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April 18, 2015

April 18, 2015

Posted by **Henri** on Monday, April 20, 2009 at 2:30pm.

Toggle1 length:520mm

Force applied:230N

Toggle2 lenght:445mm

Force applied:300.557N

Mean pitch of thread:13mm

S.J. efficiency:35%

My answer:

Assuming T2 to be 520mm long, the equivalent force would be:

0.445mX300.557N=0.52m X F2

F2=133.748/0.52

F2=257.2N

Assuming T1 as lifting the entire load:

Velocity Ratio of SJ=Circ of leverage/pitch of thread

V.R.=2pi0.52/0.013

V.R.=251.3

then:

Efficiency=Mech. adantage/V.R.

M.A.=251.3X0.35

M.A.=87.955

The load lifted by the Jack is:

M.A.=Load/effort

Load lifted=M.A.*(F1+F2)

Load=87.955*(230+257.2)

Load=42851.8/9.81

Load=4372.62kg/1000

Load=4.373tonnes

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