Mechanical Engineering - Engineering Mechanics - Discussion

I think its quite different than we are thinking.

If I consider the acceleration of the list is "a" while moving upward and also downward, The upward force is "Fu" and downward force is "Fd",.

Resultant acceleration while moving downward is (g-a) and while moving upward {g-(-a)}=(g+a).

According to the condition Fd=Fu/2,
m(g-a)=m(g+a)/2,
a=g/3.

And hear is the tricky part, we don't have to find out the value of "a", read the question, actually, we have to find out resultant acceleration which is (g-a)
and (g-a)= g-g/3
(g-a)= 2g/3.

And the correct answer is 2g/3 (option D).

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.

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.

It should be g/3.

Answer Must be g/3.

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

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.

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.

Option B, +g/3 is the correct answer.

a = g/3.

The right answer is B.