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# Edexcel AS Mechanics 1 watch

1. Seem to be struggling with this question, could anyone help out?
The centre of mass of a non-uniform beam AB, of weight W and length 4m, is at the point G. The beam rests horizontally on two supports X and Y, where AX = 1m and YB = 2m.
A weight W/4 is suspended from the beam at B. Given that the magnitude of the force exerted on the beam by support Y is three times the magnitude of the force exerted on the beam by support X, calculate the distance AG.

Thanks
2. (Original post by Rashina)
Seem to be struggling with this question, could anyone help out?
The centre of mass of a non-uniform beam AB, of weight W and length 4m, is at the point G. The beam rests horizontally on two supports X and Y, where AX = 1m and YB = 2m.
A weight W/4 is suspended from the beam at B. Given that the magnitude of the force exerted on the beam by support Y is three times the magnitude of the force exerted on the beam by support X, calculate the distance AG.

Thanks
What have you tried? Post your working.
3. Where did you get this question from?
4. (Original post by Keira Larkin)
Where did you get this question from?
5. (Original post by Rashina)
Oh I just thought I might be able to find a worked solution if it is from the textbook.
6. (Original post by notnek)
What have you tried? Post your working.
I don't even have an idea of how to do it. Explain me how to do it and I will figure it out.
7. (Original post by Keira Larkin)
Oh I just thought I might be able to find a worked solution if it is from the textbook.
Its from a revision guide which does not have any solution banks or worked solutions.
8. I got these equations so far:

W + W/4 = x + 3x
2W/4 + x = W*z
W(1-z) + 3W/4 = 3x

z is the distance between support y and the centre of mass
9. (Original post by Rashina)
Its from a revision guide which does not have any solution banks or worked solutions.
Oh okay.
10. (Original post by Keira Larkin)
Oh okay.
Anyway thanks for trying to help me
11. (Original post by Rashina)
Anyway thanks for trying to help me
Well, now I want to know how to solve this as well. There are too many variables in my equations. Have you made any progress?
12. (Original post by Keira Larkin)
Well, now I want to know how to solve this as well. There are too many variables in my equations. Have you made any progress?
just calculated the resultant force upwards:
R+3R=(W+W/4)g
Wg= 16x/5
13. (Original post by Rashina)
just calculated the resultant force upwards:
R+3R=(W+W/4)g
Wg= 16x/5

WORKING:

So I don't think you multiply W by g because W is weight, which is a force. Therefore, W= 16x/5
Take moments about A to give the equation:
x + (2 * 3x) = Wz + W
7x = W * (z + 1)
7x = 16x/5 (z + 1)
7x * 5/16x = (z + 1)
19/16 = z
14. (Original post by Keira Larkin)
Yes! How did you find it?
15. (Original post by Rashina)
Yes! How did you find it?
16. (Original post by Keira Larkin)
The book only has final answers so I found this answer on the book. How did you find it?
17. Basically, the trick was not to take moments about support x or y. I took the moments about point A. Obviously not B because there is a force acting on point B. Refer to the second worked example in the M1 textbook in the section about non uniform planks.

When I do tricky moment questions, I just try taking moments about a point and if that doesn't work, I try another point.
18. (Original post by Keira Larkin)

WORKING:

So I don't think you multiply W by g because W is weight, which is a force. Therefore, W= 16x/5
Take moments about A to give the equation:
x + (2 * 3x) = Wz + W
7x = W * (z + 1)
7x = 16x/5 (z + 1)
7x * 5/16x = (z + 1)
19/16 = z
I wrote the working.
19. (Original post by Keira Larkin)
I wrote the working.
Thanks a lot for helping me.
20. (Original post by Rashina)
Thanks a lot for helping me.
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