Help with a pulley problem with 3 masses

• nuclearfireball_42
In summary, the figure shows that if the string that passes around the right pulley doesn't slip through, then the total string length will be constant and the acceleration of the total string length will be found by differentiating the combination of x1, x2, x3.
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.

Crystal037
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 '

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?

BvU
nuclearfireball_42 said:
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.

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

Crystal037 said:
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##.

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

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

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

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

Crystal037 said:
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.

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

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

Can you elaborate

Crystal037 said:
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.

Crystal037 said:
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?

Delta2 said:
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

haruspex said:
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

Crystal037 said:
Total sting length will be equal to x1+x2+x3
Right.
Crystal037 said:
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?
Crystal037 said:
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.

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

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 said:
a3 = 6m/s2

nuclearfireball_42 said:
a3 = 6m/s2

haruspex said:
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?

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

nuclearfireball_42 said:
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.)

nuclearfireball_42 said:
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?

Delta2 said:
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.

nuclearfireball_42
haruspex said:
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"

Delta2 said:
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.

Delta2

1. How do I calculate the mechanical advantage of a pulley system with 3 masses?

The mechanical advantage of a pulley system is calculated by dividing the output force by the input force. In this case, the output force would be the combined weight of the 3 masses, and the input force would be the force applied to the pulley system.

2. What is the equation for determining the tension in the rope of a pulley system with 3 masses?

The equation for determining the tension in the rope of a pulley system with 3 masses is T = (m1 + m2 + m3)g, where T is the tension, m1, m2, and m3 are the masses, and g is the acceleration due to gravity (9.8 m/s^2).

3. How does the number of pulleys in a system affect the mechanical advantage?

The more pulleys in a system, the greater the mechanical advantage. This is because each additional pulley reduces the amount of force needed to lift the load. For a pulley system with 3 masses, the mechanical advantage would be equal to the number of pulleys in the system, which in this case is 3.

4. Can a pulley system with 3 masses have a mechanical advantage greater than 3?

No, a pulley system with 3 masses cannot have a mechanical advantage greater than 3. This is because the mechanical advantage of a pulley system is equal to the number of pulleys in the system, and in this case, there are only 3 pulleys.

5. How does friction affect the mechanical advantage of a pulley system with 3 masses?

Friction can decrease the mechanical advantage of a pulley system by reducing the efficiency of the system. This means that more input force is needed to lift the same load. However, the impact of friction on the mechanical advantage of a pulley system with 3 masses would depend on the specific design and materials used in the system.

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