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 1 of 4.
Abhilash said:
3 years ago
Option B, +g/3 is the correct answer.
Kunal said:
4 years ago
The Correct Answer is D.
Tu = m(g+a), Td= m(g-a)/2; solving the 2 equations the 'Uniform Acceleration' a=g/3
Now the acceleration of the lift (in terms of g),
Upward g+a therefore it will be 4g/3,
Downward g-a therefore it will be 2g/3,
Thus Tu = 2Td.
Tu = m(g+a), Td= m(g-a)/2; solving the 2 equations the 'Uniform Acceleration' a=g/3
Now the acceleration of the lift (in terms of g),
Upward g+a therefore it will be 4g/3,
Downward g-a therefore it will be 2g/3,
Thus Tu = 2Td.
Chandu said:
5 years ago
Tension due to up word movement of lift is m(g+a).
Tension due to downward is m(g-a).
Since tension in the downward 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.
Here, (-g/3 is Retardation not Acceleration),
So, the answer is D.
Tension due to downward is m(g-a).
Since tension in the downward 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.
Here, (-g/3 is Retardation not Acceleration),
So, the answer is D.
(2)
Divashu Guleria said:
5 years ago
@All.
Don't get confused about relative acceleration. The supposed 'a' is actually the final acceleration of the lift and the effect of 'g' is already taken into account.
The answer is g/3.
Don't get confused about relative acceleration. The supposed 'a' is actually the final acceleration of the lift and the effect of 'g' is already taken into account.
The answer is g/3.
(1)
MIKKU said:
5 years ago
Let the tension in cable moving downwards is T1,
And the tension in cable moving upward is T2.
And T1 = 1/2 *( T2).
Case 1 when lift is moving downwards.
A/q to newton's second law.
mg - T1= m a.
T1 = mg-ma= m ( g-a) ----> (1)
Case 2 when lift is moving upward.
A/q to newton's second law.
T2- mg = ma.
T2= ma + mg = m ( g+a).
Now T1 = 1/2*(T2).
2*T1=T2.
2*{m(g-a)} = {m(g + a)}.
2g - g = a + 2a.
g = 3a.
Therefore a = g/3 ans. ( b)
And the tension in cable moving upward is T2.
And T1 = 1/2 *( T2).
Case 1 when lift is moving downwards.
A/q to newton's second law.
mg - T1= m a.
T1 = mg-ma= m ( g-a) ----> (1)
Case 2 when lift is moving upward.
A/q to newton's second law.
T2- mg = ma.
T2= ma + mg = m ( g+a).
Now T1 = 1/2*(T2).
2*T1=T2.
2*{m(g-a)} = {m(g + a)}.
2g - g = a + 2a.
g = 3a.
Therefore a = g/3 ans. ( b)
(1)
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.
1/2 upward = Downward.
1/2 m(g+a) = mg-ma,
mg+ma = 2mg-2ma,
mg =3ma,
g = 3a,
a = g/3.
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.
Deepak yadav said:
6 years ago
Answer Must be g/3.
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.
Td = tension in downword.
Tu = tension in upward.
Hence , 2Td =Tu.
2M(g+a) = M(g-a),
3a = g,
a = +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.
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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