Question

a 31 - kg child is sitting on a playground teeter - totter, 1.8 m from the pivot. what is the minimum distance, on the other side of the pivot, such that a 420 - n force will make the child lift off the ground?
a) 0.87 m
b) 1.0 m
c) 1.8 m
d) 1.3 m
e) 0.93 m
f) 1.5 m

Explanation:

Step1: Calculate the torque due to the child

Torque formula is $\tau = rF$. The force due to the child is $F = mg$, where $m = 31\,\text{kg}$ and $g= 9.8\,\text{m/s}^2$. So $F=31\times9.8 = 303.8\,\text{N}$. The distance of the child from the pivot is $r_1 = 1.8\,\text{m}$. The torque due to the child is $\tau_1=r_1F_1=1.8\times303.8$.

Step2: Calculate the distance for the given force

For the child to lift off, the torque due to the $420 - N$ force ($\tau_2=r_2F_2$, where $F_2 = 420\,\text{N}$) must be at least equal to the torque due to the child ($\tau_1=\tau_2$). So $r_2=\frac{\tau_1}{F_2}=\frac{1.8\times303.8}{420}$.
Calculating $\frac{1.8\times303.8}{420}=\frac{546.84}{420}\approx1.3\,\text{m}$.

Answer:

d) $1.3\,\text{m}$