Risermax
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#1
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#1
Can someone help me with this mechanics question, I will reply to this with a pic of the question.
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Risermax
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#2
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#2
(Original post by Risermax)
Can someone help me with this mechanics question, I will reply to this with a pic of the question.
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mqb2766
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#3
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#3
(Original post by Risermax)
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A couple consists of a pair of equal/opposite parallel forces which produce pure rotation (no translation), so what do you think?
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Risermax
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#4
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For (i) I know that the resultant force = 0 but then I have to find the moment aswell, I know what the forces are but how about the distance if I was to take clockwise moments about A.
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mqb2766
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#5
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#5
(Original post by Risermax)
For (i) I know that the resultant force = 0 but then I have to find the moment aswell, I know what the forces are but how about the distance if I was to take clockwise moments about A.
If you have to find the moment, find the point about which it will rotate and the moment due to one of the forces, then double it.
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Risermax
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#6
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I'm a bit confused, can you please reply with a step by step process of how you would do it.
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mqb2766
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#7
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#7
(Original post by Risermax)
I'm a bit confused, can you please reply with a step by step process of how you would do it.
If they form a couple, the point of rotation will be the centre of the two points. Then just find the moment as usual:
force * perpendicular distance
or
perpendicular force * distance
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Risermax
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#8
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(Original post by mqb2766)
If they form a couple, the point of rotation will be the centre of the two points. Then just find the moment as usual:
force * perpendicular distance
or
perpendicular force * distance
Is it possible you can draw your thought process out for me please
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mqb2766
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#9
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#9
(Original post by Risermax)
Is it possible you can draw your thought process out for me please
How about uploading your pic / working for the forces for one of the questions?
Make sure you join the dots with a line and mark on the centre.

The easiest way to work out the moment due to one force is to consider the moments due to each of the x and y force components, then combine (add) them.
Last edited by mqb2766; 3 months ago
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Risermax
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#10
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#10
(Original post by mqb2766)
How about uploading your pic / working for the forces for one of the questions?
Make sure you join the dots with a line and mark on the centre.

The easiest way to work out the moment due to one force is to consider the moments due to each of the x and y force components, then combine (add) them.
I'm sorry I'm still confused, if this was a beam or plank question this would've been way easier, but I dont think I can apply the same thing here because I don't even know the perpendicular distance.
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mqb2766
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#11
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#11
(Original post by Risermax)
I'm sorry I'm still confused, if this was a beam or plank question this would've been way easier, but I dont think I can apply the same thing here because I don't even know the perpendicular distance.
Join the two dots (where the force is applied) with a line
The centre is the point of rotation
Then consider the moment due to the horizontal forces. So horizontal force * perpendicular distance, where the perpendicular distance is 1/2 the vertical distance between the two points.
Then consider the moment due to the vertical forces. So vertical force * perpendicular distance, where the perpendicular distance is 1/2 the horizontal distance between the two points.
Add them, making sure their signs (directions) are appropriate.

Note that in each case you have two vertical and two horizontal forces, so you could simplify the calculation slightly, by considering the distance (horizontal and vertical) between each point.

There is nothing complicated in sketching the above, have a go.
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Risermax
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#12
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#12
(Original post by mqb2766)
Join the two dots (where the force is applied) with a line
The centre is the point of rotation
Then consider the moment due to the horizontal forces. So horizontal force * perpendicular distance, where the perpendicular distance is 1/2 the vertical distance between the two points.
Then consider the moment due to the vertical forces. So vertical force * perpendicular distance, where the perpendicular distance is 1/2 the horizontal distance between the two points.
Add them, making sure their signs (directions) are appropriate.

Note that in each case you have two vertical and two horizontal forces, so you could simplify the calculation slightly, by considering the distance (horizontal and vertical) between each point.

There is nothing complicated in sketching the above, have a go.
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This is what I drew after I found the centre im confused about the perpendicular distance. On the answers it says 6×2-3×4=0 which proves that it's not a couple and it's in equilibrium since the total clockwise moments about A is zero and I'm guessing 6 and 3 is the distance but I dont get how they got that.
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mqb2766
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#13
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#13
The forces are applied at (2,5) and (-1,-1). The center is at (0.5,2) when you join those two points.

The force at (2,5) is (2,4). The horizontal force 2 provides a clockwise moment and the vertical force 4 provides an anticlocikwise moment. The moment about the center for (2,4) is therefore
3*2 - 1.5*4
as 3 is the vertical distance from the centre and 1.5 is the horizontal distance from the centre. You double it because its a "couple" so
6*2 - 3*4
which as you say is zero so its not a couple as the moment is zero.

Note the above is a verbose way of calculating it. The quicker way is to just calculate the moment of one point about the other, so the distances are doubled and you can write down
6*2 - 3*4
where youre taking the moment of the force (2,4) at (2,5) about the point (-1,-1). The horizontal and vertical forces are 2 and 4, and the vertical and horizontal perpendicular distances are 6 and 3, respectively.
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Risermax
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#14
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#14
(Original post by mqb2766)
The forces are applied at (2,5) and (-1,-1). The center is at (0.5,2) when you join those two points.

The force at (2,5) is (2,4). The horizontal force 2 provides a clockwise moment and the vertical force 4 provides an anticlocikwise moment. The moment about the center for (2,4) is therefore
3*2 - 1.5*4
as 3 is the vertical distance from the centre and 1.5 is the horizontal distance from the centre. You double it because its a "couple" so
6*2 - 3*4
which as you say is zero so its not a couple as the moment is zero.

Note the above is a verbose way of calculating it. The quicker way is to just calculate the moment of one point about the other, so the distances are doubled and you can write down
6*2 - 3*4
where youre taking the moment of the force (2,4) at (2,5) about the point (-1,-1). The horizontal and vertical forces are 2 and 4, and the vertical and horizontal perpendicular distances are 6 and 3, respectively.
Oh ok I understand thank you. Also for the (ii) would I have to the long way to do it or can I still use the short way?
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mqb2766
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#15
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#15
(Original post by Risermax)
Oh ok I understand thank you. Also for the (ii) would I have to the long way to do it or can I still use the short way?
Either should be fine.
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Risermax
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#16
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#16
(Original post by mqb2766)
Either should be fine.
Ok thanks but one more thing the moment for ii is 21nm and they did 6×3+3×1 for it but I just don't see it.
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mqb2766
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#17
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#17
(Original post by Risermax)
Ok thanks but one more thing the moment for ii is 21nm and they did 6×3+3×1 for it but I just don't see it.
The first question is a hint about how to do it. If you consider the forces as given, its hard. However, can you reexpress them so that they're in
horizontal + vertical
form as per the first question. Then proceed as per the first question.
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Risermax
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#18
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#18
(Original post by mqb2766)
The first question is a hint about how to do it. If you consider the forces as given, its hard. However, can you reexpress them so that they're in
horizontal + vertical
form as per the first question. Then proceed as per the first question.
The horizontal forces for A are -2i and -i so the resultant horizontal force would be -3i and the vertical forces are -2j and 3j so the resultant force would be j. The horizontal forces for b are i and 2i so the resultant would be 3i and the vertical forces would be -3j and 2j so the resultant force would be -j. Then I know that the perpendicular distance are 6 and 3 but I dont know what to do from there
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mqb2766
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#19
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#19
(Original post by Risermax)
The horizontal forces for A are -2i and -i so the resultant horizontal force would be -3i and the vertical forces are -2j and 3j so the resultant force would be j. The horizontal forces for b are i and 2i so the resultant would be 3i and the vertical forces would be -3j and 2j so the resultant force would be -j. Then I know that the perpendicular distance are 6 and 3 but I dont know what to do from there
When you add the two component forces at the point (2,5) you get (3,-1). Its a displacement of (3,6) from the other point. Both the horizontal and vertical force components produce a clockwise moment so
6*3 + 3*1
which is simply force*perpendicular distance for each component as before.

It would probably be worth redrawing the grid for this question part and explicitly marking on the horizontal and vertical components as well as the perpenedicular distances.
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Risermax
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#20
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#20
(Original post by mqb2766)
When you add the two component forces at the point (2,5) you get (3,-1). Its a displacement of (3,6) from the other point. Both the horizontal and vertical force components produce a clockwise moment so
6*3 + 3*1
which is simply force*perpendicular distance for each component as before.

It would probably be worth redrawing the grid for this question part and explicitly marking on the horizontal and vertical components as well as the perpenedicular distances.
I kind of understand it but why is it not 3×-1 since the resultant is (3,-1).
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