A force of 314 N is exerted on the end of a wrench in order to apply a torque of 40.0 N·m to a bolt head. The point of application of the f

Question

A force of 314 N is exerted on the end of a wrench in order to apply a torque of 40.0 N·m to a bolt head. The point of application of the force is 21.5 cm from the center of the bolt. What angle does the force make with respect to the wrench handle?

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Thành Công 1 month 2021-08-16T02:53:18+00:00 2 Answers 0 views 0

Answers ( )

    0
    2021-08-16T02:54:57+00:00

    Answer:

    Explanation:

    Force, F = 314 N

    Torque, τ = 40 Nm

    distance, r = 21.5 cm = 0.215 m

    Let the angle is θ.

    τ = r x F x Sin θ

    40 = 0.215 x 314 x Sin θ

    Sin θ = 0.5925

    θ = 36.3°

    0
    2021-08-16T02:55:08+00:00

    Answer:

    36.16 degree

    Explanation:

    We are given that

    Force,F=314 N

    Torque=T=40 Nm

    Distance,r=21.5 cm=21.5\times 10^{-2} m

    1 m=100 cm

    We have to find the angle made by force with respect to the wrench handle.

    We know that

    Torque,T=rfsin\theta

    Using the formula

    40=21.5\times 10^{-2}\times 314sin\theta

    sin\theta=\frac{40}{314\times 21.5\times 10^{-2}}

    sin\theta=0.59

    \theta=sin^{-1}(0.59)

    \theta=36.16 degree

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