A woman holds a book by placing it between her hands such that she presses at right angles to the front and back covers. The book has a mass of m = 1.7 kg and the coefficient of static friction between her hand and the book is μs = 0.52. What is the minimum force she must apply with each of her hands Fmin in Newtons, to keep the book from falling?

4.3316 N

Explanation:

Using,

F' = μsR ... Equation 1

Where F' = Friction force, μs = Coefficient of static friction, R = normal reaction.

But,

R = mg ... Equation 2

Where m = mass of the book, g = acceleration due to gravity.

substitute equation 2 into equation 1

F' = μsmg ... Equation 3

Given: m = 1.7 kg, μs = 0.52, g = 9.8 m/s²

Substitute into equation 3

F' = 1.7 (9.8) (0.52)

F' = 8.6632 N

Since she is using both hands, we assume that the force from each hands are equal,

The force applied on each hand = 8.6632/2 = 4.3316 N.

Hence the minimum force applied on each hand to keep the book from falling = 4.3316 N

8.66 N

Explanation:

Parameters given:

Mass, m = 1.7 kg

Coefficient of static friction, μs = 0.52

The minimum force required to keep the book from falling is given as:

Fmin = μs * N

Where N = normal force

N = m * g

=> Fmin = μs * m * g

Fmin = 0.52 * 1.7 * 9.8

Fmin = 8.66 N