Math moments

Can someone help me with this question I am not very good at moments? I have drawn a diagram that I think is accurate but I do not know what to do next.
1)A non-uniform rod AB of length 3m and mass 4kg rests horizontally on two supports at C and D where AC = 0.5m and BD = 1m. Given that the centre of mass of the rod is at M where AM = 1.75m calculate the magnitudes of the reactions at the supports.

Reply 1

mqb2766

Original post by Martha Jones

You have two unknowns (reactions) to calculate and could take moments about points and/or balance forces vertically. Try writing down the various equations and see which look the most useful and upload?

Note if you take moments about a point where an unknown force acts, then it is effectively eliminated from that equation.

Reply 2

Martha Jones

Do you mean like take moments at C for instance and then use the answer to work out D?

Reply 3

mqb2766

Original post by Martha Jones

Do you mean like take moments at C for instance and then use the answer to work out D?

Sure, and the other way round would be the easiest way to work out the two reactions.

Reply 4

Martha Jones

Does this look right so far?

Reply 5

mqb2766

Original post by Martha Jones

Does this look right so far?

Looks about right. To get the two reactions either

•

Take moments about C and D (or M)

•

Take moments about C (or D or M) and use the Rc+Rd=4g.

Either is fine.

(edited 8 months ago)

Is this correct?

Original post by Martha Jones

Is this correct?

Your diagram is a bit confusing with respect to the com and its not 0.5m from D (thats the midpoint, so the com if the rod was uniform).

Also, often you leave the forces as multiples of "g", but it can be question/answer dependent.

(edited 8 months ago)

Reply 8

Martha Jones

If M is the midpoint and the whole thing is 3m then would wherever M is not be 1.5m from A or B

Reply 9

mqb2766

Original post by Martha Jones

If M is the midpoint and the whole thing is 3m then would wherever M is not be 1.5m from A or B

The question states the rod is non-uniform and AM=1.75m.

(edited 8 months ago)

What about now?

Original post by Martha Jones

What about now?

Looks about right but as before Id have left them in terms of g so 2/3 g and 3 1/3 g.

Note an additional way to check (or even do the question) would be to "take moments" about M, but as there are only the two reactions involved, the force (reaction) ratio must be equal to the distance ratio so 1.25 : 0.25 or 5:1. So one reaction is 5 times the other or equivalently one reaction is 1/6 (4g) and the other is 5/6 (4g)

(edited 8 months ago)

Reply 12

Martha Jones

Original post by mqb2766

Looks about right but as before Id have left them in terms of g so 2/3 g and 3 1/3 g.
Note an additional way to check (or even do the question) would be to "take moments" about M, but as there are only the two reactions involved, the moment ratio must be equal to the distance ratio so 1.25 : 0.25 or 5:1. So one reaction is 5 times the other or equivalently one reaction is 1/6 (4g) and the other is 5/6 (4g)