Chapter 6, Problem 22A

Boxes of various masses are on a friction-free level table.

Rank each of the following from greatest to least.

a. the net forces on the boxes

b. the accelerations of the boxes

To determine

To rank: The net force on the box of various masses from greatest to least.

Answer to Problem 22A

The ranking of forces from greatest to least is, $\text{D}>\text{A}=\text{B}=\text{C}$

Explanation of Solution

Given:

The given boxes on a friction-free level table are shown below.

  

Formula used:

Net force on an object, if two forces are travelling in same direction is,

  $F_{net}=F_1+F_2$

Net force on an object, if two forces are travelling in same direction is,

  $F_{net}=F_1-F_2$

Calculation:

Right side is considered as positive and left side is considered as negative.

Consider case (A)

Horizontal force acting along positive x-direction = F_x = 10 N

Horizontal force acting along negative x-direction F_-x = 5 N

The net force acting on the box is,

  $\begin{array}{c}{\text{F}}_{\text{net}}{\text{=F}}_{\text{x}}-{\text{F}}_{\text{-x}}\\ {\text{F}}_{\text{net}}\text{=}\text{10}\text{N}-\text{5}\text{N}\\ {\text{F}}_{\text{net}}\text{=}\text{5}\text{N}\end{array}$

Net force acting on box A is 5 N along positive direction of x-axis.

Consider case (B)

Horizontal force acting along positive x-direction = F_x = 15 N

Horizontal force acting along negative x-direction F_-x = 10 N

The net force acting on the box is,

  $\begin{array}{l}{\text{F}}_{\text{net}}{\text{=F}}_{\text{x}}-{\text{F}}_{\text{-x}}\\ {\text{F}}_{\text{net}}\text{=}\text{15}\text{N}-10\text{N}\\ {\text{F}}_{\text{net}}\text{=}\text{5}\text{N}\end{array}$

Net force acting on box is 5 N along positive direction of x-axis.

Consider case (C).

Horizontal force acting along positive x-direction= F_x = 15 N

Horizontal force acting along negative x-direction F_-x = 10 N

The net force acting on the box is,

  $\begin{array}{l}{\text{F}}_{\text{net}}\text{=}\text{15}\text{N}-10\text{N}\\ {\text{F}}_{\text{net}}\text{=}\text{5}\text{N}\end{array}$

Net force acting on box is 5 N along positive direction of x-axis.

Consider case (D).

Horizontal force acting along positive x-direction= F_x = 15 N

Horizontal force acting along negative x-direction F_-x = 5 N

The net force acting on the box is,

  $\begin{array}{l}{\text{F}}_{\text{net}}{\text{=F}}_{\text{x}}-{\text{F}}_{\text{-x}}\\ {\text{F}}_{\text{net}}\text{=}\text{15}\text{N}-5\text{N}\\ {\text{F}}_{\text{net}}\text{=}10\text{N}\end{array}$

Net force acting on box is 10 N along positive direction of x-axis.

Conclusion:

Therefore, from greatest to least, the net force is $\text{D}>\text{A}=\text{B}=\text{C}$

To determine

To rank: The net acceleration of the box of various masses from greatest to least.

Answer to Problem 22A

Ranking of acceleration from greatest to least is, Case (A) = Case(C) > Case (B) = Case (D)

Explanation of Solution

Given:

Given boxes on the friction less lever table is shown below.

  

Formula used:

Acceleration of box is,

  $\therefore \text{a}\text{=}\frac{{\text{F}}_{\text{net}}}{\text{M}}$

Where, F_net is the net force acting on the box

Calculation:

Consider Case (A):

Net force acting on box is 5 N along positive direction of x-axis.

Acceleration of 5 kg box along horizontal direction is,

  $\begin{array}{l}\therefore \text{a}\text{=}\frac{{\text{F}}_{\text{net}}}{\text{M}}\left(\because {\text{F}}_{\text{net}}\text{=}\text{Ma}\right)\\ \therefore \text{a}\text{=}\frac{\text{5}\text{}\text{N}}{5\text{kg}}\\ \therefore \text{a}\text{=}1{\text{m/s}}^{\text{2}}\end{array}$

Acceleration of 5 kg box along positive x-direction is 1 m/s²

Consider Case (B):

Net force acting on box 5 N along positive direction of x-axis.

Acceleration of 10 kg box along horizontal direction is,

  $\begin{array}{l}\therefore \text{a}\text{=}\frac{{\text{F}}_{\text{net}}}{\text{M}}\left(\because {\text{F}}_{\text{net}}\text{=}\text{Ma}\right)\\ \therefore \text{a}\text{=}\frac{\text{5}\text{}\text{N}}{10\text{kg}}\\ \therefore \text{a}\text{=}0.5{\text{m/s}}^{\text{2}}\end{array}$

Acceleration of 10 kg box along positive x-direction is 0.5 m/s²

Consider Case (C):

Net force acting on box is 5 N along positive direction of x-axis.

Acceleration of 5 kg box along horizontal direction is,

  $\therefore \text{a}\text{=}\frac{{\text{F}}_{\text{net}}}{\text{M}}$

  $\begin{array}{l}\therefore \text{a}\text{=}\frac{{\text{F}}_{\text{net}}}{\text{M}}\left(\because {\text{F}}_{\text{net}}\text{=}\text{Ma}\right)\\ \therefore \text{a}\text{=}\frac{\text{5}\text{}\text{N}}{5\text{kg}}\\ \therefore \text{a}\text{=}1{\text{m/s}}^{\text{2}}\end{array}$

Acceleration of 5 kg box along positive x-direction is 1 m/s²

Consider Case (D):

Net force acting on box is 10 N along positive direction of x-axis.

Acceleration of 20 kg box along horizontal direction is,

  $\begin{array}{l}\therefore \text{a}\text{=}\frac{{\text{F}}_{\text{net}}}{\text{M}}\left(\because {\text{F}}_{\text{net}}\text{=}\text{Ma}\right)\\ \therefore \text{a}\text{=}\frac{10\text{}\text{N}}{\text{20}\text{kg}}\\ \therefore \text{a}\text{=}0.5{\text{m/s}}^{\text{2}}\end{array}$

Acceleration of 20 kg box along positive x-direction is 0.5 m/s²

Conclusion:

Therefore, acceleration from greatest to least, is

Case (A) = Case(C) > Case (B) = Case (D)