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    A car of mass 920kg is driven over a hump backed bridge at a speed of 18 ms. The road surface of the bridge forms a circular arc of radius 50m.a) calculate the push of the car on the road.my answer : 5962 (rounded)b) what is the greatest speed at which the car may be driven over the bridge if the wheels are not to lose contact with the road?my answer: 22.2 msI used the equation a = v^2/r and used acceleration of gravity If I get anything wrong can you guys tell me where I did go wrong? Thanks in advance!
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    (Original post by Zevo)
    A car of mass 920kg is driven over a hump backed bridge at a speed of 18 ms. The road surface of the bridge forms a circular arc of radius 50m.a) calculate the push of the car on the road.my answer : 5962 (rounded)b) what is the greatest speed at which the car may be driven over the bridge if the wheels are not to lose contact with the road?my answer: 22.2 msI used the equation a = v^2/r and used acceleration of gravity If I get anything wrong can you guys tell me where I did go wrong? Thanks in advance!

    How did you arrive at the answer 5962N?
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    (Original post by Stonebridge)
    How did you arrive at the answer 5962N?
    I got it from the equation : F = mv^2 / r
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    (Original post by Zevo)
    I got it from the equation : F = mv^2 / r

    At the top of the curve (the question doesn't actually specify where on the curve) there are two forces on the car
    • it's weight downwards (mg)
    • the reaction force of the road upwards (R)

    The centripetal force, mv² / r is downwards and is the resultant of those two forces.

    So

    mg - R = mv² / r

    You need to find R. This is the force of the road on the car. By Newton's 3rd law this is also the force of the car on the road.
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    (Original post by Stonebridge)
    At the top of the curve (the question doesn't actually specify where on the curve) there are two forces on the car
    • it's weight downwards (mg)
    • the reaction force of the road upwards (R)

    The centripetal force, mv² / r is downwards and is the resultant of those two forces.

    So

    mg - R = mv² / r

    You need to find R. This is the force of the road on the car. By Newton's 3rd law this is also the force of the car on the road.
    Okay I see. I shall do that and post the answer!
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    http://www.thestudentroom.co.uk/show....php?t=2568965

    That explanation Stonebridge gave to me that day is still stuck with me today, treat C, R and W as vectors, give them a direction and solving it becomes simple!

    Also the way I think about it; resultant force=centripetal force. Only R and W affect it.


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    (Original post by Jaydude)
    http://www.thestudentroom.co.uk/show....php?t=2568965

    That explanation Stonebridge gave to me that day is still stuck with me today, treat C, R and W as vectors, give them a direction and solving it becomes simple!

    Also the way I think about it; resultant force=centripetal force. Only R and W affect it.


    Posted from TSR Mobile

    Thanks Jaydude. I knew this was out there somewhere but couldn't find it.
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    Okay I tried doing it the way you have told me so I calculated the centripetal force which gave me 5961.6N and the force that that which gravity pulls the car to the ground and it gave me 9025.2N. I subtracted the centripetal force from the force at which the gravity is pulling the car to give me 3063.6N

    for the speed part I calculated that the greatest speed at which the car can go is 12.9ms and I did that by re-arranging the centripetal force equation substituted 3063.6N and the rest.
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    (Original post by Zevo)
    Okay I tried doing it the way you have told me so I calculated the centripetal force which gave me 5961.6N and the force that that which gravity pulls the car to the ground and it gave me 9025.2N. I subtracted the centripetal force from the force at which the gravity is pulling the car to give me 3063.6N
    That's the correct value for the force. Give the answer to 3 significant figures.

    for the speed part I calculated that the greatest speed at which the car can go is 12.9ms and I did that by re-arranging the centripetal force equation substituted 3063.6N and the rest.
    No. The maximum speed is where the wheels lose contact with the ground. At this point R=0
    In that case the centripetal force = mg as the value of R in the formula is zero.
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    (Original post by Stonebridge)
    That's the correct value for the force. Give the answer to 3 significant figures.



    No. The maximum speed is where the wheels lose contact with the ground. At this point R=0
    In that case the centripetal force = mg as the value of R in the formula is zero.
    Okay I think I've got it now. I used the same rearrangement to find v from centripetal force equation. substituted 920 x g into it instead of 3063.6 because as you said you can only get the greatest speed if R=0 therefore I need to make W - C = 0. So anyway I got v = 22.15ms
    I substituted that into C equation to check if my W - C = 0
    and it did.
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    (Original post by Zevo)
    Okay I think I've got it now. I used the same rearrangement to find v from centripetal force equation. substituted 920 x g into it instead of 3063.6 because as you said you can only get the greatest speed if R=0 therefore I need to make W - C = 0. So anyway I got v = 22.15ms
    I substituted that into C equation to check if my W - C = 0
    and it did.
    Yes that's correct now.
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    Thank you for help.
 
 
 
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