Question

two children push on opposite sides of a door during play. both push horizontally and perpendicular to the door. one child pushes with a force of 165 n at a distance of 0.700 m from the hinges, and the second pushes at a distance of 0.400 m. what force (in n) must the second exert to keep the door from moving? assume friction is negligible. n resources reading

Explanation:

Step1: Recall the torque equilibrium condition

For the door to not move, the torques about the hinges must be equal. Torque formula is \(\tau = rF\) (since the force is perpendicular to the door, \(\sin\theta = 1\)). Let \(F_1 = 165\space N\), \(r_1=0.700\space m\), \(r_2 = 0.400\space m\), and \(F_2\) be the unknown force. At equilibrium \(\tau_1=\tau_2\), so \(r_1F_1=r_2F_2\).

Step2: Solve for \(F_2\)

Rearrange the equation \(F_2=\frac{r_1F_1}{r_2}\). Substitute \(r_1 = 0.700\space m\), \(F_1=165\space N\), and \(r_2 = 0.400\space m\) into the formula: \(F_2=\frac{0.700\times165}{0.400}\).
Calculate \(0.700\times165 = 115.5\), then \(F_2=\frac{115.5}{0.400}=288.75\space N\).

Answer:

\(289\space N\) (rounded to three significant figures)