Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    0
    ReputationRep:
    A train on a fair ground is initially stationary before it descends through a height of 45 m into a dip that has a radius of curvature of 78 m.

    Calculate the speed of the train at the bottom of the dip? Answer = 29.7 ms^-1. I'm fine with this part.

    Calculate centripetal force at the bottom? Answer = 11.3 ms^-2. Fine with this as well.

    Calculate the extra support force on a person of weight 600 N in the train. Answer = 690 N. I can't quite get to the answer, these are my workings:

    S = support force

    S - mg = (mv^2) / r

    S = (mv^2) / r + mg

    S = ((600/9.81)29.7^2) / 78 + 600

    S = 689.84 + 600

    Where am I getting the extra 600 N from? Thanks help would be appreciated.
    Offline

    20
    ReputationRep:
    (Original post by mtig)
    A train on a fair ground is initially stationary before it descends through a height of 45 m into a dip that has a radius of curvature of 78 m.

    Calculate the speed of the train at the bottom of the dip? Answer = 29.7 ms^-1. I'm fine with this part.

    Calculate centripetal force at the bottom? Answer = 11.3 ms^-2. Fine with this as well.
    isn't that an acceleration?

    Calculate the extra support force on a person of weight 600 N in the train. Answer = 690 N. I can't quite get to the answer, these are my workings:

    S = support force

    S - mg = (mv^2) / r

    S = (mv^2) / r + mg

    S = ((600/9.81)29.7^2) / 78 + 600

    S = 689.84 + 600

    Where am I getting the extra 600 N from? Thanks help would be appreciated.
    was there a diagram?

    I'm guessing the track curves vertically upwards and you're asked to calculate the extra force required to accelerate the passenger upwards I.E. the additional force on top of the force (mg) that'd be there if they were just sitting stationary.

    (689.84 IS 690 to 3sf)
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by Joinedup)
    isn't that an acceleration?


    was there a diagram?

    I'm guessing the track curves vertically upwards and you're asked to calculate the extra force required to accelerate the passenger upwards I.E. the additional force on top of the force (mg) that'd be there if they were just sitting stationary.

    (689.84 IS 690 to 3sf)

    The second part of the question asked for the centripetal acceleration not the force, I typed the question wrong sorry.

    Yes there is a diagram, it's descending downwards like the shape of the diagram below.

    Thanks.


 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    Would you like to hibernate through the winter months?
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

    Quick reply
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.