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If the torque required to loosen a nut holding a wheel on a car is 48 N · m, what force must be exerted at the end of a 0.23 m lug wrench to loosen the nut when the angle between the force and the wrench is 41◦ ? Answer in units of N.

Respuesta :

Answer:

F = 318.1 N

Explanation:

As we know that torque to open the nut is given by formula

[tex]\tau = \vec r \times \vec F[/tex]

so we can write it as

[tex]\tau = rFsin\theta[/tex]

now we know that

[tex]\tau = 48 Nm[/tex]

r = 0.23 m

angle between force and the wrench is 41 degree

so we have

[tex]48 = (0.23)Fsin41[/tex]

[tex]F = \frac{48}{(0.23)sin41}[/tex]

[tex]F = 318.1 N[/tex]

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