Question

To conserve power and to reduce wear on braking equipment, the counterweight A is attached to the hosting drum assembl 3 which operates mine cage C (a) If frictional resistances are neglected, what should be the weight W of the counterweight A in order to balance the 2500 lb weight of the empty cage C when at rest? (b) if a weighs 5500 lb and the equivalent coefficient of friction is 0.1, what should the total weight We of the loaded cage be in order to gain a velocity of 20 ft/s after it has dropped 35 ft. from rest? Bearing and cage frictions are negligible 1 = 40 ft-lb-s? 2 ft. radius 3 ft. radius fA=0.1

Answer 1

ANSWER

Given that
To conserve POWOT and 20 gleduce weay on braking Esuipment, the counter weight ☆ is attached to the hoisting - yom) Q8 Sembly B. Given situation is as follows:- KB - I = 40 ft to g. sino = 3/3 COS0- 415 U W A frictional Resistances wie neglected Free. Body Diagram!- X 9V WACOSO 250016 Wasino cft WA FOY Equilibrium Ti = WA $100 12. = 250016 Taxt = 79X3Ft

Then WA sing X2 = 3500 x3lb. WA #315) x2 = 9500 43 16 WA = 3500X3 16 : (315) 2 WA= 7500 16 12 WA = 6250 16. (b) If a weighs 550016 and the Eulvalent cocfficient of friction f is 01 Hege AA'=x2 ce! = 0x3 (AA) VA = 2 ve. then AA'= a (c . In our case, į moves 3514 ,ie cel =35ft Then AM = 70 ft Thus, a moves by Toft

3394 then ha = 70 sinoft ha =14ft Work Energy theorem! Final kinetic Energy - Initial kinetic Energy = work done by all the forces - => ☆ hieve? + ha VA the Int = (wex33– kpx14 -WA COS® X0-1) 9 Substitute the value's >> * Wex (2012+ 4 25500 x (x20] + x40x(99) : = (wcx33 –5500X14–5500 x 4x01) 32.9. Cheze wz ve for yop slip & the thread]. => { wc (400) + 1 x Sgoo x ( 1x400) + 4 x 90 x 90 00 = (x33– 77000 – 5500 X0.08) 32-2-

२००00 + 2750 (1777) +1x40 x 44.49 = (330c -Tboo— 440) 32.2 २००७८ + 4888675 + 888-8 = (33000-71440) 32-2- २००७e +48956-3 - 1062.6W -2493568 1062 600 -२५१3568 -200100 - 489756-300 W0862-6- २१४3324-320 862.6wc = २१४33243 - wc = २१833243 862-6 40-3458-521b,