Help with a pulley problem with 3 masses

[nuclearfireball_42]

Homework Statement

In the system shown in the figure, there is no friction, the blocks are weightless, the thread is weightless and inextensible, m 1 = 2 kg, m 2 = 4 kg, m 3 = 1 kg. Find the modulus and direction of acceleration of the load with mass m 3 .

Relevant Equations

F = ma

The figure :

What I understand from the figure :
T₁=m₁a₁
T₂=m₂a₂
T₃-m₃g=m₃(-a₃)

- The three masses given all have different mass so each of them has different acceleration
- How do one substitute to obtain the answer for a₃?
- I've tried to substitute to find the value of a₃, but it seems that I keep getting the answer in terms of the other two terms a₂ and a₁. The same thing happens when I try to solve for a₂ and a₁.

 

[BvU]

Hi,

You need another equation is basically what you are saying. Observe that you haven't exhausted the given information yet: ' the thread is weightless and inextensible '

 

[Delta2]

At first you are saying that the blocks are weightless and then you are saying $T_3-\mathbf{m_3g}=…$. Perhaps you meant that the pulleys are weightless/massless?

 

[haruspex]

The three masses given all have different mass so each of them has different acceleration

That would follow if each were subject to the same net force, but they are not.
Use @BvU's hint in post #2 to find the relationships between the accelerations. You also need equations relating the tensions.

 

[Crystal037]

Isn't T1=T2=2T and T3=T

 

[BvU]

Isn't T1=T2=2T and T3=T

What is T ?

Anyway: no. As @haruspex says: the forces on the masses are not equal

You find a relation between the tensions if you make a free body diagram for the right pulley (which happens to be massless).

Apparently my hint in #2 needs to be laid up a bit thicker: introducing $x_1,x_2,x_3$ with $a_1 = {d^2 x_1\over dt^2}$, etc. , what do you know of $a_3$ as a function of $a_1$ and $a_2$ ?

However, all this yields us five equation in total, with six unknowns ... so still one more to go

 

[Delta2]

I think we can benefit with two equations from the fact that the right pulley is massless (and hence moment of inertia-less), one from torque balance from which we can infer that $T_1=T_3$ and one from force balance from which we can infer that $T_1+T_3=T_2$.

 

[BvU]

And with that we have six equations with six unknowns. Handed to you on a silver platter, @Crystal037 !

 

[Crystal037]

I am not able to visualise how T1=T3 and T1 +T3=T2 and how are x1, x2, x3 known

 

[BvU]

Have you drawn a free body diagram for the right pulley yet ?

 

[Crystal037]

Doesn't the free body diagram of right pulley has only one force and i.e. tension

 

[BvU]

how are x1, x2, x3 known

They are not. But there is a relationship that you can differentiate twice to get a relationship between the $a_i$. And differentiating gets rid of initial conditions: it only concerns differences.

If that's awkward: If $m_1$ moves 1 cm and $m_2$ is fixed, how much does $m_3$ move ?
Same question 2 and 1 reversed. Just like you always do with pulley problems. Comes down to the given: wire inextensible.

 

[Crystal037]

m3 will also move 1cm, then how will we come to know about x1, x2, x3

 

[Physics lover]

I think you need to use constraint and find relation between a1,a2 and a3.

 

[Crystal037]

Can you elaborate

 

[Delta2]

Doesn't the free body diagram of right pulley has only one force and i.e. tension

No, it also has the forces of tension of $T_1$ and $T_3$ IF we assume that the string that passes around the right pulley does not slip through.

 

[haruspex]

Can you elaborate

I'm not sure how @BvU is defining x₁ etc. I suggest taking x₁ as the horizontal distance from m₁ to the centre of the right hand pulley, x₂ as the horizontal distance between the pulley centres, and x₃ as the vertical distance from m₃ to the centre of the left hand pulley.
What combination or combinations of these distances form constants? Think about the total string lengths. What do you get by differentiating such a combination twice?

 

[Crystal037]

No, it also has the forces of tension of $T_1$ and $T_3$ IF we assume that the string that passes around the right pulley does not slip through.

Why do we assume that the pulley doesn't slip through

 

[Crystal037]

I'm not sure how @BvU is defining x₁ etc. I suggest taking x₁ as the horizontal distance from m₁ to the centre of the right hand pulley, x₂ as the horizontal distance between the pulley centres, and x₃ as the vertical distance from m₃ to the centre of the left hand pulley.
What combination or combinations of these distances form constants? Think about the total string lengths. What do you get by differentiating such a combination twice?

Total sting length will be equal to x1+x2+x3 and after differentiating such a combination will find acceleration but I can't understand which combination will give acceleration of which system

 

[haruspex]

Total sting length will be equal to x1+x2+x3

Right.

after differentiating such a combination will find acceleration

Yes, the acceleration of the total string length. But that length is constant, right? So what are its derivatives equal to?

which combination will give acceleration of which system

Next is to consider how individual $\ddot x_i$, or other combinations, relate to the accelerations of the masses.

 

[Crystal037]

But the relative position would change and that would give velocity and acceleration

 

[nuclearfireball_42]

Thanks everybody for the reply :) . I've understood how to get the equation for acceleration of the bodies. I've also understood that T₁ is equal in magnitude to T₃. But the answer that I got doesn't seem to match the one given. Is it true for me to assume that the tension in the string connected to m₂ ( T₂ ) is 2 times the value of T and is in the same direction as the acceleration of m₂?

 

[Crystal037]

What is the answer?

 

[nuclearfireball_42]

What is the answer?

a₃ = 6m/s²

 

[haruspex]

a₃ = 6m/s²

I get a lower value. Please post your working.

 

[nuclearfireball_42]

I get a lower value. Please post your working.

Here's my working :

- The answer I get this time is the same with the one given. But is this a right way of solving this question?

 

[Physics lover]

I get a lower value. Please post your working.

hey haruspex,what are you getting Is it $2.5 m/s^2.$

 

[haruspex]

Here's my working :

View attachment 251293

- The answer I get this time is the same with the one given. But is this a right way of solving this question?

Looks good. (I found my mistake and now also get 6m/s².)

 

[Delta2]

Here's my working :

View attachment 251293

- The answer I get this time is the same with the one given. But is this a right way of solving this question?

I understand all the equations you use in the image except perhaps the one involving the accelerations which seems to be $-2a_2-a_1+a_3=0$ can you or someone else elaborate on this equation?

 

[haruspex]

I understand all the equations you use in the image except perhaps the one involving the accelerations which seems to be $-2a_2-a_1+a_3=0$ can you or someone else elaborate on this equation?

It seems each acceleration is being taken as positive in the likely direction of movement: 1 to the right, 2 to the left, 3 down.
For each unit of distance 1 advances, 3 drops 1 unit.
For each unit 2 advances, 3 drops 2 units.

 

[Delta2]

It seems each acceleration is being taken as positive in the likely direction of movement: 1 to the right, 2 to the left, 3 down.
For each unit of distance 1 advances, 3 drops 1 unit.
For each unit 2 advances, 3 drops 2 units.

Thanks @haruspex but I still don't understand
if "For each unit of distance 1 advances, 3 drops 1 unit" this means that $|\vec{a_1}|=|\vec{a_3}|$ right?
Also I don't understand how the following holds:
"For each unit 2 advances, 3 drops 2 units"

 

[haruspex]

Thanks @haruspex but I still don't understand
if "For each unit of distance 1 advances, 3 drops 1 unit" this means that $|\vec{a_1}|=|\vec{a_3}|$ right?

Only if 2 stays put. Similarly, 3 dropping 2 for each unit 2 moves is on the assumption that 1 stays put.
So in algebra it's $\frac{\partial a_3}{\partial a_1}=1$ etc.