Mechanical Engineering - Engineering Mechanics - Discussion
Discussion Forum : Engineering Mechanics - Section 4 (Q.No. 2)
2.
If tension in the cable supporting a lift moving downwards is half the tension when it is moving upwards, the acceleration of the lift is
Discussion:
38 comments Page 4 of 4.
Vanraj Gajarotar said:
9 years ago
Upward tension T = m(g + a).
Downward tension T = m(g - a).
Lift moving downwards is half the tension when it is moving upwards,
m(g + a) = [m(g - a)]/2.
2m(g + a)=m(g - a).
a = -g/3.
Therefore right answer is "D".
Downward tension T = m(g - a).
Lift moving downwards is half the tension when it is moving upwards,
m(g + a) = [m(g - a)]/2.
2m(g + a)=m(g - a).
a = -g/3.
Therefore right answer is "D".
Gaurav said:
9 years ago
@Gurdeep.
You are right, the answer is option D.
You are right, the answer is option D.
Keyur said:
9 years ago
2m(g-a)=m(g+a)
2g-2a=g+a
g=3a
a=g/3
2g-2a=g+a
g=3a
a=g/3
Gurdeep gill said:
1 decade ago
Tension due to up word movement of lift is m(g+a).
Tension due to downward is m(g-a).
Since tension in down word is half the up word lifting mean up word double the downward.
2m(g+a) = m(g-a).
2g+2a = g-a.
g = -3a.
a = -g/3.
And due to - sign answer is D.
Tension due to downward is m(g-a).
Since tension in down word is half the up word lifting mean up word double the downward.
2m(g+a) = m(g-a).
2g+2a = g-a.
g = -3a.
a = -g/3.
And due to - sign answer is D.
Kenmar Basar said:
1 decade ago
m(g+a) = 2m(g-a).
2g-2a = g+a.
g = 3a.
a = g/3.
This is what I got instantly. I think @Shyam your are right but u need to explain
2g-2a = g+a.
g = 3a.
a = g/3.
This is what I got instantly. I think @Shyam your are right but u need to explain
Prakash said:
1 decade ago
I also agree that acceleration should be g/3.
Gaurav said:
1 decade ago
I think the answer should be g/3.
1st case: When lift is moving down with constant acceleration 'a'.
mg-T1 = ma.
2nd case: When lift is moving up with constant acceleration 'a'.
T2-mg = ma.
T2 = 2T1.
Solving both equations we will get a = g/3.
1st case: When lift is moving down with constant acceleration 'a'.
mg-T1 = ma.
2nd case: When lift is moving up with constant acceleration 'a'.
T2-mg = ma.
T2 = 2T1.
Solving both equations we will get a = g/3.
Shyam said:
1 decade ago
Acceleration will be constant which is g=9.81 mm/s^2
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