A 0.75 kg physics book with dimensons of 24.0 cm by 20.0 cm is on a table...?

Question

A. What force does this book apply to the table?

B. What pressure does the book apply?

10 pts to whoever can explain this best using correct mathematical reasoning

Answer 1

The book is not accelerating in the vertical direction, therefore the net force acting on it must be 0. Since the force of Earth's gravity is acting on every object on or near Earth's surface, there must be a force equal in magnitude but opposite in direction that balances the force of gravity on the book. This is the normal force exerted by the table on the book.

Now, Newton's third law says that for every force, there must exist a reaction force equal in magnitude and opposite in direction. Therefore, if the table is exerting a normal force on the book, the book must be exerting a force equal in magnitude but opposite in direction (down) on the table. Since the magnitude of the normal force is the same as the magnitude of the force of gravity on the book, the magnitude of the book pushing on the table will be the same, Fg = mg, where m is the mass of the book and g is the acceleration due to gravity on earth, 9.8 m/s^2 [down]. Solving for Fg we get 7.35 N [down]. The normal force is therefore Fn = 7.35 N [up], and the force of the book pushing on the table is the opposite of this, 7.35 N [down].

The area of the book is given by the formula for area of a rectangle, A = lw. Subbing in 24.0 cm for length and 20.0 cm for width we get A = 480 cm^2 (or 0.048 m^2).

Pressure is defined as force exerted per unit area. The formula for it, then, is P = F / A. Subbing in 7.35 N [down] for force and 480 cm^2 for area, we get approximately 0.0153 N/cm^2, or 153 N/m^2 (153 Pascals).

Answer 2

The force is the book's weight, given by w = m*g. (This will work out as having units of kg*m/s^2. That's the same as Newtons.)

Pressure is force / area, so the pressure the book applies is m*g/480 cm^2. The result will be in Newtons per cm^2.