Civil Engineering - UPSC Civil Service Exam Questions - Discussion
Discussion Forum : UPSC Civil Service Exam Questions - Section 4 (Q.No. 48)
48.
Two bodies of masses m1 and m2 are communicated by a light inextensible string passing over a small smooth fixed pulley; m1 > m2. What is the acceleration of the system ?
Discussion:
1 comments Page 1 of 1.
Vikal Burman said:
8 years ago
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A tension force induces in the string on both the masses in an upward direction.
Let tension force is T.
T = m1g-m1a.........(1)
T = m2g+m2a........(2)
Acceleration is same for both masses but it will be in opposite direction.
Thus;
a2 = -a1.
(1)=(2).
m1g-m1a=m2g+m2a,
m1g - m2g = m1a+m2a,
g(m1-m2) = a(m1+m2),
a = g(m1-m2)/(m1+m2).
A tension force induces in the string on both the masses in an upward direction.
Let tension force is T.
T = m1g-m1a.........(1)
T = m2g+m2a........(2)
Acceleration is same for both masses but it will be in opposite direction.
Thus;
a2 = -a1.
(1)=(2).
m1g-m1a=m2g+m2a,
m1g - m2g = m1a+m2a,
g(m1-m2) = a(m1+m2),
a = g(m1-m2)/(m1+m2).
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