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).

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)

Ans D is correct.

Proof whenever lift moving downwards tension is less compare to upwards tension . Hence here given tension for downwards 1/2 men's 0.5 . Upwards men's more than 0.5 is correct but given option A. g/2 = 0.5. Option b. g/3 = 0.33 option c. g/4 = 0.25.

So answer D is correct.

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 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.

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.

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.

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".

@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.

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.