Question

Question: To crack a certain nut in a nutcracker, forces with magnitudes of at least 40 N must act on its shell from both sides. For the nutcracker of Figure, with distances L =12 cmand D = 2.6 cm , what are the force components F_$\perp$ (perpendicular to the handles) corresponding to that 40 N?

Short answer

Answer:

$F_{p\perp }=8.7\text{N}$

Step-by-step solution

Understanding the given information

L = 12 cm

d = 2.6 cm

Concept and formula used in the given question

Using the concept of static equilibrium and writing the equation for torque in terms of force and distance, you can find the required component of the force.
The formula is given below.

Applied torque can we written as,

$\overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F}$

Calculation for the force components  

Applied torque on the nut by the nut cracker will be same as the torque caused by the perpendicular force, so you have

$$\begin{gathered} \left(L\times F_{\perp }\right)=\left(d\times 40\right) \\ \left(12\times F_{\perp }\right)=\left(2.6\times 40\right) \\ {F}_{\perp }=8.7\text{N} \end{gathered}$$

Hence, $F_{p\perp }=8.7\text{N}$