Asked by Joseph
If the rope pulling on the leg exerts a 120-N force on it, how massive should be the block hanging from the rope that passes over the pulley?
Answers
Answered by
nnn
Given
force F = 120 N
the angle theta = 20 degrees
here the vertical component of the tension in the sting is T sin theta which is equal to the force applied
that is T sin theta = 120
T = 120/si20 N
t = 350.85 N
now the weight mg = T
m = T / g
m = 350.85 /9.8 kg
m = 35.80 kg
the mass of the block should be hang the block hanging from the rope that passes over the pulley is 35.80 kg
force F = 120 N
the angle theta = 20 degrees
here the vertical component of the tension in the sting is T sin theta which is equal to the force applied
that is T sin theta = 120
T = 120/si20 N
t = 350.85 N
now the weight mg = T
m = T / g
m = 350.85 /9.8 kg
m = 35.80 kg
the mass of the block should be hang the block hanging from the rope that passes over the pulley is 35.80 kg
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