A force of 6.0 N applied at 30^ to a wrench at a distance of 8 cm from the nut is just sufficient to loosen the nut. Determine the force F that would be required to loosen it if it acts perpendicularly to the wrench at

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Quantrex Academy · Accepted Answer

Correct answer: D. 1.5 N

The torque required to loosen the nut remains constant in both cases.

The torque

\tau

produced by a force is given by

\tau = r F \sin \theta

.

For the first case:

F_1 = 6.0 \text{ N}

r_1 = 8 \text{ cm}

\theta_1 = 30^\circ

\tau_1 = 8 \times 6.0 \times \sin 30^\circ = 48 \times \dfrac{1}{2} = 24 \text{ N cm}

For the second case:

F_2 = F

r_2 = 16 \text{ cm}

\theta_2 = 90^\circ

\tau_2 = 16 \times F \times \sin 90^\circ = 16F \text{ N cm}

Equating the torques:

\tau_1 = \tau_2

24 = 16F

F = \dfrac{24}{16} = 1.5 \text{ N}

A force of 6.0 N applied at 30^ to a wrench at a distance of 8 cm...

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