Question

Part A.)Six boxes held at rest against identical walls.
Rank the boxes on the basis of the magnitude of the normal force acting on them.
Rank from largest to smallest. To rank items as equivalent, overlap them.

1.)130N--->7kg

2.)150N--->1kg

3.)150N--->7kg

4.)120N--->3kg

5.)140N--->5kg

6.)140N--->3kg

(Since the boxes are at rest, Newton's 2nd law dictates that the horizontal forces on each box must add up to zero. You can use this information to determine the normal forces. If two boxes are both pushed against the wall by the same force, then they should experience the same normal force.)

Part B.) Rank the boxes on the basis of the frictional force acting on them.

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

1.)130N--->7kg

2.)150N--->1kg

3.)150N--->7kg

4.)120N--->3kg

5.)140N--->5kg

6.)140N--->3kg

Answer 1

Concepts and reason

The concept required to solve the given question is Newton’s Second Law.

Initially, the ranking of boxes on the basis of normal force and frictional force acting on them can be done by applying Newton’s Second Law and equilibrium condition for force.

Later, since, the box is stationary, the horizontal forces acting on the box will sum up to zero, thus giving the value of normal force acting on the box.

Since, the box is stationary, the vertical forces acting on the box will sum up to zero, thus finally giving the value of frictional force acting on the box.

Fundamentals

According, to the Newton’s first law, when a particle is at rest or is moving with constant velocity in an inertial frame of reference, the net force acting on it that is the vector sum of all the forces acting on it must be zero. In mathematical form this is expressed as,

$\sum {\vec F} = 0$

Here, $F$ is the force.

The Newton’s second law states that the net force on an object is the product of mass of the object and final acceleration of the object. The equation of the Newton’s second law is,

$\sum {\vec F = m\vec a}$

Here, $\sum {\vec F}$ is the net force on the object, $m$ is mass of the object, and $\vec a$ is the acceleration of the object.

The equilibrium condition for the force gives,

$\Sigma F = 0$

Here, $F$ is force.

(A)

The normal force will be applied on the box in the horizontal direction that is along the x axis due to the surface supporting the box.

Hence, the horizontal force acting on the box will be,

$F - N = 0$

Here, $F$ is the force acting on the box and $N$ is the normal force.

Rearrange the above equation.

$N = F$

The forces in descending order is $150{\rm{ N}} > {\rm{140 N}} > 130{\rm{ N}} > 120{\rm{ N}}$.

Since, the normal force is equal to the force acting on the box, the same order will apply for ranking of normal forces acting on the box.

(B)

The net vertical force acting on the box will sum up to zero.

$\sum {{F_{\rm{y}}}} = 0$

The weight of the box will act downwards that is towards the center of earth and the frictional force acts opposite to the direction in which the box moves to oppose its motion.

The vertical force acting on the box will be,

$mg - f = 0$

Here, $f$ is the frictional force, $m$ is the mass and $g$ is the acceleration due to gravity.

Rearrange the above equation.

$f = mg$

Here, $g$ is a constant. Hence the magnitude of frictional force will only depend on the mass of the object.

The mass of the boxes in descending order is $7{\rm{ kg}} > 5{\rm{ kg}} > 3{\rm{ kg}} > 1{\rm{ kg}}$.

The above given order will apply to the ranking of boxes on the basis of frictional forces acting on them.

Ans: Part A

The ranking of the normal force acting on the boxes will be $\begin{array}{c}\\150{\rm{ N}}\left( {1{\rm{ kg}}} \right) = 150{\rm{ N}}\left( {7{\rm{ kg}}} \right) > 140{\rm{ N}}\left( {5{\rm{ kg}}} \right)\\\\140{\rm{ N}}\left( {5{\rm{ kg}}} \right) = 140{\rm{ N}}\left( {3{\rm{ kg}}} \right) > 130{\rm{ N}}\left( {7{\rm{ kg}}} \right) = 130{\rm{ N}}\left( {1{\rm{ kg}}} \right) > 120{\rm{ N}}\left( {3{\rm{ kg}}} \right)\\\end{array}$