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.
ANup said:
9 years ago
While going up,
T1-mg=ma ------------- (1)
While going down,
Mg-t2=ma --------------- (2)
Solving 1 & 2,
2(mg-ma)-mg=ma.
Or, mg-2ma=ma.
Or, mg=3ma.
Or, a=g/3.
T1-mg=ma ------------- (1)
While going down,
Mg-t2=ma --------------- (2)
Solving 1 & 2,
2(mg-ma)-mg=ma.
Or, mg-2ma=ma.
Or, mg=3ma.
Or, a=g/3.
Purushottam said:
9 years ago
@Gurdeep @Gaurav.
Problem shows that;
m(g-a) = m(g+a)/2.
Problem shows that;
m(g-a) = m(g+a)/2.
Kafeel Ahmad khan said:
8 years ago
mg-T= ma ----> (1)
2T-mg=ma ----> (2).
Putting the value of T in equation (2).
2(mg-ma)-mg = ma,
2mg-2ma-mg = ma,
mg = 3ma,
a = g/3.
2T-mg=ma ----> (2).
Putting the value of T in equation (2).
2(mg-ma)-mg = ma,
2mg-2ma-mg = ma,
mg = 3ma,
a = g/3.
Badan singh said:
8 years ago
Yes, the correct answer is g/3.
Dheeraj kumar said:
8 years ago
g/3 if the acceleration is constant otherwise, option D.
Rk reddy said:
8 years ago
g/3 is the answer.
A.jay said:
8 years ago
Answer is g/3. The question doesn't asks about direction of lift so there is no point of negative g/3.
Guin said:
8 years ago
If the tension is being half on moving downwards then surely be double on upward.
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