Question

Belt Friction - Problem 7-52. Shaft A is driven counterclockwise and has two turns of rope wound on it (Figure P7-52). The rope slips on the shaft if no tension Tis applied. Determine the minimum T required to cause the rotating shaft to pull block B up the slope. u=0.2 200 kg 0.3 55 FIGURE P7-52 ANS: T=157.5 N Click to add notes Non- 1920 PM

Answer 1

pgnor A solution : Given that's → Shaft A is doiven Countee clock wise and two twens ao scope wound The rope slips on shaft no Tension T a applied The minimum T required to cause rotating shaft to pull block B up slope. M20.2 - ooka B M=0.3 » Weight of block. B w = (200) (9.8) 1969 N Tuin = 1962N

19.102 > Namal reaction force on block B JVE wcos (55°) where, W 1962 Sub, = W 05735 = 1962*0.5735 = 1125.207 V = 1125.207 → fructional force on block is Uv = 0.3 XV whege V=125.2 = 0.3 X 1125.2 = 337.6 Luv- 337.61 The face balance along with Prolined slope. TP-ein-wsin (550) 0] 1!= 1944-8 N

> wraps pg .No 13 ů angle on shaft A - (a) (360) = 720 → Convent value to 790 = 180 = 12.5664° = 2.5133 1944.8 9.5133 e where, T = 157.53N 19444408 2:5133 157.53 = 12.3456 in the minimum value of Tension T ů 157.5N