Question #186389

A screw jack has a two-start thread of pitch 5mm. An effort of 40N is applied tangentially to the tommy bar, at a radius of 350 mm, to lift a load of 2200 N.


  1. Calculate the velocity ratio and the mechanical advantage of the jack.
  2. Calculate the work done in overcoming friction when the load is raised a distance of 75 mm.
  3. Evaluate the efficiency of the screw jack and discuss what effect it has on the likelihood of overhauling and the law of the machine.
1
Expert's answer
2021-04-29T07:27:30-0400

Pitch, p=5 mm=5×103 mp=5\space mm=5\times10^{-3}\space m

Radius, r=350 mm=350×103 mr=350\space mm=350\times10^{-3}\space m

Effort, E=40 NE=40\space N

Load, L=2200 NL=2200\space N


(1) Velocity Ratio, (V.R)=2πrp=2×3.14×350×1035×103=439.6(V.R)=\dfrac{2\pi r}{p}=\dfrac{2\times3.14\times350\times10^{-3}}{5\times10^{-3}}=439.6

Mechanical Advantage, (MA)=LoadEffort=220040=55(MA)=\dfrac{Load}{Effort}=\dfrac{2200}{40}=55


(2) Height raised, h=75 mm=75×103 mh=75\space mm=75\times10^{-3}\space m

Efficiency, η=MAVR=55439.6×100=12.511%\eta=\dfrac{MA}{VR}=\dfrac{55}{439.6}\times100=12.511\%

Work Done =mghη=2200×75×1030.125=1320 J=\dfrac{mgh}{\eta}=\dfrac{2200\times75\times10^{-3}}{0.125}=1320\space J


(3) Efficiency, η=12.511%Efficiency, \space \eta=12.511\%



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Comments

David
11.06.23, 22:33

Great and efficient work, clear

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