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17 April, 20:10

The safe load, L, of a wooden beam supported at both ends varies jointly as the width, w, the square of the depth, d, and inversely as the length, l. A wooden beam 3in. wide, 6in. deep, and 11ft long holds up 1213lb. What load would a beam 6in. wide, 3in. deep and 12ft long of the same material support? (Round off your answer to the nearest pound.)

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  1. 17 April, 20:31
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    L' = 555.95 lb

    Explanation:

    Analyzing the given conditions in the question, we get

    The safe load, L is directly proportional to width (w) and square of depth (d²)

    also,

    L is inversely proportional length (l) i. e L = k/l

    combining the above conditions, we get an equation as:

    L = k (wd²/l)

    now, for the first case we have been given

    w = 3 in

    d = 6 in

    l = 11 ft

    L = 1213 lbs

    thus,

    1213 lb = k ((3 * 6²) / 11)

    or

    k = 123.54 lbs / (ft. in³)

    Now,

    Using the calculated value of k to calculate the value of L in the second case

    in the second case, we have

    w = 6 in

    d = 3 in

    l = 12 ft

    Final Safe load L' = 123.54 * (6 * 3²/12)

    or

    L' = 555.95 lb
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