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
1m
• Use static equilibrium net Torque = 0
• Choose the pivot point at the lower hinge
• Gate torque at its center of mass
Fup
1.2m
d1
1m
FG
Pivot point
NET =   = FG d1 – Fup (1m) = 0
• Find d1
• d12 = (1.2)2 +(0.5)2
• d1 = 1.3 m
Fup
1.2m
FG
1m
d1
Pivot point
Find FG
• Find angle 
1.2m

Cos  = 1.2/1.3 = 0.923
1.3 m
 = cos-1(0.923) = 22.620°
Pivot point
• Find FG and component perpendicular to
d1
1
2
FG

FG = 51 kg 9.81 m/s2 = 500 N
 + 2 = 90 and 1 +2 = 90
= 1
FG = FG cos  = 500N 1.2/1.3 = 461.8 N
NET =   = FG d1 – Fup (1m) = 0
• D1 = 1.3 m
• FG = 461.8 N
FG d1 = Fup (1m)
Fup = 461.8 N 1.3m/1m
= 600 N
Find Fdn
• Apply equilibrium conditions to forces
FNET =  F = 0
F is a vector so
FX = 0
 FY = 0
600 N
1.2m
d1
1m
500 N
For lower hinge FX = - 600 N
FY = 500 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