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.
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.
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Original post
by Martha JonesCan 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.
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.
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.
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Do you mean like take moments at C for instance and then use the answer to work out D?
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by Martha JonesDo 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.
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Reply 5
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by Martha JonesLooks 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.
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Reply 7
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Original post
by Martha JonesYour 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.
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If M is the midpoint and the whole thing is 3m then would wherever M is not be 1.5m from A or B
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Original post
by Martha JonesIf 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.
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Reply 11
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Original post
by Martha JonesLooks 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 2 years ago)
Reply 12
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Original post
by mqb2766Looks 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)
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)
Thank you for your help
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