Question

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

• A. $6\text{N}$
• B. $3\text{N}$
• C. $1.5\text{N}$
• D. No perpendicular force is sufficient.

Answer 1

The correct option is C $1.5\text{N}$



As given in situation $1$,
$r=8\text{cm}$ $\theta ={30}^{\circ }$ $F=6\text{N}$
Torque required to loosen the nut $\tau =rF\sin \theta$
$\tau =\frac{8}{100}\times 6\times \sin 30$

$\tau =\frac{8}{100}\times 6\times \frac{1}{2}=\frac{24}{100}\text{N-m}$

For loosening the nut, same value of torque will be required on the nut.
Hence,
As given in situation $2$,
$r^{\prime }=16\text{cm}$, ${\theta }^{\prime }={90}^{\circ }$

$\tau =r^{\prime }{F}^{\prime }\sin {\theta }^{\prime }$
$\tau =r^{\prime }{F}^{\prime }\sin {90}^{\circ }$

$\frac{24}{100}=\frac{16}{100}\times F^{\prime }\times \sin {90}^{\circ }$
$\Rightarrow F^{\prime }=\frac{24}{16}=1.5\text{N}$