Question

2 masses are suspended by a cord that passes of rape Ollie with negligible mass. The court also has negligible mass. One the masses M1 has a mass of 7.0 kg and the other mass and to has a massive 3.0 kg. The pulley turns on a shaft through the center of the pulley, which supports the pulley and all the masses. The vertical force of the shaft on the pulley is

a. 49N upward
b. 49N downward
c. 82N upward
d. 82N downward
e. 98N upward

Answer 1

Page Ans: The situation descsibed in problem is shown below in figuse - Light Pulley Light string | Cord M2 Mi With My = 7 kg and M₂ = 3kg = Light_pulley means its its Mass is negligible Tension in string and on both sides i's same. we = Light / Massless string means - Tension throughout the string length is same. Now - As know that Block My moves downward and block M₂ moves upward, with the accelsation of same magnitude. Let a be the magnitude of this accelsation New Fosces acting on mi - W Ming Vestically downwasds (22 Ivestically upwards, due to string As the block movec downward so Mig - I = Me

Page Fosces acting (2) I Continued com My area Mag Vestically downwards vestically upwards due to string. As the block M₂ moves upwards- T-M2g = Ma Now- the forces acting on the pulley are SO- (0) T downwards due to left side String (2) I downwesds due to hight side string (3) N Vestically upwards due to shaft. As the pulley does not more N=T+T N = 8T Using equation and equation ④ we can calculate I Multiply equation @ by M₂ - M, M₂ g - MT = M, M, a @ Multiply equation © by M MT - MM M, M a Subtract equation from equation -

Continued MIT - M, Mag - (M. Mag - M₂T) = M, M₂ a M, M, a MIT - M.Mag-MM29 +MI = 0 (M +M) T = 2MM24 Putting this T= 2 M, M 2 g MI+M2 value of I in equation ③ Upwards N = 4M, M₂ g (M+M2) N = 4X7X3x9.8 1o = 82.32 N Upwards OR N=82 N Upwards. Hence- Vestical Force of the shaft on the pulley = 82N Upward Ane
Problem is solved using basics of forces and first and second law of Newton.