A pivoted beam has mass mi suspended from one end and an Atwood\'s machine suspended from the other with masses m2 and ma suspended on either side. The frictionless pulley has negligible mass and size (a) Find the relation between mm2, m, and l2 which will ensure that the beam has no tendency to rotate just after the masses are released (b) What would you predict the relation would be in the case that all three masses are equal? Does your answer from part (a) agree with that prediction? 713
Solution
a) For beam to be at rest, torque on beam must be zero.
Taking torque about point of support, we get
T1*L1 = T2*L2
T1 is tension in the string attached to mass m1. Since mass m1 is at rest
T1 = m1 g
T2 is tension in the string between pulley and beam. Since pulley is massless,
T2 = 2*T , where T is tension in the string joining m2 and m3
T = 2 m2 m3 g/(m2 + m3)
Hence T2 = 4 m2 m3 g/(m2 + m3)
using values of T1 and T2 in first equation we get
m1 g L1 = 4 m2 m3 g L2/(m2 + m3)
m1 L1 = 4 m2 m3 L2/(m2 + m3)
b) With all three masses equal L1 should be 2 times L2, as mass on right side is 2 times mass on left and all masses are at rest.
Using m1 = m2 = m3 = m, we get
m L1 = 4m 2 L2 /2m
L1 = 2L2
.
A pivoted beam has mass mi suspended from one end and an Atwood-'s mac.docx
1. A pivoted beam has mass mi suspended from one end and an Atwood's machine suspended from
the other with masses m2 and ma suspended on either side. The frictionless pulley has negligible
mass and size (a) Find the relation between mm2, m, and l2 which will ensure that the beam has
no tendency to rotate just after the masses are released (b) What would you predict the relation
would be in the case that all three masses are equal? Does your answer from part (a) agree with
that prediction? 713
Solution
a) For beam to be at rest, torque on beam must be zero.
Taking torque about point of support, we get
T1*L1 = T2*L2
T1 is tension in the string attached to mass m1. Since mass m1 is at rest
T1 = m1 g
T2 is tension in the string between pulley and beam. Since pulley is massless,
T2 = 2*T , where T is tension in the string joining m2 and m3
T = 2 m2 m3 g/(m2 + m3)
Hence T2 = 4 m2 m3 g/(m2 + m3)
using values of T1 and T2 in first equation we get
m1 g L1 = 4 m2 m3 g L2/(m2 + m3)
m1 L1 = 4 m2 m3 L2/(m2 + m3)
b) With all three masses equal L1 should be 2 times L2, as mass on right side is 2 times mass on
left and all masses are at rest.
Using m1 = m2 = m3 = m, we get
m L1 = 4m 2
L2 /2m
L1 = 2L2
