1. Homework Statement
The door has a mass of 30 kg, is 200 cm long and 90 cm wide. The hinges are installed at 25 cm from the bottom (R2) and 25 cm from the top (R1). What are is the magnitude of the force exerted by each hinge on the door? The door is in a state of static equilibrium.
Torque = rFsinθ
∑Torque = 0 Nm
∑F = 0
Fg = 30kg*9.8m/(s^2) = 294 N
The Attempt at a Solution
a) Place pivot midway between hinges.
⇒F1 and F2 exert no torque (Parallel to axis or rotation).
⇒ F1+F2 = Fg = 294 N. No way to find each force individually.
⇒Lever arm for R1 = 0.75 m, θ = 90°; same for R2
Since ∑Torque = 0 Nm; Torque by R1 = Torque by R2 (opposite sign), so they must exert the same force (opposite sign) on the door.
STUCK. A hint please? Thanks!