A block of 966 N shown below is suspended from the pulley B and the sag...

Question

Question

Answer 1

Answer #1

here,

as B is at the middle of A and C

the force in both the strings is equal

d = 0.15 m

weight of block , w = 966 N

angle of BC with vertical , theta = arctan( (0.4/2) / d)

theta = 53.13 degree

equating the forces

2 * t * cos(theta) = w

2 * t * cos(53.13) = 966

t = 805 N

the force in BC is 805 N

Answer 2

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