Mechanical Engineering - Engineering Mechanics - Discussion
Discussion Forum : Engineering Mechanics - Section 2 (Q.No. 19)
19.
A block of mass m1, placed on an inclined smooth plane is connected by a light string passing over a smooth pulley to mass m2, which moves vertically downwards as shown in the below figure. The tension in the string is
m1/m2
m1.g sin α
m1.m2/m1+m2
Discussion:
13 comments Page 2 of 2.
Kaushal said:
6 years ago
How! Can we see derivative?
Prashanth said:
5 years ago
Solution;
(m1+m2)a=m2g-m1gsinα.
a=(m2g-m1gsinα)/ (m1+m2) --------> (eq1).
T-m1gsinα=m1a.
T=m1a+m1gsinα.
T=m1(a+gsinα) --------> (eq2)
Substitute eq1 in ( eq2).
T=[m1.g.m2(1.sinα)]/(m1+m2).
(m1+m2)a=m2g-m1gsinα.
a=(m2g-m1gsinα)/ (m1+m2) --------> (eq1).
T-m1gsinα=m1a.
T=m1a+m1gsinα.
T=m1(a+gsinα) --------> (eq2)
Substitute eq1 in ( eq2).
T=[m1.g.m2(1.sinα)]/(m1+m2).
Freind said:
4 years ago
For mass m2 : m2g-T=m2a ------> 1.
For mass m1 : T-m1gsinx=m2a -----> 2.
Find out acceleration first, then put 'a' in any one of the equation then you will get it.
For mass m1 : T-m1gsinx=m2a -----> 2.
Find out acceleration first, then put 'a' in any one of the equation then you will get it.
(2)
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