Chapter 13.1, Problem 4PP

To determine

To Calculate:The net forcethatwill be exerted on a front door to a house and its direction.

Answer to Problem 4PP

  $\text{2}{\text{.7×10}}^{\text{4}}\text{N}$

Directed from inside of the house outward.

Explanation of Solution

Given data:

Height of a door, $l=195\text{cm}\text{(1}\text{.95}\text{m)}$

Width of a door, $w=91\text{cm}\text{(0}\text{.91}\text{m)}$

The pressure drops by $15\%$ .

Formula used:

A force (per unit area) that acts on any object is defined by pressure. Force to calculate pressure difference of a particular area, formula is

  $F=P_{diff}A=P_{diff}lw$

Calculation:

The pressure difference across the door is $P_{diff} =\left(15\%\right)\left(P_{atm}\right)$

Here, $P_{atm}$ is atmospheric pressure which is $1.0\times {10}^5 Pa$

  $\begin{array}{c}{\text{P}}_{diff}\text{=}\left(\text{0}\text{.15}\right)\text{(1}{\text{.0×10}}^{\text{5}}\text{Pa)}\\ \text{=1}{\text{.5×10}}^4\text{ Pa}\end{array}$

  $F=P_{diff}lw$ ,

  $\text{F}\text{=}\text{(1}{\text{.5×10}}^{\text{4}}\text{ Pa) }\left(\text{1}\text{.95 m}\right)\left(\text{0}\text{.91 m}\right)\text{= 2}{\text{.7×10}}^{\text{4}}\text{ N}$

Thus, the exerted force is directed from inside of the house outward.

Conclusion:

The force $\text{2}{\text{.7×10}}^{\text{4}}\text{ N}$ is directed from inside of the house outward.