Chapter 7, Problem 21A

Three sets of double boxes rest on a table.

Rank the following from greatest to least.

a. the normal force that the table exerts on the sets.

b. the normal force exerted by the bottom block on the top block

To determine

The rank of the normal force that the table exerts on the sets from greatest to least.

Answer to Problem 21A

The rank of the normal force that the table exerts on the sets from greatest to least is A=B=C.

Explanation of Solution

Formula used:

The expression for the normal force as follows:

  $F=W$

Here, $W$ is the weight.

Calculation:

Consider the acceleration due to gravity as $g=9.8\text{m}/{\text{s}}^{\text{2}}$ .

Sketch the Free Body diagram of the sets as shown in figure.

  

Refer to figure:

The normal force on set A as follows:

  $\begin{array}{l}F=W\\ F_{\text{NA}}=mg\\ F_{\text{NA}}=\left(1.0\text{ kg}\times 9.8\text{m}/{\text{s}}^{\text{2}}\right)+\left(1.4\text{ kg}\times 9.8\text{m}/{\text{s}}^{\text{2}}\right)\\ F_{\text{NA}}=23.5\text{ N}\end{array}$

The normal force on set B as follows:

  $\begin{array}{l}F=W\\ F_{\text{NB}}=mg\\ F_{\text{NB}}=\left(1.4\text{ kg}\times 9.8\text{m}/{\text{s}}^{\text{2}}\right)+\left(1.0\text{ kg}\times 9.8\text{m}/{\text{s}}^{\text{2}}\right)\\ F_{\text{NB}}=23.5\text{ N}\end{array}$

The normal force on set C as follows:

  $\begin{array}{l}F=W\\ F_{\text{NC}}=mg\\ F_{\text{NC}}=\left(1.2\text{ kg}\times 9.8\text{m}/{\text{s}}^{\text{2}}\right)+\left(1.2\text{ kg}\times 9.8\text{m}/{\text{s}}^{\text{2}}\right)\\ F_{\text{NC}}=23.5\text{ N}\end{array}$

Conclusion:

Thus, the rank of the normal force that the table exerts on the sets from greatest to least is A=B=C.

To determine

The rank of the normal force exerted by the bottom block on the top block from greatest to least.

Answer to Problem 21A

The rank of the normal force exerted by the bottom block on the top block from greatest to least is $B>C>A$ .

Explanation of Solution

Formula used:

The expression for Newton’s second law of acceleration as follows:

  $F=ma$

Here, $m$ is the mass and $a$ is the acceleration.

Calculation:

Consider the acceleration due to gravity as $g=9.8\text{m}/{\text{s}}^{\text{2}}$ .

Sketch the Free Body diagram of the force by the bottom block on the top block as shown in Figure.

  

For the two block system:

Refer to Figure:

The normal force on set A as follows:

  $\begin{array}{l}F=W\\ F_{\text{NA}}=mg\\ F_{\text{NA}}=\left(1.0\text{ kg}\times 9.8\text{m}/{\text{s}}^{\text{2}}\right)\\ F_{\text{NA}}=9.8\text{ N}\end{array}$

The normal force on set B as follows:

  $\begin{array}{l}F=W\\ F_{\text{NB}}=mg\\ F_{\text{NB}}=\left(1.4\text{ kg}\times 9.8\text{m}/{\text{s}}^{\text{2}}\right)\\ F_{\text{NB}}=13.7\text{ N}\end{array}$

The normal force on set C as follows:

  $\begin{array}{l}F=W\\ F_{\text{NC}}=mg\\ F_{\text{NC}}=\left(1.2\text{ kg}\times 9.8\text{m}/{\text{s}}^{\text{2}}\right)\\ F_{\text{NC}}=11.8\text{ N}\end{array}$

Conclusion:

Thus, the rank of the normal force exerted by the bottom block on the top block from greatest to least is $B>C>A$ .