Mechanical Engineering - Engineering Mechanics - Discussion
Discussion Forum : Engineering Mechanics - Section 2 (Q.No. 40)
40.
Two bodies of masses m1,and m2 are hung from the ends of a rope, passing oyer a frictionless pulley as shown in the figure below. The acceleration of the string will be
Discussion:
3 comments Page 1 of 1.
K YUGANDHAR said:
6 years ago
At Mass m1.
m1g-T=m1a - - - >1.
At Mass m2
T-m2g=m2a.
In this
T= m2a+m2g sub in eq1, and Get a.
m1g-T=m1a - - - >1.
At Mass m2
T-m2g=m2a.
In this
T= m2a+m2g sub in eq1, and Get a.
(1)
PANKAJ KUMAR said:
9 years ago
ACCORDING TO NEWTON'S SECOND LAW OF MOTION.
(M1 + M2) A = M1G - M2G.
A = (M1G-M2G)/(M1 +M2).
(M1 + M2) A = M1G - M2G.
A = (M1G-M2G)/(M1 +M2).
(1)
Manoj said:
1 decade ago
Consider both the masses as FBD separately.
Then equating tensions from both the cases.
Then equating tensions from both the cases.
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