Question

Below are eightcrates of different mass. (Intro1 figure) The crates are attached to massless ropes, asindicated 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 thesurface on which it slides is the same for all eight crates.

Rank the ropes on the basis of the force eachexerts on the crate immediately to its left.

Rank from largest to smallest. Torank items as equivalent, overlap them.

Answer 1

Concepts and reason

The concept used to solve this problem is kinetic frictional force, normal force.

Calculate the co-efficient of kinetic friction between each crate and the surface on which it slides. After that rank the ropes based on force exert from largest to smallest.

Fundamentals

The kinetic friction force is defined as the product of kinetic friction co-efficient to the normal force. It acts between the moving surfaces, and is the amount of retarding force.

The expression of kinetic friction is,

${f_k} = {\mu _k}N$

Here, ${f_k}$ is frictional force, ${\mu _k}$ is frictional force co-efficient and $N$ is normal force.

And the normal force is,

$N = mg$

Here, $m$ is mass, and $g$ is gravitational acceleration.

So,

${f_k} = {\mu _k}mg$

Calculate the kinetic friction force.

The expression of kinetic friction force is,

${f_k} = {\mu _k}mg$

Here, ${f_k}$ is frictional force, ${\mu _k}$ is frictional force co-efficient,$m$ is mass, and $g$ is gravitational acceleration.

It means the kinetic friction force is directly proportional to the mass.

${f_k} \propto m$

Calculate the kinetic friction for rope C.

Substitute $1{\rm{ kg}}$ for $m$.

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

Calculate the kinetic friction for rope F.

Substitute $3{\rm{ kg}}$ for $m$.

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

Calculate the kinetic friction for rope G.

Substitute $3{\rm{ kg}}$ for $m$.

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

Calculate the kinetic friction for rope H.

Substitute $5{\rm{ kg}}$ for $m$.

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

The expression of kinetic friction force is,

${f_k} = {\mu _k}mg$

Here, ${f_k}$ is frictional force, ${\mu _k}$ is frictional force co-efficient,$m$ is mass, and $g$ is gravitational acceleration.

Calculate the kinetic friction for rope B,

At rope B, the kinetic friction force due to both the caret B and C because both are connected and move in same direction.

So,

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

Substitute $3{\rm{ kg}}$ for $m$.

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

Calculate the kinetic friction for rope A,

At rope A, the kinetic friction force due to all the caret A, B and C because all are connected and move in same direction.

So,

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

Substitute ${\rm{6 kg}}$ for $m$.

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

Calculate the kinetic friction for rope E,

At rope E, the kinetic friction force due to both the caret E and F because both are connected and move in same direction.

So,

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

Substitute ${\rm{5 kg}}$ for $m$.

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

Calculate the kinetic friction for rope D.

At rope D, the kinetic friction force due to all the caret D, E and F because all are connected and move in same direction.

So,

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

Substitute ${\rm{6 kg}}$ for $m$.

$\begin{array}{c}\\{f_{k{\rm{D}}}} = {\mu _k}\left( {{\rm{6 kg}}} \right)g\\\\ = 6{\mu _k}g\\\end{array}$

Ans:

The largest to smallest kinetic friction force rank is,

${\rm{A}} = {\rm{D}} > {\rm{E}} = {\rm{H}} > {\rm{B}} = {\rm{F}} = {\rm{G}} > {\rm{C}}$