Question

two children playing on a frictionless garden gate invent a new game called \gate\. the idea is that they will get on opposite sides of the gate and each push such that the gate does not move. if they both push horizontally and perpendicular to the gate and one child pushes with a force of 160 n at a distance of 0.740 m from the hinges, determine the force the second child must exert in order to keep the gate from moving if she pushes at a distance of 0.340 m from the hinges. n resources reading

Explanation:

Step1: Recall torque formula

Torque \(\tau = rF\) (where \(r\) is the distance from the pivot and \(F\) is the force). For the gate not to move, the torques must balance: \(\tau_1=\tau_2\).

Step2: Substitute values into torque - balance equation

Let \(F_1 = 160\space N\), \(r_1=0.740\space m\), \(r_2 = 0.340\space m\). From \(\tau_1=\tau_2\) (i.e., \(r_1F_1=r_2F_2\)), we can solve for \(F_2\).

$$F_2=\frac{r_1F_1}{r_2}$$

Substitute \(r_1 = 0.740\space m\), \(F_1=160\space N\), \(r_2 = 0.340\space m\) into the formula:

$$F_2=\frac{0.740\times160}{0.340}$$

$$F_2=\frac{118.4}{0.340}\approx348.24\space N$$

Answer:

\(348\space N\)