A girl pushes a door handle with a force of 50 N at a perpendicular distance of 1.0 m from the hinges, while a man pushes at the centre of the same door with a force of 90 N. Who produces the larger turning effect (moment) and wins?

What decides who "wins" when pushing a door?

A door turns based on the moment (turning effect) about the hinges, not just on who pushes with the bigger force. The moment depends on both the force and the perpendicular distance from the pivot (the hinges).

Compute each person’s moment about the hinges

The moment (torque) is $$ \tau = F\times d $$ where $d$ is the perpendicular distance from the hinge line.

Girl (at the handle): $$ \tau_g = 50\,\text{N}\times 1.0\,\text{m} = 50\,\text{N\, m} $$

Man (at the centre of the door): If the handle is $1.0\,\text{m}$ from the hinges, the door width is $1.0\,\text{m}$, so the centre is at $$ d = \frac{1.0}{2} = 0.5\,\text{m} $$ Then $$ \tau_m = 90\,\text{N}\times 0.5\,\text{m} = 45\,\text{N\, m} $$

Compare and decide

Since $50\,\text{N\, m} > 45\,\text{N\, m}$, the girl produces the larger turning effect, so she would "win" (the door would turn her way).