When a force of 6.0 , N is exerted at 30^{cir | vedclass

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(D) The torque $	au$ required to loosen the nut is given by $	au = r F \sin(	heta)$.In the first case,$r_1 = 8 \, cm$,$F_1 = 6.0 \, N$,and $	heta_

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$1.5$

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(D) The torque $	au$ required to loosen the nut is given by $	au = r F \sin(	heta)$.In the first case,$r_1 = 8 \, cm$,$F_1 = 6.0 \, N$,and $	heta_1 = 30^{\circ}$.Therefore,$	au = 8 \, cm 	imes 6.0 \, N 	imes \sin(30^{\circ}) = 8 	imes 6.0 	imes 0.5 = 24 \, N \cdot cm$.In the second case,we need to apply a force $F_2$ perpendicularly $(	heta_2 = 90^{\circ})$ at a distance $r_2 = 16 \, cm$.Since the required torque remains the same,$	au = r_2 F_2 \sin(90^{\circ})$.$24 = 16 	imes F_2 	imes 1$.$F_2 = \frac{24}{16} = 1.5 \, N$.

When a force of $6.0 \, N$ is exerted at $30^{\circ}$ to a wrench at a distance of $8 \, cm$ from the nut,it is just able to loosen the nut. What force $F$ would be sufficient to loosen it,if it acts perpendicularly to the wrench at $16 \, cm$ from the nut (in $, N$)?

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