Answer
If the measured length of the can pull tab were $r1=1.8cm$ and $r2=0.7cm$. The force act on the can is $F2=25.7N$.
Work Step by Step
We can approach this problem the same way we do with the lever problem. We have two forces acting on two ends of a lever, which is the can pull tab.
On our hand end we have $F1 = 10N$ and $r1$
On the can end we have $F2 =?$ and $r2$
To determine the value of $F2$, we use the equation below:
$F2=\frac{r1}{r2} F1 =\frac{r1}{r2} 10N$
Now, we need to measure the can pull tab length. We measure the can pull tab length to be $r1=1.8cm$ and $r2=0.7cm$. So,
$F2 = \frac{1.8}{0.7} 10N = 25.7N$
