If blocks A and B of mass 10 kg and 6 kg respectively, are placed on the inclined plane and released, determine the force developed in the l

Question

If blocks A and B of mass 10 kg and 6 kg respectively, are placed on the inclined plane and released, determine the force developed in the link. The coefficients of kinetic friction between the blocks and the inclined plane are mA = 0.1 and mB = 0.3. Neglect the mass of the link.

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Latifah 4 years 2021-09-04T17:48:15+00:00 1 Answers 67 views 0

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  1. taiduc
    0
    2021-09-04T17:50:08+00:00
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    Answer:

    The force developed in the link is 6.36 N.

    Explanation:

    Given that,

    Mass of block A = 10 kg

    Mass of block B = 6 kg

    Coefficients of kinetic friction \mu_{A}= 0.1

    Coefficients of kinetic friction \mu_{B}= 0.3

    Suppose the angle is 30°

    We need to calculate the acceleration

    Using formula of acceleration

    a=\dfrac{m_{A}g\sin\theta+m_{B}g\sin\theta-\mu_{A}m_{A}g\cos\theta-\mu_{A}m_{A}g\cos\theta}{m_{A}+m_{B}}

    Put the value into the formula

    a=\dfrac{10\times9.8\sin30+6\times9.8\sin30-0.1\times10\times9.8\times\cos30-0.3\times6\times9.8\times\cos30}{16}

    a=3.415\ m/s^2

    We need to calculate the force developed in the link

    For block A,

    Using balance equation

    ma=m_{A}g\sin\theta-\mu m_{A}g\cos\theta-T

    T=ma+\mu m_{A}g\cos\theta-m_{A}g\sin\theta

    Put the value into the formula

    T=10\times3.415+0.1\times10\times9.8\times\cos30-10\times9.8\times\sin30

    T=-6.36\ N

    Negative sign shows the opposite direction of the force.

    Hence, The force developed in the link is 6.36 N.

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