A block of mass m is placed on a smooth wedge of inclination θ . The whole system is accelerated horizontally so that the block does not sli

Question

A block of mass m is placed on a smooth wedge of inclination θ . The whole system is accelerated horizontally so that the block does not slip on the wedge. The force exerted by the wedge on the block has a magnitude: (A) mg tan θ (B) mg sin θ (C) mg cos θ (D) mg sec θ

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Khoii Minh 5 years 2021-08-23T10:56:46+00:00 1 Answers 21 views 0

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  1. mit
    0
    2021-08-23T10:58:40+00:00
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    Answer: mg/Cosθ

    Explanation:

    Taking horizontal acceleration of wedge as ‘a’

    FCosΘ = FsinΘ

    F = mass(m) × acceleration(a) = ma

    For horizontal resolution g = 0

    Therefore,

    Horizontal = Vertical

    maCosΘ = mgSinΘ

    aCosΘ = gSinΘ

    a = gSinΘ/CosΘ

    Recall from trigonometry :

    SinΘ/Cosθ = tanΘ

    Therefore,

    a = gtanΘ

    Normal force acing on the wedge:

    mgCosΘ + maSinΘ – – – – (y)

    Substitute a = gtanΘ into (y)

    mgCosΘ + mgtanΘsinΘ

    tanΘ = sinΘ/cosΘ

    mgCosΘ + mgsinΘ/cosΘsinΘ

    mgCosΘ + mgsin^2Θ/cosΘ

    Factorizing

    mg(Cosθ + sin^2Θ/cosΘ)

    Taking the L. C. M

    mg[(Cos^2θ + sin^2Θ) /Cosθ]

    Recall: Cos^2θ + sin^2Θ = 1

    mg[ 1 /Cosθ]

    mg/Cosθ

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