Question

find the two forces.
keisha:
343n
allie:
421.4n
define keisha’s distance in terms of the length of the seesaw and allie’s distance.
when there is no rotation, the sum of the torques is zero.
solve for $r_a$ and $r_k$.

Explanation:

Step1: Set up torque equilibrium

Let $L$ = length of seesaw, $r_K$ = Keisha's distance from pivot, $r_A$ = Allie's distance from pivot. For rotational equilibrium:
$$F_K r_K = F_A r_A$$
Substitute $F_K=343\,\text{N}$, $F_A=421.4\,\text{N}$:
$$343 r_K = 421.4 r_A$$

Step2: Relate distances to seesaw length

Assume pivot is at the midpoint? No, general relation: $r_K + r_A = L$, so $r_K = L - r_A$.

Step3: Solve for $r_A$

Substitute $r_K = L - r_A$ into torque equation:
$$343(L - r_A) = 421.4 r_A$$
$$343L = 421.4 r_A + 343 r_A$$
$$343L = 764.4 r_A$$
$$r_A = \frac{343}{764.4}L \approx 0.449L$$

Step4: Solve for $r_K$

Use $r_K = L - r_A$:
$$r_K = L - 0.449L = 0.551L$$
Or from torque ratio:
$$r_K = \frac{421.4}{343}r_A \approx 1.229r_A$$

Answer:

Keisha's distance in terms of seesaw length and Allie's distance: $r_K = L - r_A$ or $r_K = \frac{421.4}{343}r_A \approx 1.229r_A$
$r_A \approx 0.449L$, $r_K \approx 0.551L$