, and for making the effort to explain why 'FL sin theta' is wrong. However I am sorry to say that in taking into account your explanations above and giving the matter much thought I am no nearer understanding how we get either ..
a) two solutions for this problem or b) how from the equation ... 'Fr sin theta' to the solutions given in the book, which are ...
x = (l/2) (1 + [2h/l]), for l > 2h and x = l , for l > 2h
I am afraid, if anything, I am even more confused than I was when I first posted this. Sorry.
Jackthehat

There are two ways we can get the correct expression for the torque.

1. Use a bit of geometry to find the distance from O to the (projected) line BP. This would be the correct lever arm for the force F.

2. Resolve the force F into components parallel and normal to the line OBA. Normal to OBA the component is F sin(θ).
We can take its point of action as anywhere we choose along F's line of action, i.e. BP. It will be convenient to choose point B. So we have these two forces applied at point B and we need the torque from each. Can you take it from there?

 

[jackthehat]

Hi Haruspex,
That makes things a bit clearer .. it gives me a sense of where to go from here. I will give it a go and see where it takes me. Thank you for all of your help.
Jackthehat