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27 May, 13:04

Assume the following values: d1 = 0.880 m, d2 = 1.11 m, d3 = 0.560 m, d4 = 2.08 m, F1 = 510 N, F2 = 306 N, F3 = 501 N, F4 = 407 N, and MA = 1504 N⋅m. Express the Cartesian components of the resultant force and the couple moment in newtons and newton-meters to three significant figures separated by commas.

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  1. 27 May, 16:30
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    = 2630.6 N. m

    Explanation:

    (FR) x = ΣFx = - F4 = - 407 N

    (FR) y = ΣFy = -F1-F2 - F3 = - 510 - 306 - 501 = - 1317 N

    (MR) B = ΣM + Σ (±Fd)

    = MA + F1 (d1 + d2) + F2d2 - F4d3

    = 1504 + 510 (0.880+1.11) + 306 (1.11) - 407 (0.560)

    = 2630.64 N. m (counterclockwise)
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