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

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