Turn on thread page Beta
    • Thread Starter
    Offline

    0
    ReputationRep:
    Hey, I have no idea how to do this:

    A lorry of weight 40000N crosses a 20m long bridge of weight 100000N. What are the forces supplied by each bridge support when it is 5m from one support?


    Thanks
    Offline

    1
    ReputationRep:
    Total force 'down' must equal total force up

    20/5 = 4 therefore 4 supports.


    40000 + 100000 = 140000

    140000/4 = 35000 each support.
    Offline

    0
    ReputationRep:
    Think of the principle of moments.
    Offline

    0
    ReputationRep:
    (Original post by Hancock orbital)
    Total force 'down' must equal total force up

    20/5 = 4 therefore 4 supports.


    40000 + 100000 = 140000

    140000/4 = 35000 each support.
    I think you misunderstood, unless I did :p:

    Two supports, one at each end. The truck is 5 m from one end.

    OP, think torque. Net torque and net force must be zero for the system to remain in equilibrium.
    Offline

    1
    ReputationRep:
    (Original post by D-Day)
    I think you misunderstood, unless I did :p:

    Two supports, one at each end. The truck is 5 m from one end.

    OP, think torque. Net torque and net force must be zero for the system to remain in equilibrium.
    yeah - four supports does seem strange for a 20m bridge. duh.
    • Thread Starter
    Offline

    0
    ReputationRep:
    I'm not sure if this is right but i got 80,000N and 87,000N.....
    Offline

    9
    ReputationRep:
    Got to add up to 140000 - so no its not right
    • Thread Starter
    Offline

    0
    ReputationRep:
    Hmm....

    Not very sure how to do it then...

    i did 40kNx5m=200KNm+ (100knx10m)= 1200kNm= anticlockwise momentum

    clockwise= 20mx= 1200kNm
    clockwise=1200kNm/20m=60,000N

    then i did the 140,000-60,000= 80,000N

    and then i did something similar to the other side and got 87,000N....


    So then is the answer 60,000N and 80,000N, if it has to add up to 140kN?
    • Thread Starter
    Offline

    0
    ReputationRep:
    HELP ~ What have i done wrong?
    Offline

    0
    ReputationRep:
    Net torque is zero. Assume that the pivot point is on one of the supports. This eliminates an unknown.

    (40,000 N)(5 m)+(100,000 N)(10 m)=(F2)(20 m)

    Now move the pivot point to the other support.

    (100,000 N)(10 m)+(40,000 N)(15 m)=(F1)(20 m)
    • Thread Starter
    Offline

    0
    ReputationRep:
    Well so....
    F2=600,000N
    F1=800,000N

    Then is that the answer?
    Offline

    0
    ReputationRep:
    Check your work. Those are off by a factor of ten. Should be 80,000 and 60,000.
    • Thread Starter
    Offline

    0
    ReputationRep:
    Ok, thanks alot
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: November 20, 2008

3,036

students online now

800,000+

Exam discussions

Find your exam discussion here

Poll
Should predicted grades be removed from the uni application process

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

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