CIE M1 help!Watch

#1
Anyone please help me in solving the third part to the attached question. This thing is which vertical forces would I take equal to obtain the reaction force, and which horizontal forces would I take equal to get frictional component. Plus, please explain why, and how we took them equal. Many people weren't able to answer it, so please if anyone can do it, then please reply. I'll be very thankful for the answer.

Edit: See the third post to look how did I resolve the forces.
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8 years ago
#2
The ring is smooth so the tension in BR and AR are equal.
Have you done a force diagram for the bead?
Can you resolve the tension T horizintally and vertically?
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#3
See the attachment.
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8 years ago
#4
To find the tension T, resolve vertically the forces acting on the ring:
Tcos60 + Tcos60 = 0.5g, hence T=5N (g=10)
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#5
(Original post by vc94)
To find the tension T, resolve vertically the forces acting on the ring:
Tcos60 + Tcos60 = 0.5g, hence T=5N (g=10)
Yeah, I have already found the T. I'm just not able to find R.
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8 years ago
#6
On B:
resolve horizontally to get F=Tsin(60) = 4.33
resolve vertically to get R = 0.3g + Tcos(60) = 5.5
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#7
(Original post by vc94)
On B:
resolve horizontally to get F=Tsin(60) = 4.33
resolve vertically to get R = 0.3g + Tcos(60) = 5.5
Well, that's the statement of mark scheme. You saw my previous attachment in which forces were resolved, right? And in that R and Tcos(60) are in same direction and both opposite to the weight of bead, i.e 0.3g. Don't we take the opposite directed forces equal?
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8 years ago
#8
Tcos60 is acting downwards in the same direction as the weight of the bead, i.e. opposing the normal reaction R.
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#9
So have I drawn the Vertical component of T wrong? =s Wasn't it like that when some wire was attached to a wall or fixed point, the Tension would be in direction towards wall? And applying head to tale rule, the Tcos(60) occurs in same direction as R? Please see my previous attachment again, and if it's wrongly resolved, then please explain how? Thanks~
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8 years ago
#10
Looking at your diagram, friction F arrow should be pointing left (opposing potential motion).
Tension T is acting on the bead B, so an arrow should be directed away from B.
1
#11
First, please explain why frictional force should be at left? Secondly as wire is also attached to A, so would tension be directed towards A? And as it's one wire, so that's why the tension in BR away from B? =O
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8 years ago
#12
The potential motion of the bead is to the right, since friction opposes motion F must be directed to the left.

Your 3 forces acting on A are correct. However, from Newton's third law (every force has an equal and opposite reaction?), the tension in the string acting on B cannot be zero!
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#13
What is potential motion, and how does question tells that potential motion of B is towards right? o_O
And so, as T is acting on B, so would B exert an equal and opposite T on rope, and we'll take horizontal and vertical components of 'that' force to find F and R, right?
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8 years ago
#14
The system is in equilibrium so the bead is on the point of moving to the right.
Since T is acting on B, you take horizontal and vertical components to find F and R.
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#15
Anyways, thanks a lot for your explanation. But I am not able to understand that why the bead will move towards right, it can even move towards left, if system is in equilibrium?! =(
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8 years ago
#16
(Original post by Zishi)
Anyways, thanks a lot for your explanation. But I am not able to understand that why the bead will move towards right, it can even move towards left, if system is in equilibrium?! =(
The weight of the ring downwards will encourage motion of the bead to the right.
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#17
Hey, thanks a lot again! I was being such a dumb! o_O I've finally understood that! =)
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