Question

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.

Answer 1

Answer:

F = 318.1 N

Explanation:

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

$\tau = \vec r \times \vec F$

so we can write it as

$\tau = rFsin\theta$

now we know that

$\tau = 48 Nm$

r = 0.23 m

angle between force and the wrench is 41 degree

so we have

$48 = (0.23)Fsin41$

$F = \frac{48}{(0.23)sin41}$

$F = 318.1 N$