Question

Anita holds her physics textbook and complains that it is too heavy. Andrew says that her hand should exert no force on the book because the atmosphere pushes up on it and balances the downward pull of Earth on the book (the book's weight). Jim disagrees. He says that the atmosphere presses down on things and that is why they feel heavy. Approximately how large is the force that the atmosphere exerts on the bottom of the book? Suppose that the dimensions of the book are 0.22 m times 0.28m, the mass of the book is 3 kg and pressure is 1.0 times 10^5 N/m^2.

Answer 1

The force that the atmosphere exerts on the bottom of the book is about 6160 Newton

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Further explanation

The basic formula of pressure that needs to be recalled is:

Pressure = Force / Cross-sectional Area

or symbolized:

$\large {\boxed {P = F \div A} }$

P = Pressure (Pa)

F = Force (N)

A = Cross-sectional Area (m²)

Let us now tackle the problem !

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Given:

Area of the book = A = 0.22 × 0.28 = 0.0616 m²

Mass of the book = m = 3 kg

Atmospheric Pressure = P = 1.0 × 10⁵ N/m²

Asked:

Force on the bottom of the book = F = ?

Solution:

$P = F \div A$

$F = P \times A$

$F = 1.0 \times 10^5 \times 0.0616$

$F = 6160 \texttt{ Newton}$

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The force that the atmosphere exerts on the bottom of the book is approximately equal to the force that the atmosphere exerts on the top of the book. Therefore these forces will be cancel out. If Anita holds the book , she only has to exert force equal to the weight of the book.

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Learn more

• Minimum Coefficient of Static Friction : brainly.com/question/5884009
• The Pressure In A Sealed Plastic Container : brainly.com/question/10209135
• Effect of Earth’s Gravity on Objects : brainly.com/question/8844454

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Answer details

Grade: High School

Subject: Physics

Chapter: Pressure

Answer 2

Answer:

The force exerted is 6189.4 N

Solution:

As per the question:

Area of the book, A = $0.22\times 0.28 = 0.0616\ m^{2}$

Mass of the book, m = 3 kg

Pressure of the book, P = $1.0\times10^{5}\ N/m^{2}$

Now,

Pressure, P = $\frac{F}{A}$

Force, F = PA = $1.0\times10^{5}\times 0.0616 = 6160\ N$

Force on mass of 3 kg, F' = mg = $3\times 9.8 = 29.4\ N$

Now, the force at the bottom is given by:

F'' = F + F' = 6160 + 29.4 = 6189.4 N