Homework Help: Is my math correct(2)

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

What load can be lifted by a screw jack given:
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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