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1. A metal rod 2m, is lying on a smooth ice rink. The centre of mass of the rod is 0.8m from one end.the rod is picked up by that end, which is held 1m above the ice with a force of 15N. Find the weight of the rod........

2. (Original post by joyoustele)
A metal rod 2m, is lying on a smooth ice rink. The centre of mass of the rod is 0.8m from one end.the rod is picked up by that end, which is held 1m above the ice with a force of 15N. Find the weight of the rod........

Please just stick to posting in the Maths forum for these.

Anyway, have you drawn a diagram for the forces?
3. (Original post by RDKGames)
Please just stick to posting in the Maths forum for these.

Anyway, have you drawn a diagram for the forces?
I thought this was the maths forum?

Yh i drew a diagram, the bit that is confusing me is the bit that says, " the rod is picked up by the end, which is held 1m above the ice, with a force of 15N"
4. (Original post by joyoustele)
I thought this was the maths forum?

Yh i drew a diagram, the bit that is confusing me is the bit that says, " the rod is picked up by the end, which is held 1m above the ice, with a force of 15N"
You posted it in the A-Level forum and I moved it.

Unless I'm being thick, I don't see how this is relevant unless it comes of use in some later part of that Q. That being said, just take the pivot to to be the other end of the rod and solve for the mass.
5. (Original post by RDKGames)
You posted it in the A-Level forum and I moved it.

Unless I'm being thick, I don't see how this is relevant unless it comes of use in some later part of that Q. That being said, just take the pivot to to be the other end of the rod and solve for the mass.
Okay i will try that, here is my diagram, it's missing the 1m height and 0.8 centremass distance, i dont know why
6. (Original post by joyoustele)
Okay i will try that, here is my diagram
Oh I just thought the whole thing was picked up by one hand on one end, not just one end lifted off the surface of the ice, but yes in your scenario the 1m height does make an appearance when you want to find the horizontal distance from the other end of the rod. I.e. it's lifted at some angle theta, which you can work out, hence use this theta to work out the horizontal distance between the right end of the rod to the Mg force.

A different approach is to consider the s.f. between the smaller and larger triangles to work out the side length.
7. (Original post by RDKGames)
You posted it in the A-Level forum and I moved it.

Unless I'm being thick, I don't see how this is relevant unless it comes of use in some later part of that Q. That being said, just take the pivot to to be the other end of the rod and solve for the mass.
It worked, Thank you, i have learned something. I wish to be a genius like you one day.
8. there will be another vertical force at the right hand end where the rod meets the ice.
9. (Original post by joyoustele)
It worked, Thank you, i have learned something. I wish to be a genius like you one day.
Oh, alright then, ignore my last post then!
10. Have i drawn this right? Cyclists sits on a bike,mass of bike is 14kg, [and the line of action of its weight is the perpendicular bisector of OfOr] where Of and Or are the centres of the front and rear wheels respectively. The mass of the cyclists is 80kg and her weight acts in a line through the saddle. Given that the distance OfOr is 1020mm and that the vertical line through the saddle is 240mm in front of Or, Find Magnitude of the force exerted by the ground.

Attachment 698222

RDKGamesthe beaɾ

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