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Please Help With This Physics Problem. by concord129(m): 7:03am On Nov 01, 2019 |
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). |
Re: Please Help With This Physics Problem. by DaudaAbu(m): 7:49am On Nov 01, 2019 |
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 1 Like |
Re: Please Help With This Physics Problem. by DaudaAbu(m): 8:15am On Nov 01, 2019 |
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 1 Like |
Re: Please Help With This Physics Problem. by concord129(m): 8:38am On Nov 01, 2019 |
DaudaAbu: Thanks for the solutions. I really appreciate it. |
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