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18 September, 06:26

A 95 N force exerted at the end of a 0.50 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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Answers (2)
  1. 18 September, 07:19
    0
    18.4°

    Explanation:

    Force, F = 95 N

    Torque, τ = 15 Nm

    length, r = 0.5 m

    τ = r F Sinθ

    where, θ be the angle between distance vector and force vector

    So, 15 = 0.5 x 95 x Sinθ

    Sin θ = 0.3159

    θ = 18.4°

    Thus, the angle is 18.4°.
  2. 18 September, 10:24
    0
    angle = 18.40 degree

    Explanation:

    given data

    force = 95 N

    distance = 0.50 m

    torque = 15 N · m

    to find out

    angle between the wrench handle and the direction of the applied force

    solution

    we will apply here torque equation that is express as

    torque = distance * force * sin (θ) ... 1

    put here value we will get angle that is

    15 = 0.50 * 95 * sin (θ)

    sin (θ) = 0.315789

    θ = 18.40 degree
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