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A screw jack whose pitch is 4.4mm is used to raise a car of mass 800kg through a height of 40cm. The lenght of the tommy bar of the Jack is 70cm, if the efficiency of the jack is 60%. Calculate (a) velocity ratio of the jack (b) M.A of the jack (c) effort required in raising of the body (g=10 m/s pi=22/7).

A screw jack whose pitch is 4.4mm is used to raise a body of mass 8000kg through a height of 20cm. The length of the Tommy bar of the jack is

70cm. If the efficiency of the jack is 80%,

calculate

(i) the velocity ratio of the jack

(ii) mechanical advantage of the jack

(iii) effort required in raising the body

(iv) work done by the effort in raising the body [=10ms 2 ; pi= 22 / 7 ] WAEC 1993 El

Solution

Pitch, p = 4.4mm = 0.0044m; mass, m=8000kg; g= 1 0m/s 2

Load, L= mg = 8000 x 10 = 80000N; distance load moved, l = 20cm = 0.2m;

length of Tommy bar, r = 70cm = 0.70m; efficiency, £ = 80%

2nr 2 x 22 / 7 x 0.70 4.4

0) The velocity ratio, V.R= — = 5 ^ = 00044 = 1000

M.A

(ii) From £ = x 100% we obtain

V J V. K

£x V.R 80 X 1000

Mechanical advantage, M. A = = — = 800

For a screw,

work done = work ouput

Effort x distance moved by effort = load x distance moved by load

Effort x circumference of handle = load x pitch

E x 27ir = L x p

Hence, the velocity ratio, V. R = P

For a perfect or frictionless screw, M.A = V.R
2nr

Hence, M.A = V.R = P

For a screw jack, r ts the length of the handle or Tommy bar of the screw jack.

https://archive.org/stream/AllInclusiveCalcsInPhysics/All-Inclusive%20Calculations%20in%20Physics_djvu.txt

DaudaAbu:
https://archive.org/stream/AllInclusiveCalcsInPhysics/All-Inclusive%20Calculations%20in%20Physics_djvu.txt

Thanks for the solutions. I really appreciate it.