# Moments: Problem with question itself

1. Apr 2, 2006

### GregA

This question is bugging me and though from my point of view it doesn't make sense I'm hoping that someone else can see my thinking and tell me whether they agree with it or not (because this is about the fourth question I've had to say that I can't solve this because the book is talking BS).

The question:
A non uniform rod of weight W and length 2a is suspended by a string attached to the midpoint of the rod. The rod is horizontal when a weight 3W hangs from one end of the rod. If this weight is removed, find the supporting force that is needed at the opposite end to maintain the rod in it's horizontal position.

My Working:
The diagram below shows what forces I think are acting on the rod and where (roughly) they are acting...as the rod weighs W and a weight at one end weighs 3W then the tension in the string must be 4W...also because the string is supporting the rod at dead centre ie at a length 'a' from either end then the weight of the rod is acting somewhere on the opposite side to which the 3W weight is hanging

By taking clockwise moments at the midpoint of the rod I get:
$$-W(x-a) +3W(a) = 0$$
$$Wa - Wx + 3Wa = 0$$
$$4a = x$$
but the length of the rod is only 2a...in addition to this though if x = 4a then the point where the weight W acts is 3 times further from the pivot than 3W acts (the distance where the weight acts from the 'midpoint' is x-a)...which should be correct for zero rotation...How the heck am I supposed to proceed with this when the question just seems like garbage?
Am I correct to just let this question go and carry on with another or is there something I'm missing? (there have been a few questions that I've had to give up on for similar reasons)

Basically what I'm trying to say is that if this rod had all its weight acting right at the other end of the rod then the moment about the midpoint is still only -W(a) yet at the other end the moment is 3W(a)...and it's horizontal
by the way...the answer at the back of the book is W (I can only justify this if the supporting weight is held at a distance 3a from the midpoint...but this is not the end of the rod)

http://img408.imageshack.us/img408/3346/drawing12tq.th.gif [Broken]

Last edited by a moderator: May 2, 2017
2. Apr 2, 2006

### nrqed

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hi...

I am as confused as you. This seems to be a typo in the question because as it stands it does not make sense. As you say, even if the mass would be almost all at one extremity, the force you would need to apply at the other extremity would be *at most* the weight of the object, W. The question does not make sense to me.
The other questions worked out ok? It seems like you do understand pretty well the concepts involved.

Patrick

Patrick

Last edited by a moderator: May 2, 2017
3. Apr 2, 2006

### GregA

Thanks for the reply Patrick ...glad someone else is confused by this one as well...I'm doing ok on the other questions ..got a few more to complete the chapter. (thankfully I'm over two thirds the way into this book and I'll be glad when I can finish it)

4. Apr 2, 2006

### arildno

This question, and the answer given is just BS, as far as I can see.
Since the maximal torque the rod's weight can do about the midpoint is of magnitude Wa (i.e, C.M at the endpoint of the rod), it follows that this cannot balance the applied torque 3Wa.

Thus, the exercise is just nonsense.