The iron door of a building is 2.4 m broad. It can be opened by applying a force of 10 kgf at the middle of the door. What is the least force required to open the door ? where the force should be applied ?
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Answered by
8
torque balance
F1r1=F2r2
10×1.2=2.4×F2
F2=5kgf(minimum)
force should be applied at the handle ..
F1r1=F2r2
10×1.2=2.4×F2
F2=5kgf(minimum)
force should be applied at the handle ..
Answered by
14
Hey user!
Here's the answer:
Middle of the door is = 2.4m÷2
Middle of the door is = 1.2 m away from the hinge.
Hence moment of force required to open the door= F× r
= 10 kgf × 1.2 m
=12 kgf × m.
》For minimum force the distance of application of the force should be maximum. Hence the force should be applied at the farthest point from the highest i.e. at the free end of the door.
Thus,
12 kgf × m = F× 2.4 m
F= (12 kgf × m)÷2.4 m
F= 5 kgf.
Here's the answer:
Middle of the door is = 2.4m÷2
Middle of the door is = 1.2 m away from the hinge.
Hence moment of force required to open the door= F× r
= 10 kgf × 1.2 m
=12 kgf × m.
》For minimum force the distance of application of the force should be maximum. Hence the force should be applied at the farthest point from the highest i.e. at the free end of the door.
Thus,
12 kgf × m = F× 2.4 m
F= (12 kgf × m)÷2.4 m
F= 5 kgf.
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