# M1 Static rigid bodies questionWatch

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#1
A uniform rod AB of length 2a metres and mass mkg is smoothly hinged at A. It is maintained in equilibrium by a horizontal force of magnitude P acting at B. The rod is inclined at 30 degrees to the horizontal with B below A.

a) Show that P=sqrt3/2 mgN

Can someone help with the diagram.. I have no idea how to draw it and also how to solve the problem.
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11 months ago
#2
(Original post by dont know it)
A uniform rod AB of length 2a metres and mass mkg is smoothly hinged at A. It is maintained in equilibrium by a horizontal force of magnitude P acting at B. The rod is inclined at 30 degrees to the horizontal with B below A.

a) Show that P=sqrt3/2 mgN

Can someone help with the diagram.. I have no idea how to draw it and also how to solve the problem.

The rod is smoothly hinged at A which means it can swing from left to right. The force P is holding it in equilibirum. If you were to release this force then the rod would swing to the right. Can you picture this now?

I haven't included the reaction force that acts on the rod by the hinge. Often it has unknown magnitude/direction so the best thing to do is show a vertical force V and a horizontal force H acting at A. In this case it may be obvious what the forces are. These two forces may not be required in the working of your question (they're not for part a) anyway).

Any thoughts on how to start?
Last edited by Sir Cumference; 11 months ago
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11 months ago
#3
(Original post by Notnek)

The rod is smoothly hinged at A which means it can swing from left to right. The force P is holding it in equilibirum. If you were to release this force then the rod would swing to the right. Can you picture this now?

I haven't included the reaction force that acts on the rod by the hinge. Often it has unknown magnitude/direction so the best thing to do is show a vertical force V and a horizontal force H acting at A. In this case it may be obvious what the forces are. These two forces may not be required in the working of your question (they're not for part a) anyway).

Any thoughts on how to start?
Wouldn't there simply be only a horizontal force at the hinge as its smoothly hinged
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11 months ago
#4
(Original post by Physikoi)
Wouldn't there simply be only a horizontal force at the hinge as its smoothly hinged
Yeah, but if as Notnek said it doesn't matter to find P
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11 months ago
#5
(Original post by Physikoi)
Wouldn't there simply be only a horizontal force at the hinge as its smoothly hinged
You're thinking of a wall. A hinge just acts to maintain equilibirum so it will balance out other forces acting on the rod. In this case the hinge force must contain horizontal and vertical components to balance the horizontal force P and the vertical force mg.

For the OPs question (part a) you don't need to worry about the hinge forces.
Last edited by Sir Cumference; 11 months ago
1
11 months ago
#6
(Original post by Notnek)
You're thinking of a wall. A hinge just acts to maintain equilibirum so it will balance out other forces acting on the rod. In this case the hinge force must contain a horizontal and vertical forces to balance the horizontal force P and the vertical force mg.

For the OPs question (part a) you don't need to worry about the hinge forces.
So is there no such thing as a smooth hinge? What does that mean?
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11 months ago
#7
(Original post by bingbong654)
So is there no such thing as a smooth hinge? What does that mean?
A smooth hinge just means that there is no friction when the rod pivots about it. If there was friction from the hinge in your model then the question would become more complicated. In A Level the hinge will always be smooth.
Last edited by Sir Cumference; 11 months ago
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11 months ago
#8
(Original post by Notnek)
A smooth hinge just means that there is no friction when the rod pivots about it. If there was friction in your model then the question would become more complicated. In A Level the hinge will always be smooth.
Oh, so like a frictional torque like how you can rotate yourself on a swivel chair when without it the conservation of angular momentum would mean that you would go nowhere?
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11 months ago
#9
(Original post by bingbong654)
Oh, so like a frictional torque like how you can rotate yourself on a swivel chair when without it the conservation of angular momentum would mean that you would go nowhere?
I'm not sure we're talking about the same thing but it's possible. A smooth hinge provides horizontal/vertical reaction forces but provides no reaction to moment.
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11 months ago
#10
(Original post by Notnek)
I'm not sure we're talking about the same thing but it's possible. A smooth hinge provides horizontal/vertical reaction forces but provides no reaction to moment.
It's okay I get it
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#11
(Original post by Notnek)

The rod is smoothly hinged at A which means it can swing from left to right. The force P is holding it in equilibirum. If you were to release this force then the rod would swing to the right. Can you picture this now?

I haven't included the reaction force that acts on the rod by the hinge. Often it has unknown magnitude/direction so the best thing to do is show a vertical force V and a horizontal force H acting at A. In this case it may be obvious what the forces are. These two forces may not be required in the working of your question (they're not for part a) anyway).

Any thoughts on how to start?
Yep thanks I solved it but 1 more thing. For questions like this(Q4): http://pmt.physicsandmathstutor.com/...%20Edexcel.pdf

I can't seem to understand or visualise why there would be a non zero vertical component and non zero horizontal. So for example, in ladder problems the reaction from the ground would be perpendicular to the ground meaning there's no horizontal component. But in this case, the reaction can be split into components.
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11 months ago
#12
(Original post by dont know it)
Yep thanks I solved it but 1 more thing. For questions like this(Q4): http://pmt.physicsandmathstutor.com/...%20Edexcel.pdf

I can't seem to understand or visualise why there would be a non zero vertical component and non zero horizontal. So for example, in ladder problems the reaction from the ground would be perpendicular to the ground meaning there's no horizontal component. But in this case, the reaction can be split into components.
Try holding a pencil in your hand with your fingers loosely at the top and let it swing from left to right. As it swings, your fingers will produce forces with different directions to prevent the pencil from falling. Your fingers are acting as a hinge. For most cases with a hinge, the reaction that the hinge produces has an unknown magnitude/direction so the best thing to do is just assume the reaction has a vertical/horizontal component. You may find that one of these is 0 in the working if it's not obvious in the question.

A ladder resting against a wall is completely different. The wall will always produce a normal reaction force and there will also be a horizontal force if the wall is rough.

When faced with a hinge, if you just include an unknown V and H force at the hinge then most of the time you should be fine.
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