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M1 Moments Questions

Hiya there,

I'm doing a Mechanics M1 question but I can't seem to get my head around it and would like some help or method on how to go around it since i've been stuck on it for some time now.

The question is:
A uniform bar AB of length 2.8 m and weight 80 N has loads of 20 N and 40 N attached at A and B respectively. If the bar balances in a horizontal position when smoothly supported at C, find the distance of C from A.

This is my current diagram (If its correct?)
Thanks:smile:
(edited 9 years ago)
m1.png1.8KB
Original post by Choco-Omar
Hiya there,

I'm doing a Mechanics M1 question but I can't seem to get my head around it and would like some help or method on how to go around it since i've been stuck on it for some time now.

The question is:
A uniform bar AB of length 2.8 m and weight 80 N has loads of 20 N and 40 N attached at A and B respectively. If the bar balances in a horizontal position when smoothly supported at C, find the distance of C from A.

This is my current diagram (If its correct?)
Thanks:smile:


You do not have the support C in your diagram
Reply 2
Avatar for Choco-Omar
Choco-Omar
OP
2
Original post by TenOfThem
You do not have the support C in your diagram


Yeah, I think it should be like this:
m1.png2.2KB
Original post by Choco-Omar
Yeah, I think it should be like this:


Yes
You need to label the distance AC as x or something .... Then take some moments
Reply 4
Avatar for Choco-Omar
Choco-Omar
OP
2
Original post by TenOfThem
Yes
You need to label the distance AC as x or something .... Then take some moments


Right.. So if the distance from AC is x, then the distance from C to the centre is (1.4-x) .

The moments calculations would be around C correct?
So it would be..
20(x) + 80(1.4-x) + 40(2.8-x) ?

Sorry, Im kind of getting confused with which calculations to be taken.
Original post by Choco-Omar
Right.. So if the distance from AC is x, then the distance from C to the centre is (1.4-x) .

The moments calculations would be around C correct?
So it would be..
20(x) + 80(1.4-x) + 40(2.8-x) ?

Sorry, Im kind of getting confused with which calculations to be taken.


Yes for lengths
Yes, take moments about C
You need to consider direction for your moments and you need an equation
Reply 6
Avatar for keeno25
keeno25
17
I am still stuck on how to do this question. Can anyone help?
Reply 7
Avatar for niam004
niam004
2
so as moments clockwise=moments anti-clockwise
you get this equation:

20x=80(1.4-x) 40(2.8-x)

expand and simplify to get
140x=224
x= 1.6m
(edited 3 years ago)
Reply 8
Avatar for davros
davros
16
Original post by niam004
so as moments clockwise=moments anti-clockwise
you get this equation:

20x=80(1.4-x) 40(2.8-x)


This is a 6-year-old thread and the last post was 6 months ago - I suspect they're sorted now :biggrin:
Reply 9
Avatar for niam004
niam004
2
Original post by davros
This is a 6-year-old thread and the last post was 6 months ago - I suspect they're sorted now :biggrin:

Other people may be looking for the same question as there was no clear solution posted. As you can see 8 months ago someone was still looking for a solution.
Reply 10
Avatar for davros
davros
16
Original post by niam004
Other people may be looking for the same question as there was no clear solution posted. As you can see 8 months ago someone was still looking for a solution.

Usually better to start your own thread in those circumstances - otherwise people start quoting TSR users who've departed many years ago :biggrin:

Also, it's not appropriate to post full solutions - check out the sticky post at the top of the forum for posting guidelines :smile:

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