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A screw jack 25% efficient and having a screw pitch of 0.4cm is used to raise a load through a certain height. If the process...

      

A screw jack 25% efficient and having a screw pitch of 0.4cm is used to raise a load through a certain height. If the process the handle turns through a circle of radius 40cm. Calculate the
1. Velocity ratio of the machine
2. Mechanical advantage of the machine
3. Effort required to raise a load of 1000N with the machine ( take pi =3 1/4)

  

Answers


CHARLES
i) Velocity ratio (V.R) of a screw = Effort distance/Load distance
Since the handle turns through a circle, then;
Effort distance = 2*pi*R
= 2*3.14*0.4
=2.512

Load distance = Pitch of the screw
=0.004

Therefore, V.R=2.512/0.004
= 628
ii) Since efficiency is the ratio of mechanical advantage to the velocity ratio, then:
Mechanical Advantage,M.A = Efficiency*V.R
= 0.25*628
= 157
3) M.A = Load/Effort, therefore;
Effort=Load/M.A
= 1000/157
=6.369 N
Charcher answered the question on March 2, 2018 at 18:05


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