A door has mass M, width W, and height H. It is attached to a door frame by only two hinges: one at 5H/6 (the top hinge) and one at H/6 (the bottom hinge). Calculate the force that the door frame exerts on each hinge.
Imagine the points: A and B are the bottom and top hinges, respectively, O is the center of the door.
Use Lami’s Theorem(If three forces acting on a particle keep it in equilibrium, each is proportional to the sine of the angle between the other two, and the lines of forces are intersected in the same point).
Examine two similar triangles: OAB and the force triangle with following sides: gravity force (directed downwards), and reactions of bottom and top hinges,R(A) and R(B) (along the lines AO and BO, respectively):
mg/AB =R(A)/OA= R(B)/OB
OA = OB (as the hinges are located symmetrically)
AB = 2H/3
OA = sqrt ((H/3)^2 +(W/2)^2).
Then R(A) = R(B) = OA• mg/AB = sqrt ((H/3)^2 +(W/2)^2)• mg/(2H/3)
R(A) directed along AO (from A)
R(B) directed along AO (to B)
To calculate the force that the door frame exerts on each hinge, we need to analyze the forces acting on the door.
1. Gravity: The force due to gravity acts vertically downward, with a magnitude of Fg = Mg, where M is the mass of the door and g is the acceleration due to gravity (approximately 9.8 m/s^2).
2. Hinge Forces: The door is connected to the door frame by two hinges, one at the top and one at the bottom. The hinges provide a normal force to keep the door in equilibrium.
Let's start with the top hinge:
The distance from the top hinge to the top of the door is H/6, and the distance from the top hinge to the center of mass of the door is 5H/6. Since the door is in equilibrium, the sum of the moments around the top hinge must be zero.
Sum of moments around the top hinge = Weight of the door x Distance from the top hinge to the center of mass - Force exerted by the hinge x Distance from the top hinge to the top of the door.
Mg x (5H/6) - F_top x (H/6) = 0
Simplifying the equation, we get:
F_top x (H/6) = Mg x (5H/6)
F_top = (Mg x (5H/6)) / (H/6)
F_top = 5Mg
Therefore, the force that the door frame exerts on the top hinge is 5 times the weight of the door (5Mg).
Now let's find the force exerted on the bottom hinge:
The distance from the bottom hinge to the bottom of the door is H/6, and the distance from the bottom hinge to the center of mass of the door is 5H/6. Using the same principle of equilibrium, the sum of the moments around the bottom hinge must be zero.
Sum of moments around the bottom hinge = Weight of the door x Distance from the bottom hinge to the center of mass - Force exerted by the hinge x Distance from the bottom hinge to the bottom of the door.
Mg x (5H/6) - F_bottom x (H/6) = 0
Simplifying the equation, we get:
F_bottom x (H/6) = Mg x (5H/6)
F_bottom = (Mg x (5H/6)) / (H/6)
F_bottom = 5Mg
Therefore, the force that the door frame exerts on the bottom hinge is also 5 times the weight of the door (5Mg).
In conclusion, the force that the door frame exerts on each hinge is 5 times the weight of the door (5Mg).