Chapter 4, Problem 11P

To determine

a)

The weight and normal force on the box kept at the table.

Answer to Problem 11P

Solution:

Both weight and Normal force are.

Explanation of Solution

Given:

$m=20\text{ kg}$………………………………………………………………………………………………………………(Mass of the box)

$a=0{\text{ m/s}}^{\text{2}}$………………………………………………………………………………………………………………(Rest box has no acceleration)

Formula Used:

$\left(I\right)W=mg$

$(II){{\displaystyle \sum }}^{\text{}}F =ma$

Calculation:

(I)

$\begin{array}{l}W=mg \\ W=\left(20\right)\left(9.8\right) \\ W=196\text{ N}\end{array}$

(II)From, Free body diagram of the box.

$\begin{array}{l} {\displaystyle \sum F} =ma \\ N–W=m\left(0\right)\\ \begin{array}{l} N-196=0 \hfill \\ N=196\text{ N}\hfill \end{array}\end{array}$

To determine

b)

The normal force on the box by the table and Normal force on the box by $\text{20 kg}$box.

Answer to Problem 11P

Solution:

The Normal forces on lower and upper boxes are 294 N and 98 N respectively.

Explanation of Solution

Givens:

$m_1 =20\text{ kg}$………………………………………………………………………………………………………………(Mass of the lower box)

$m_2 =\text{ 1}0\text{ kg}$……………………………………………………………………………………………………………… (Mass of the box)

$a=0{\text{ m/s}}^{\text{2}}$………………………………………………………………………………………………………………(Rest boxes have no acceleration)

Formula:

${\displaystyle \sum F} =ma$

Calculations:

$N_1$$m_{1}g$

(I)From, Free body diagram of lower mass.

$\begin{array}{l}{\displaystyle \sum F} =ma \\ N_2-m_{2}g=m(0)\\ N_2-\left(10\right)\left(9.8\right)=0\\ N_2=98\text{N}\end{array}$

(II)From, Free body diagram of Upper mass.

$\begin{array}{l}{{\displaystyle \sum }}^{\text{}}F =ma \hfill \\ N_2-m_{2}g=m(0)\hfill \\ N_2-\left(10\right)\left(9.8\right)=0\hfill \\ N_2=98\text{N}\hfill \end{array}$

$m_2$g