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
g/2
g/3
g/4
none of these
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
38 comments Page 2 of 4.

Varadraj said:   7 years ago
Since we have initially considered 'a' as the acceleration. So what we 'a' actually comes out is the acceleration.

I also think B is the correct answer.

Naveen said:   6 years ago
Since they have asked the just acceleration, not the direction/relative acceleration, option 'B' will be right like most of the guys solved above.

Kuldeep tyagi said:   6 years ago
Right answer is option( b), +g/3.
Td = tension in downword.
Tu = tension in upward.

Hence , 2Td =Tu.
2M(g+a) = M(g-a),
3a = g,
a = +g/3.

Dhiman said:   7 years ago
No it not B,

Read the question carefully we have to find out relative acceleration which is;
(g-a)= g- g/3= 2g/3 and the answer is D.

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.

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

Joyer said:   9 years ago
The correct answer is +g/3.

@Vanraj and @Gurdeep, you have made a mistake in the condition of tension!

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.

Pratik said:   6 years ago
It's C.

1/2 upward = Downward.
1/2 m(g+a) = mg-ma,
mg+ma = 2mg-2ma,
mg =3ma,
g = 3a,
a = g/3.

Vinay BEL said:   5 years ago
It's B.

1/2 upward = Downward.
1/2 m(g+a) = mg-ma,
mg+ma = 2mg-2ma,
mg =3ma,
g = 3a,
a = g/3.


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