Fross877
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Take moments about X to determine the size of the vertical force F acting on the tyre at Y.

Why does the force at Y act anticlockwise? I thought it would be in the same direction as the force at X?
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Joinedup
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(Original post by Fross877)
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Take moments about X to determine the size of the vertical force F acting on the tyre at Y.

Why does the force at Y act anticlockwise? I thought it would be in the same direction as the force at X?
You're taking moments around point X

The force acting on the bike at X has no moment around X (cos it passes through X it's perpendicular distance from X is zero)

I think you probably need to revise moments
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Eimmanuel
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(Original post by Fross877)
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Take moments about X to determine the size of the vertical force F acting on the tyre at Y.

Why does the force at Y act anticlockwise? I thought it would be in the same direction as the force at X?
Same direction of force does not imply same moment of force.

Think in terms of a see-saw. Two people sitting at the two ends (opposite to each other) of the plank, the weight of the two people are pointing downward. But the moment of weight about the pivot is different.

Again think in terms of a couple. If you don't know what is couple look at the link below.
http://physicsnet.co.uk/a-level-phys...anics/moments/

The two forces that are acting on in a couple system, are in opposite but they gives rise to same moment of force.

Like what Joinedup has said, please revise back moment of force.
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Fross877
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(Original post by Eimmanuel)
Same direction of force does not imply same moment of force.

Think in terms of a see-saw. Two people sitting at the two ends (opposite to each other) of the plank, the weight of the two people are pointing downward. But the moment of weight about the pivot is different.

Again think in terms of a couple. If you don't know what is couple look at the link below.
http://physicsnet.co.uk/a-level-phys...anics/moments/

The two forces that are acting on in a couple system, are in opposite but they gives rise to same moment of force.

Like what Joinedup has said, please revise back moment of force.
I mean the force at Y looking at the diagram should be acting down? It doesn't make sense to me that the force at Y is acting up.
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Eimmanuel
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(Original post by Fross877)
I mean the force at Y looking at the diagram should be acting down? It doesn't make sense to me that the force at Y is acting up.
There are two forces at Y. The road exerts a force on the tyre and the tyre exerts a force on the road.
Which are you referring to?
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Fross877
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(Original post by Eimmanuel)
There are two forces at Y. The road exerts a force on the tyre and the tyre exerts a force on the road.
Which are you referring to?
I just assumed that the force at Y was the weight of the tyre? But reading the question again, is the force at Y the reaction force? If so then it would make sense that it is anticlockwise.
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Eimmanuel
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(Original post by Fross877)
I just assumed that the force at Y was the weight of the tyre? ...
No

(Original post by Fross877)
...But reading the question again, is the force at Y the reaction force?...
Can be more specific about the reaction force. Not sure what are you referring to.
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Fross877
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(Original post by Eimmanuel)
No



Can be more specific about the reaction force. Not sure what are you referring to.
The normal force?
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Eimmanuel
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(Original post by Fross877)
The normal force?
I am assuming that you are talking about action-reaction force based on Newton's 3rd law.

The normal force is NOT the reaction force to the weight of the tyre as it is acting on the same body.
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(Original post by Fross877)
The normal force?
The normal force is what the question is expecting you to use.
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(Original post by Eimmanuel)
I am assuming that you are talking about action-reaction force based on Newton's 3rd law.

The normal force is NOT the reaction force to the weight of the tyre as it is acting on the same body.
(Original post by Eimmanuel)
The normal force is what the question is expecting you to use.
Thank you, for taking the time to help, I also just researched normal and reaction forces and what I meant to say at the beginning, was the normal force and not the reaction force. Originally I didn't know the difference.
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