Start by putting $5$ kg at the $2$ m mark on the left of the seesaw. At what position do you predict that you will need to put a $10$ kg weight to balance the seesaw?

$$ Correct. The $5$ kg mass at $2$ m applies a torque with a magnitude of $10$ N $$ m$.$ A comparable torque must be applied on the right side of the seesaw to balance it$.$ m

Now, clear all masses on the seesaw and put $5$ kg at the $2$ m mark on the left of the seesaw. At what position do you predict that you will need to put a $20$ kg weight to balance the seesaw?

$$ Correct. The $5$ kg mass at $2$ m applies a torque with a magnitude of $10$ N $$ m$.$ A comparable torque must be applied on the right side of the seesaw to balance it$.$ m

Weights are added to the right side of the seesaw in the following way: $20$ kg at $2$ m$,10$ kg at $1\mathrm{.}75$ m$,10$ kg at $1\mathrm{.}5$ m$,10$ kg at $1\mathrm{.}25$ m$,$ and $15$ kg at $1$ m$.$ What is the torque $($in N $$ m$)$ on the seesaw?

$$ How do the applied force and the distance from the object relate? If multiple forces apply torque to an object, how do you determine the net torque? N $$ m