Find the tension force in each pair

[Adris]

Homework Statement

You are presented with a circumstance in which three children are playing on a frozen pond. The three small children of mass 20.00 kg, 24.00 kg, and 16.00 kg, respectively, hold hands, and are pulled across a smooth frozen pond by a larger boy on skates, who pulls a horizontal rope being held by the first child. The skater pulls on the rope with a force of 135 N.

Example:
[iii: 16kg]----[ii: 24kg]----[i: 20kg]-------->135N

instead of being on the surface of a frozen pond, the same three children were being towed on a frozen grass surface, offering a friction coefficient μ = 0.17 with the soles of their feet. In this case, what would be the acceleration of the children? And what would the force with which each pair of children should hold hands, in order to ensure the chain is not broken?

Homework Equations

I got the friction and the acceleration.

Friction force : Ff = 0.17(60kg * 9.81) = 100N
acceleration: a = 135N - 100N / 60kg = 0.58m/s^2

The Attempt at a Solution

I use the following equations but i think it could be wrong.

pair 1: (135N-100N)-(20kg*0.58m/s)=23.4N
pair 2: (23.4)-(24kg*0.58m/s)=9.48N

I am unsure if these are right.

 

[haruspex]

pair 1: (135N-100N)-(20kg*0.58m/s)=23.4N

Please explain your reasoning. (It seems most unlikely that the tension so swiftly drops from 135N to 23.4N, no?)

 

[Adris]

Please explain your reasoning. (It seems most unlikely that the tension so swiftly drops from 135N to 23.4N, no?)

The 135N is the force(F) applied, the 100N is the friction(Ff). I did F-Ff to get 35. But that is what i saw online and was unsure as to whether it is correct.
I calculated the tension without friction and with 2.25m/s acceleration, which gave me these...Tension without friction for pair1: 90N; pair2: 36N

I want to know the formula/guidance to get the correct result.

 

[haruspex]

The 135N is the force(F) applied, the 100N is the friction(Ff). I did F-Ff to get 35.

Yes, that is the net force on the three children, but it does not explain the thinking behind your equation.
An explanation would look like: "Consider the subsystem consisting of (a particular child, or two adjacent children); the horizontal forces are ... The ΣF=ma equation for this subsystem is..."

 

[Adris]

Yes, that is the net force on the three children, but it does not explain the thinking behind your equation.
An explanation would look like: "Consider the subsystem consisting of (a particular child, or two adjacent children); the horizontal forces are ... The ΣF=ma equation for this subsystem is..."

The horizontal force pulling is 135N. I am trying to find out tension force between the children holding hands, which has 2 pairs.
The system is a larger child pulling a rope , which consists of three adjacent children.

 

[haruspex]

The system is a larger child pulling a rope , which consists of three adjacent children.

A system is whatever you decide. Yes, the total system is as you say, but you can identify (sub)systems consisting of a rigid body, or any number of rigid bodies that behave as one.
For the chosen subsystem, there is a total mass and a net applied force (in some chosen direction), leading to the equation ΣF=ma for that direction. For this purpose, you can ignore forces internal to the subsystem since they consist of equal and opposite pairs that cancel. This is the standard way to solve all such problems. Writing down other equations that just seem right to you is a quick way to go wrong.

In the present problem, consider, e.g. the last of the towed children. This is a good place to start because there are fewer forces. What are the horizontal forces on that child? What is the acceleration? What equation can you write?

 

[Adris]

A system is whatever you decide. Yes, the total system is as you say, but you can identify (sub)systems consisting of a rigid body, or any number of rigid bodies that behave as one.
For the chosen subsystem, there is a total mass and a net applied force (in some chosen direction), leading to the equation ΣF=ma for that direction. For this purpose, you can ignore forces internal to the subsystem since they consist of equal and opposite pairs that cancel. This is the standard way to solve all such problems. Writing down other equations that just seem right to you is a quick way to go wrong.

In the present problem, consider, e.g. the last of the towed children. This is a good place to start because there are fewer forces. What are the horizontal forces on that child? What is the acceleration? What equation can you write?

Acceleration = 0.58m/s
For the last child who has a mass of 16kg. I got... Horizontal force = (135-100)-(16*0.58) =25.72N

Equation for above... Tension =(F-Ff) - (ma)

Please understand, I am fairly new to this.

 

[haruspex]

For the last child who has a mass of 16kg. I got... Horizontal force = (135-100)-(16*0.58)

No, you need just the forces that act directly on the third child. In ΣF=ma, never include forces that do not directly act on the mass m.