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Particles of mass 3m and 5m hang one at each end of a light inextensible string which passes over a pulley.the system is released from rest with the hanging parts taut and vertical.during the subsequent motion the resultant force exerted by the string on the pulley is of magnitude​

Respuesta :

Answer:

36.8 N

Explanation

tension in string when both bodies move vertically is given as

T = (m1 × m2 × g) / ( m1 +m2)

T = (5 × 3 × 9.81)/(5+3)

T = 147.15/8

T = 18.39375

SINCE TENSIONS ARE ON BOTH SIDES OF PULLEY AND ON PULLEY BOTH ARE DIRECTED DOWNWARD

HENCE FORCE ON PULLEY = 2 × T

                                               = 2×18.39375

                                               = 36.7875

                                                 = 36.8 N

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