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Below are eight crates of different mass. (Figure 1) The crates are attached to massless ropes, as indicated in the picture, where the ropes are marked by letters. Each crate is being pulled to the right at the same constant speed. The coefficient of kinetic friction between each crate and the surface on which it slides is the same for all eight crates.

Rank the ropes on the basis of the force each exerts on the crate immediately to its left. Rank from largest to smallest. To rank items as equivalent, overlap them.

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Revelant knowledge

*= F*

_{k} = μ_{k}F_{n}
The net force on each of the crates is 0 since they are all travelling at the same constant velocity (it doesn’t have to be equal, just constant). The tension in the pulling rope is thus perfectly matched by the frictional force in the horizontal direction, that is F = μ

_{k}mgThe tension in the rope is proportional to the mass that the rope is pulling because the coefficient of kinetic friction is the same and g = 9.81 m / s is constant. When we divide the rope tension by the corresponding letter, we get:

A=D>E=H>B=F=G>CThis problem can be solved using the concept of normal force and kinetic frictional force.

Calculate the coefficient of kinetic friction between each container and the surface where it slides. Then rank the ropes according to how much force they exert, from largest to least.

The kinetic friction force can be defined as the sum of the normal force and kinetic friction coefficient. It acts between moving surfaces and is the amount or retarding force.

Kinetic friction can be described as:

[katex]{f_k} = {\mu _k}N[/katex]

Here, [katex]{f_k}[/katex]

The normal force is:

[katex]N = mg[/katex]

Here, [katex]m[/katex]

So,

[katex]{f_k} = {\mu _k}mg[/katex]

Calculate the kinetic friction for your application.

The expression for kinetic friction force can be described as:

[katex]{f_k} = {\mu _k}mg[/katex]

Here, [katex]{f_k}[/katex]

This means that the kinetic friction force of the mass is directly proportional.

[katex]{f_k} \propto m[/katex]

Calculate the kinetic friction of rope C.

Substitute “equa_tag_8”

[katex]\begin{array}{c}\\{f_{k{\rm{C}}}} = {\mu _k}\left( {1{\rm{ kg}}} \right)g\\\\ = {\mu _k}g\\\end{array}[/katex]

Calculate the kinetic friction of rope F.

Substitute “equa_tag_10”

[katex]\begin{array}{c}\\{f_{k{\rm{F}}}} = {\mu _k}\left( {3{\rm{ kg}}} \right)g\\\\ = 3{\mu _k}g\\\end{array}[/katex]

Calculate the kinetic friction of rope G.

Substitute “equa_tag_10”

[katex]\begin{array}{c}\\{f_{k{\rm{G}}}} = {\mu _k}\left( {3{\rm{ kg}}} \right)g\\\\ = 3{\mu _k}g\\\end{array}[/katex]

Calculate the kinetic friction of rope H.

Substitute “equa_tag_14”

[katex]\begin{array}{c}\\{f_{k{\rm{H}}}} = {\mu _k}\left( {{\rm{5 kg}}} \right)g\\\\ = 5{\mu _k}g\\\end{array}[/katex]

The expression for kinetic friction force can be described as:

[katex]{f_k} = {\mu _k}mg[/katex]

Here, [katex]{f_k}[/katex]

Calculate the kinetic friction of rope B.

Rope B is subject to the kinetic friction force of both the caret C and B because they are both connected and move in the same direction.

So,

[katex]\begin{array}{c}\\{\rm{total mass}}\left( m \right) = \left( {{\rm{2 kg}}} \right) + \left( {{\rm{1 kg}}} \right)\\\\ = 3{\rm{ kg}}\\\end{array}[/katex]

Substitute “equa_tag_10”

[katex]\begin{array}{c}\\{f_{k{\rm{B}}}} = {\mu _k}\left( {{\rm{3 kg}}} \right)g\\\\ = 3{\mu _k}g\\\end{array}[/katex]

Calculate the kinetic friction of rope A.

Rope A: The kinetic friction force is due to all the carets A, B, and C. They are all connected and move in the same direction.

So,

[katex]\begin{array}{c}\\{\rm{total mass}}\left( m \right) = \left( {{\rm{2 kg}}} \right) + \left( {{\rm{1 kg}}} \right) + \left( {{\rm{3 kg}}} \right)\\\\ = 6{\rm{ kg}}\\\end{array}[/katex]

Substitute “equa_tag_22”

[katex]\begin{array}{c}\\{f_{k{\rm{A}}}} = {\mu _k}\left( {{\rm{6 kg}}} \right)g\\\\ = 6{\mu _k}g\\\end{array}[/katex]

Calculate the kinetic friction of rope E.

The kinetic friction force at rope E is due to both the caret F and E because they are both connected and move in the same direction.

So,

[katex]\begin{array}{c}\\{\rm{total mass}}\left( m \right) = \left( {{\rm{2 kg}}} \right) + \left( {{\rm{3 kg}}} \right)\\\\ = 5{\rm{ kg}}\\\end{array}[/katex]

Substitute “equa_tag_25”

[katex]\begin{array}{c}\\{f_{kE}} = {\mu _k}\left( {{\rm{5 kg}}} \right)g\\\\ = 5{\mu _k}g\\\end{array}[/katex]

Calculate the kinetic friction of rope D.

Rope D is the location where the kinetic friction force from all the carets D, E, and F occurs. They are all connected and move in the same direction.

So,

[katex]\begin{array}{c}\\{\rm{total mass}}\left( m \right) = \left( {{\rm{3 kg}}} \right)\left( {{\rm{2 kg}}} \right) + \left( {{\rm{1 kg}}} \right)\\\\ = 6{\rm{ kg}}\\\end{array}[/katex]

Substitute “equa_tag_22”

[katex]\begin{array}{c}\\{f_{k{\rm{D}}}} = {\mu _k}\left( {{\rm{6 kg}}} \right)g\\\\ = 6{\mu _k}g\\\end{array}[/katex]Ans:

This is the kinetic friction force rank ranging from the largest to the smallest.

[katex]{\rm{A}} = {\rm{D}} > {\rm{E}} = {\rm{H}} > {\rm{B}} = {\rm{F}} = {\rm{G}} > {\rm{C}}[/katex]