Particles of mass 3m and 5m hang one at each end of a light inextensible string which passes over a pulley.the system is release d 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
1 answer:
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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