1. Oct 27, 2007

### joker_900

1. The problem statement, all variables and given/known data
I don't really understand how to approach problems like these

A lift has a downward acceleration of kg (k<1). Inside the lift is mounted a pulley, of negligible friction and inertia, over which passes an inextensible string carrying two objects of masses m and 3m.

a. Let the Tension in the string be T and the accelerations of the 3m and m masses be a1 and a3. Write down the equations of motion of the two masses in terms of T.

b. Eliminate T, relating a1 and a3

c. Use the information that the string is inextensible to get another relationship between a1 and a3 (one way to use this information is to set up equations relating the motion in the inerital (shaft) frame to the non-inertial (lift) frame)

d. For k=1/3 find the acceleration of the 3m mass in the lift frame and the force exerted on the pulley by the rod that joins it to the roof of the lift.

2. Relevant equations

3. The attempt at a solution

a. 3mg - T = 3ma3 and mg - T = ma1

b. 2g = 3a3 - a1

c. In the lift frames, the accelerations are a1' and a3'

a1' = a1 - kg and a3' = a3-kg

But a1'=-a3' (as the string is inextensible)

Inserting a1'=-a3' and equating above equations:
a1-kg = kg = a3

I put this back into the equation of part (b) to get a3 in terms of k, but it was wrong. I'm sure I'm doing something fundamentally wrong, so please help!

The answers given are g/3 for the acceleration of 3m in the lift frame and 2mg for the force on the pulley.

2. Oct 27, 2007

### learningphysics

Everything you've done is exactly right. Using your results I'm getting g/3 and 2mg. only mistake I see is this:

"a1-kg = kg = a3"

which I'm guessing is just a typo. It should be:

a1-kg = kg - a3

solving this equation with

2g = 3a3 - a1

I get a3 = (2/3)g

But the queston asks for the acceleration of the 3m mass in the lift's frame. ie the non-inertial frame.

so after getting a3, use a3' = a3-kg to get a3'.
a3' = (2/3)g - kg = (2/3)g - (1/3)g = g/3

Solve for T... the force on the pulley is 2T which comes out to 2mg.

3. Oct 27, 2007

### saket

You are "fundamentally" correct!

Apart from the error in typing above equation, which should be a1-k.g = k.g - a3, I couldn't find any! And, what more, correct results are obtained. Please check your calculations. Hey, do NOT loose hope so quickly!

4. Oct 27, 2007

### joker_900

Thanks both of you! I don't know what I was doing to get the wrong value of a3'

One question, why is the force equal to 2T? Isn't the pulley accelerating?

5. Oct 27, 2007

### saket

Basis of the assumptions.

Pulley is accelerating.
Make a free body diagram (fbd) of the pulley. In the two strings, connecting to the masses, let the tensions be T1 and T2. Then, as inertia and/or friction are negligible, it can be easily shown [Use concepts of rotaional motion] that T1 = T2.
Let T1 = T2 = T.

Let T' be the tension in the string joining the pulley to the lift. Then, using Newton's 2nd Law of Motion, (T1+T2) - T' = m'.(k.g)
where, m' is the mass of the pulley and pulley is moving down with an acceleration (k.g) alongwith the lift.

Now, as inertia of wheel is to be neglected, RHS of the above equation can be set to zero.
Thereby yielding, T1+T2 = T'
=> T' = T + T = 2T.

Showing T1 = T2 is also simple.

6. Oct 27, 2007

### learningphysics

Yeah, as Saket mentioned... pulley is assumed to have 0 inertia => 0 mass. so the net force acting on the pulley is ma = 0a = 0.

The two tensions need to be equal, again as a consequence of 0 inertia. 0 inertia implies 0 torque acting on the pulley. That leads to the two tensions being equal.

7. Oct 27, 2007

### saket

Yeah, that's right.
Furthermore, just for the information of "joker_900", negligible friction in the pulley is also, like-wise, self-sufficient to show that T1 = T2.