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Circular motion question
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ella sian
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#1
Its 1aiii on this
https://pmt.physicsandmathstutor.com...n%201%20QP.pdf
on the answers it says that the total force has to be at an angle of 15-75 degrees to the horizontal can anyone explain why?
https://pmt.physicsandmathstutor.com...n%201%20QP.pdf
on the answers it says that the total force has to be at an angle of 15-75 degrees to the horizontal can anyone explain why?
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username3249896
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#2
The contact force has 2 components:
(i) the normal component which balances the weight
(ii) frictional component opposing the driving force which causes movement or a tendency to move. In this case, the frictional component opposes the driving force T. The frictional components acts opposite to T.
There is a horizontal and vertical component to the contact force, so it is at an angle to the horizontal.
(i) the normal component which balances the weight
(ii) frictional component opposing the driving force which causes movement or a tendency to move. In this case, the frictional component opposes the driving force T. The frictional components acts opposite to T.
There is a horizontal and vertical component to the contact force, so it is at an angle to the horizontal.
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ella sian
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#3
(Original post by BobbJo)
The contact force has 2 components:
(i) the normal component which balances the weight
(ii) frictional component opposing the driving force which causes movement or a tendency to move. In this case, the frictional component opposes the driving force T. The frictional components acts opposite to T.
There is a horizontal and vertical component to the contact force, so it is at an angle to the horizontal.
The contact force has 2 components:
(i) the normal component which balances the weight
(ii) frictional component opposing the driving force which causes movement or a tendency to move. In this case, the frictional component opposes the driving force T. The frictional components acts opposite to T.
There is a horizontal and vertical component to the contact force, so it is at an angle to the horizontal.
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username3249896
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#4
(Original post by Zebragirl)
Thank you so much also on the question below it says that the max force produced by the engine is 28kn why is only this used for q1bi when ur working out the acceleration with f=ma like wouldnt there be a component of the contact force opposing the forward force?
Thank you so much also on the question below it says that the max force produced by the engine is 28kn why is only this used for q1bi when ur working out the acceleration with f=ma like wouldnt there be a component of the contact force opposing the forward force?
The calculation in (b)(i) ignores any resistive forces such as friction. We only take the force produced by engine (28kN) and ignore any resistive forces.
In reality, resistive forces such as air resistance and friction will decrease the net forward force and this will be part of your answer in (b)(ii)
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