1. Torque problem 9-33
The uniform gate in the diagram below has a mass of 51.0 kg. Assume that the force exerted on the gate by the upper hinge acts in the horizontal direction. Calculate the magnitude and direction of the supporting forces exerted by each hinge
2.4 m
1 m

2. Use static equilibrium net Torque = 0
Choose the pivot point at the lower hinge
Gate torque at its center of mass
Fup 1m FG d1
1.2m
Pivot point

3. NET =   = FG d1 – Fup (1m) = 0
Find d1
d12 = (1.2)2 +(0.5)2
d1 = 1.3 m
Fup 1m FG d1
1.2m
Pivot point

4. Find FG Find angle  Find FG and component perpendicular to d1 1.2m 
Pivot point 
Cos  = 1.2/1.3 = 0.923
 = cos-1(0.923) = °
Find FG and component perpendicular to d1
FG = 51 kg 9.81 m/s2 = 500 N
 + 2 = 90 and 1 +2 = 90
= 1
FG = FG cos  = 500N 1.2/1.3 = N  2 1 FG

5. NET =   = FG d1 – Fup (1m) = 0
D1 = 1.3 m
FG = N
FG d1 = Fup (1m) Fup = N 1.3m/1m = 600 N

6. FNET =  F = 0 F is a vector so FX = 0  FY = 0
Find Fdn
Apply equilibrium conditions to forces
FNET =  F = 0 F is a vector so FX = 0  FY = 0

7. 600 N
1.2m 1m d1
500 N
For lower hinge FX = N
FY = N
Therefore the magnitude of Fdn is ((500)2 + (-600)2)1/2 = 781 N
For the direction, the opposite side is 500 N and the adjacent side is 600 N
 = tan-1(500/600) = 39.8 ° above the horizontal