Question

When a force of 6⋅0 N is exerted at 30° 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?
Figure

Answer 1

Explanation:

please go through the attachment

Answer 2

Force (F) = 1.5 N  would be sufficient to loosen it if it acts perpendicularly to the wrench at 16 cm from the nut.

Explanation:

Step 1:

Force of 6N acts at an angle of 30° on a body, now we have to solve the force of 6 N along the wrench and perpendicular to it.

                          P = 6.cos30°  

 Now, the total torque of point A = 0

Step 2:

Consequently, the force component perpendicular to the lock

                          $\begin{aligned}&P^{\prime}=6 . \sin 30^{\circ}\\&P^{\prime}=6 \times \frac{1}{2}\\&P^{\prime}=3 N\end{aligned}$

Step 3:

The torque ( nut ),

                          $=3.0 \times \frac{8}{100}$

                         $=0.24 \mathrm{N}-\mathrm{m} -----> \text { eqn }(1)$

The torque of force F (nut),

                         $=F \times \frac{16}{100} N-m$ $-----> eqn ( 2 )$

Step 4:

Equating the two that we got,

                        $F \times \frac{16}{100}=0.24$

                       $F=\frac{0.24}{16} \times 100$

                      $F=\frac{24}{16}$

                      $F=\frac{3}{2}$

                      $\mathrm{F}=1.5 \mathrm{N}$

Therefore, force ( F ) = 1.5N  would be sufficient to loosen it if it acts perpendicularly to the wrench at 16 cm from the nut.