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-   -   Physics (Current and Parallel Forces) (https://www.askmehelpdesk.com/showthread.php?t=7807)

  • Feb 15, 2005, 07:36 PM
    mdcstudent
    Physics (Current and Parallel Forces)
    Hi HelpDesk:

    Please I need help how to figure out these problems:

    1) An auto engine of mass 295Kg is located 1.0 m from one end of a 4.0
    M work-bench. If the uniform bench has a mass of 45.0Kg, what weight
    Must each end of the bench support?

    2) There is a graphic like:

    _____|_________________|_____
    10.m |A 200m |B 1.0m

    In the graphic shows 4.0m long, weighs 1550N and is made of uniform material. A weight of 245N hangs 1.00m from the end. Find the tension of each support cable.


    Any solutions or feed backs it will be appreciated it

    Thanks

    James D.
    [email protected]
  • Sep 20, 2009, 10:46 PM
    kokikiko

    my ans for number one problem:
    F1=243.75
    F2=96.25
  • Sep 20, 2009, 10:51 PM
    kokikiko

    here's my solution for problem number 1
    F1(4m)=295kg(3m)+45(2m)
    F1=295(3)+45(2) divided by 4

    F1=243.75

    for F2:

    F1+F2=F3+F4
    F2=295+45-243.75
    F2=96.25
  • Sep 21, 2009, 08:30 AM
    Unknown008

    Well... what I got for 1) is:

    Since the mass of the bench is 45 kg, it exerts a force of 450 N (taking acceleration due to gravity 10 m/s^2)

    So, both ends will firstly support 450/2 = 225 N

    Ok, now the engine.

    Take the pivot to be at the left end of the bench and find the moments. Let's assume the engine is 1 m from the left end.

    Clockwise moment = Anticlockwise moment
    (2950 x 1) = (F x 4)
    F = 2950/4 = 737.5 N

    So, total weight on the right end support = 737.5 + 225 = 962.5 N

    Now, weight of engine on the left is given by 2950-737.5 = 2212.5 N

    Total weight on left support = 2212.5+225 = 2437.5 N

    Weights: 962.5 N and 2437.5 N

    Now try the second one, using the same principle.

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