A 95 N force exerted at the end of a 0.35 m long torque wrench gives rise to a torque of 15 N · m. What is the angle (assumed to be less tha

Question

A 95 N force exerted at the end of a 0.35 m long torque wrench gives rise to a torque of 15 N · m. What is the angle (assumed to be less than 90°) between the wrench handle and the direction of the applied force?

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Hải Đăng 5 years 2021-08-14T23:08:01+00:00 1 Answers 434 views 0

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  1. MichaelMet
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    2021-08-14T23:09:15+00:00
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    Answer:

    The angle between the wrench handle and the direction of the applied force is 26.8°

    Explanation:

    Given;

    applied force, F = 95 N

    length of the wrench, r = 0.35 m

    torque on the wrench due to the applied force, τ = 15 N.m

    Torque is calculated as;

    τ = rFsinθ

    where;

    r is the length of the wrench

    F is the applied force

    θ is the angle between the applied force and the wrench handle

    Make Sin θ the subject of the formula;

    Sinθ = τ / rF

    Sinθ = 15 / (0.35 x 95)

    Sinθ = 0.4511

    θ = Sin⁻¹(0.4511)

    θ = 26.8°

    Therefore, the angle between the wrench handle and the direction of the applied force is 26.8°

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