Worked exercise — The door with two forces

Problem

A door of length $L = 2\;\text{m}$ is subject to a force $F_1 = 10\;\text{N}$ applied at the edge and a force $F_2 = 5\;\text{N}$ at the midpoint of the door, both perpendicular to the door. Find the force at the hinge for equilibrium.

Solution

Choice of pole. It is convenient to take moments about the hinge (point O): that way the hinge reaction, applied right at O, has zero lever arm and does not appear in the moment balance.

Moment balance. $F_1$ acts at the edge (lever arm $2\;\text{m}$), $F_2$ at the midpoint of the door (lever arm $1\;\text{m}$). Assuming they produce rotation in opposite senses: $\sum M_O = F_1\cdot 2 - F_2\cdot 1 = 20 - 5 = 15\;\text{N}\cdot\text{m} \neq 0$

Conclusion. The resultant moment is not zero: the door is not in rotational equilibrium and tends to rotate about the hinge. To balance it, an opposing moment of $\ev{|M| = 15\;\text{N}\cdot\text{m}}$ is required.

Note: the hinge reaction alone cannot supply this moment (it has zero lever arm about O); a third force applied at some distance from the hinge is needed. For example, a perpendicular force $F_3$ applied at the edge ($b = 2\;\text{m}$) would need to be $F_3 = 15/2 = 7{,}5\;\text{N}$, in the direction that opposes the rotation.

Links

Topics: Statics and equilibrium Concepts: Moment of a force · Static equilibrium Skills: Moment balance · Choice of pole