Fig. 3.1 presents, from above, a door together with an automatic door-closer. When the door swings open and shut, the hinge functions as the pivot. A girl opens the door by applying a force $P$ at point X. Force $P$ acts perpendicular to the surface of the door.
(a(i))[2]
Calculate the moment of force $P$ about the hinge. Force $P$ is $25\,\text{N}$ and point X lies $0.72\,\text{m}$ from the hinge.
(a(ii))[2]
The door turns through $90^\circ$ about the hinge. A circle with radius $0.72\,\text{m}$ has a circumference of $4.5\,\text{m}$. Calculate the work done on the door by force $P$.
(b)[2]
As the door opens, a force $F$ acts on the door in the direction shown in Fig. 3.1. Even though force $F$ is greater than force $P$, the door still rotates about the hinge. Explain why.
Worked solution & mark scheme
This 6-mark question has a full step-by-step worked solution and mark scheme. One marking point: “F = Fy or, equivalently, 25 × 0.72” …