That classic 3-blocks question

  • Thread starter Celestiela
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In summary, the problem involves a system of three masses connected by strings and pulled by a gigantic hand with a force of 65.4N. To find the net acceleration, we use Newton's second law and divide the force by the total mass of the system. Tension in the strings plays a crucial role in determining the forces on each mass. The force on M1 is simply the force pulling it to the right, while the force on M2 is the difference between the tension in the string and the force exerted by M1. The force on M3 is the net force on the entire system, which includes the forces from M1 and M2. In order to find the tension between M1 and M2, we
  • #1
Celestiela
24
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OK 3 blocks attached by 3 strings. stick figure picture follows:

M1---(T1)---M2---(T2)---M3---(T3)---(Gigantic Hand)

T3 = 65.4N
M1=11.2 kg
M2=24 kg
M3=30.8 kg

Ok, I'm not going to post what the question asks because I'm not really concerned with getting the homework problem right. I just need help understanding the system. I have no idea what to place where. I know I have to put ma=ma somewhere, but it's saturday and I left my brain at school. :zzz:
 
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  • #2
Hehe. Ok, I'm guessing the hand is pulling the whole system, which means the whole system has some net acceleration. This can be found directly from Newton's second law. The net acceleration of the body is the sum of the external forces acting on it divided by its mass. Clump the mass of the three together and divide by the force the hand exerts to find the net acceleration.

The most useful thing in these types of problems is to know what tension really is. It is a force exerted on a string by objects on either end. The amount of force that BOTH ends share will be the tension in the string, so for example,

If a string has a mass pulling it to the left at 10N and another mass pulling it on the right at 5N, what is the tension in the string?

Both ends of the string have a minimum of 5N acting on them, so that is the tension. The other 5N from the left will result in a net force on the entire system.

Onto your problem
You are given the tension in the third rope, and the three masses. If you read above, you can figure out that the force applied by the hand minus the force from the other three blocks is 65.4N

The net acceleration is

[tex] \vec{a} = \frac{\vec{F}}{m_{net}} = \frac{65.4}{66} = 0.99 \ or \ about \ 1m/s^2[/tex]

The force on M1 by Newton's 2nd Law is

[tex] \vec{F}_{m1} = m\vec{a} = (11.2kg)(1m/s^2) = 11.2N [/tex]

Where we rounded the acceleration. The only force acting on M1 is the force pulling it to the right. There are no forces to its left.

Force on mass 2 by Newton's second law is
edit - fixed
[tex] T_{2} - T_{1} = ma = 24N \ with \ T_1 = 11.2N [/tex]


The net force on block 2 is the sum of the hand pulling it to the right (the net force on the system) and M1 pulling it to the left.


Can you do the third mass?
 
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  • #3
So I do Fm3=(m3*a)-Fm2 right? then I get 18N.

But 18N + 11.2N + 12.8N doesn't equal 65.4? Is that ok?

And then we find the tension of the string in the middle by subtracting one force from the other? For example, what would the tension be between M1 and M2?

Thanks for helping on a saturday! :!)
 
  • #4
Remember, M3 is dragging BOTH M1 and M2.
 
  • #5
I'm still confused.

Tension is just the net force between the two bodies right? So what's the relationship between the forces acting on the bodies and the forces acting between the bodies?

I think what I don't understand is what the numbers mean. If those bodies are accelerating at 1 m/s^2, then why do I have to subtract those forces of M1 and M2? It's accelerating, it's moving regardless of how many blocks are behind it.
 
  • #6
whozum said:
If a string has a mass pulling it to the left at 10N and another mass pulling it on the right at 5N, what is the tension in the string?

Both ends of the string have a minimum of 5N acting on them, so that is the tension. The other 5N from the left will result in a net force on the entire system.

Onto your problem
You are given the tension in the third rope, and the three masses. If you read above, you can figure out that the force applied by the hand minus the force from the other three blocks is 65.4N

While I expect you know what you meant to say, there are a couple of things you did say that you need to look at again.

If a string has a mass pulling it to the left at 10N and another mass pulling it on the right at 5N, what is the tension in the string?

The only way this could happen is if there is a tension gradient in the string, and that could only happen if the string had mass, which is not the case in this problem. A massless string experiencing different forces at both ends would have infinite acceleration. In this problem, and in your calculations, the tension in each string is constant and serves as a mechanism for two masses to exert equal and opposite forces upon one another.

Onto your problem
You are given the tension in the third rope, and the three masses. If you read above, you can figure out that the force applied by the hand minus the force from the other three blocks is 65.4N

Your statement suggests that 65.4N is the difference between the force applied by the hand and forces to the left, which fits with your description of tension, but that description is not correct. I don't think you want to talk about 65.4N as being the difference between the force of the hand and other forces. 65.4 N is one force (or more correctly one action-reaction pair of forces), the force applied by the hand (and the reaction force T3), and that is all. If the "object" experiencing the force is the group of three masses, then your statement is correct, but only because the the 65.4N is the only external force acting, so the "force from the other three blocks" means external forces acting on those blocks which is zero. If other forces (friction for example) were acting on the blocks, and the blocks were considered as one object, the 65.4N would still be 65.4N and T3, but the net external force would be reduced and the acceleration would be less. In this frictionless problem, if you are looking at M3, T3 minus the force from the other blocks is not 65.4N; T3 is 65.4N and T3 - T2 is the net force acting on block M3.

I think what the OP needs to see explicitly is something that is contained in your analysis, but not so obviously stated up front, which is that each block is acted upon by one or two forces and that the blocks share a common acceleration

T1 = M1 a
T2 - T1 = M2 a
T3 - T2 = M3 a
(T3 = force applied by gigantic hand = 65.4N)

Add the first three equations to arrive at your starting point.

T3 = (M1 + M2 + M3) a

a = T3/(M1 + M2 + M3)

Then step through the first three equations to find the tensions as you have outlined.

If there were friction forces f1, f2 and f3 involved, the equations would become

T1 - f1 = M1 a
T2 - T1 - f2 = M2 a
T3 - T2 - f3 = M3 a
(T3 = force applied by gigantic hand = 65.4N)

EDIT

This is not correct

[tex] \vec{F}_{m2} = m\vec{a} - \vec{F_{m1}} = (24kg)(1m/s^2) - (11.2N) = 12.8N[/tex]

It should be

[tex] T2 - T1 = (24kg)(1m/s^2)[/tex]

[tex] T2 = (24kg)(1m/s^2) + T1 [/tex]
 
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  • #7
Ok, the net acceleration of each block as we established earlier is 1m/s^2.

The net force on each block is therefore the block's mass multiplied by 1m/s^2.

Net force on M1: 11.2N
Net force on M2: 24N
Net force on M3: 30.8N

#1 is easy since the force is just in one direction. The force on it is 11.2N, and this is teh tension in the string that is pulling it.

#2, The net force is 24N, but it has two forces pulling it, the tension from M1, and the pull from the hand. The sum of these is 24N.

[tex] F_{net} = ma = 24N [/tex]

The force experienced by M2 is due to the tension in the rope between it and M3.
[tex] T_{m32} - T_{m12} = ma = 24N [/tex]

[itex] T_{m12} = 11.2N [/itex], so [itex] T_{m3} [/itex] (tension between M2 and M3) is 35.2N

edit: I've made a mistake as OlderDan pointed out, so this information might not be correct.
 
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  • #8
See my edit in my post about T2. I got to go- I'm sure you will sort this out.
 
  • #9
Okay the mistake I made is what I thought it was initially, I was actually in the process of amending it when you posted. My post before this is the correct solution.

I apologize for my mistake.
 

Related to That classic 3-blocks question

1. What is "That classic 3-blocks question"?

"That classic 3-blocks question" refers to a famous problem in mathematics and computer science that involves arranging three blocks of different sizes in a specific order using only three pegs and following certain rules.

2. What are the rules of "That classic 3-blocks question"?

The rules of "That classic 3-blocks question" are as follows:

  • Only one block can be moved at a time.
  • Each move consists of taking the upper block from one of the pegs and placing it on top of the stack on another peg.
  • A larger block cannot be placed on top of a smaller block.

3. What is the goal of "That classic 3-blocks question"?

The goal of "That classic 3-blocks question" is to move the blocks from their starting positions on one peg to another peg, while following the rules, in the fewest number of moves possible.

4. How many moves does it take to solve "That classic 3-blocks question"?

The minimum number of moves required to solve "That classic 3-blocks question" is 7.

5. What is the significance of "That classic 3-blocks question"?

"That classic 3-blocks question" is a classic example used in teaching problem solving and algorithmic thinking. It also has applications in the fields of mathematics, computer science, and game theory.

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