A force of 100 N hangs in the center of a cable between two anchor points that are the same height off the ground and 30 m apart horizontall

Question

A force of 100 N hangs in the center of a cable between two anchor points that are the same height off the ground and 30 m apart horizontally. The object creates a sag of 2 m in the cable. Find the force exerted by each end of the cable on its anchor point.

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Khang Minh 6 months 2021-08-04T12:53:03+00:00 1 Answers 9 views 0

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    2021-08-04T12:54:38+00:00

    Answer:

    378.3N

    Explanation:

    According to Newton’s 3rd law, the force exerting on each anchor points would equal to the tension force T on each cable, which is created to support the 100N in the center

    As the system remain balanced, then vertically the tension force would support the 100N

    2T_v = 100

    2Tcos\theta = 100

    where Θ is the angle between the cable and the vertical line intersecting the force F. As it’s 15m halfway between the 2 anchor

    tan\theta = 15/2 = 7.5

    \theta = tan^{-1}7.5 = 1.44 rad

    cos\theta = cos(1.44) = 0.132

    therefore 2T0.132 = 100

    T = 100/(2*0.132) = 378.3 N

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