Q) 30-kg hanging object is connected by a light, inextensible cord
over a light, frictionless pulley to a 5.00-kg block that is sliding on a flat
table. Taking the coefficient of kinetic friction as 0.162, find the tension in
the string.
Sol:
It’s given that M1=5kg, M2=30kg, Also M1 is moving
with acceleration a, caused by the Tension in string: Ts, and the
friction, Fr, Hence aM1= Ts - uM1g, where u= coefficient
of kinetic friction.
Similarly, the mass 30kg is moving with the same above
acceleration a by the forces of weight and the Tension in the string,
Hence aM2=M2g- Ts
By adding the above two equations we get,
a= (M2g-uM1g)/ (M1+M2)
Therefore the tension in the string is given by,
Ts = M1g (M2-uM1)/ (M1+M2)
+uM1g
By substituting the values in the above equation, we
get
Ts = [5*9.8*(30-0.162*5)/
(5+30)] +0.162*5*9.8
= 48.8N
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