M2 Centre of mass / moments question

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Jian17
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Hello, could someone help me to see how the force is acting on the prism on part ii of this question? https://gyazo.com/f8f20c8cc135fae215e854f9975699a6

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mqb2766
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The force is applied at a tangent point on the arc DE, where the angle is 60 degrees to the vertical (30 to the horizontal).
Imagine the dashed radius a bit until it is 60 degrees with the horizontal. The point on the circumference is the point where the force is acting and its acts in a tangential direction.
(Original post by Jian17)
Hello, could someone help me to see how the force is acting on the prism on part ii of this question? https://gyazo.com/f8f20c8cc135fae215e854f9975699a6

Thanks
Last edited by mqb2766; 10 months ago
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Jian17
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(Original post by mqb2766)
The force is applied at a tangent point on the arc DE, where the angle is 60 degrees to the vertical (30 to the horizontal).
Imagine the dashed radius a bit until it is 60 degrees with the horizontal. The point on the circumference is the point where the force is acting and its acts in a tangential direction.
what does tangential direction mean, that it acts along the tangent?
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mqb2766
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Yes, that looks right. Imagine that the surface is very rough so no slippage and the force is acting along the surface.
(Original post by Jian17)
what does tangential direction mean, that it acts along the tangent?
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Jian17
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(Original post by mqb2766)
Yes, that looks right. Imagine that the surface is very rough so no slippage and the force is acting along the surface.
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done thanks, I got some general questions.

In part i since I was told the object is about to topple, I would assume it topples about A cause of the shape the prism is? also the way I did that question was doing centre of masses about A, and then since centre of mass is at A, my cofm would be =0 Name:  image-819cbc1f-33f6-4f9d-9d87-7a216e2729c777718237-compressed.jpg.jpeg
Views: 20
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Afterwards, in part two, does the weight of the object (cofm) still act on A? cause that's the way I did it by doing moments about B. I'm a bit confused would be glad to get this clear.
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mqb2766
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For part i) you must consider the horizontal component of the COM above A.
For part ii) you're told its in equilibrium on the point of rotating about B. The COM will still be vertically above A, but its about to rotate about B.
So it looks like you're right in both cases.
(Original post by Jian17)
Name:  image-a850265f-8b67-4734-89b6-1d9b58a8a0762026312313-compressed.jpg.jpeg
Views: 20
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done thanks, I got some general questions.

In part i since I was told the object is about to topple, I would assume it topples about A cause of the shape the prism is? also the way I did that question was doing centre of masses about A, and then since centre of mass is at A, my cofm would be =0 Name:  image-819cbc1f-33f6-4f9d-9d87-7a216e2729c777718237-compressed.jpg.jpeg
Views: 20
Size:  37.7 KB

Afterwards, in part two, does the weight of the object (cofm) still act on A? cause that's the way I did it by doing moments about B. I'm a bit confused would be glad to get this clear.
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Jian17
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So the COM of a shape never changes but the point of toppling can basically?
And what if I was told initially that if topples about B, how would I know the COM lies on A?
Sorry for being a pain , just wanna make sure I understand
(Original post by mqb2766)
For part i) you must consider the horizontal component of the COM above A.
For part ii) you're told its in equilibrium on the point of rotating about B. The COM will still be vertically above A, but its about to rotate about B.
So it looks like you're right in both cases.
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mqb2766
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The COM is an intrinsic property of the body. It is not affected by the forces that are applied to the body. The point of rotation will depend on the forces that are applied to the body. In ii) the point of tipping changes because a force is applied which changes the equilibrium position. The body has not moved, so it is still vertically above A.

All the mass of the object lies to the right of B (horizontally), so the COM will lie to the right of B, so either its in equilibirim or it topplies about A when no force is applied. The COM is calculated by considering the moment due solely to the body's mass on each side of the COM.

(Original post by Jian17)
So the COM of a shape never changes but the point of toppling can basically?
And what if I was told initially that if topples about B, how would I know the COM lies on A?
Sorry for being a pain , just wanna make sure I understand
Last edited by mqb2766; 10 months ago
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