Question

Beverage engineering. The pull tab was a major advance in the engineering design of beverage containers. The tab pivots on a central bolt in the can’s top. When you pull upward on one end of the tab, the other end presses downward on a portion of the can’s top that has been scored. If you pull upward with a$\mathbf{\text{10\hspace{0.17em}N}}$ force, approximately what is the magnitude of the force applied to the scored section? (You will need to examine a can with a pull tab.)

Short answer

The magnitude of forcethat acts on the scored sectionis$25\text{N}$ .

Step-by-step solution

Step 1: Given

i) Distances,$r_1=1.8\text{cm},r_2=0.73\text{cm}$

ii) Force,$F_1=10\text{\hspace{0.17em}N}$

Determining the concept

As the tab is in equilibrium, the net torque exerted by both forces must be equal. Therefore, by using the formula of torque, find the magnitude of force that acts on the scored section.

The formula is as follows:

${\tau }_{up}={\tau }_{down}$

Where, $\tau$ is torque

Determining themagnitude of force that acts on the scored section

For torque in equilibrium,

${\tau }_{up}={\tau }_{down}$

$F_1{r}_1=F_2{r}_2$

$10\text{\hspace{0.17em}N}\times 1.8\text{\hspace{0.17em}m}=F_2\times 0.73\text{\hspace{0.17em}m}$

$F_2=\left(\frac{1.8\text{\hspace{0.17em}cm}}{0.73\text{\hspace{0.17em}cm}}\right)\times \left(10\text{\hspace{0.17em}N}\right)$

$F_2=25\text{\hspace{0.17em}N}$

Hence, themagnitude of force that acts on the scored section is$25\text{N}$ .