This problem has two parts to this. I will type the questions for both parts along with additional info in the "

Question

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 = 1.1 kg and the coefficient of static friction between her hand and the book is µ_s = 0.53.

Part a) What is the weight of the book, F_gb in Newtons?

Part b) What is the minimum force she must apply with each of her hands, F_min in Newtons, to keep the book from falling?

Answer 1

Let's solve each part of the problem:

Part a) To find the weight of the book (F_gb), you can use the formula for weight, which is given by:

F_gb = m * g

Where:

F_gb = Weight of the book

m = Mass of the book = 1.1 kg

g = Acceleration due to gravity = 9.8 m/s²

Now, calculate F_gb:

F_gb = 1.1 kg * 9.8 m/s² ≈ 10.78 N

So, the weight of the book is approximately 10.78 Newtons.

Part b) To find the minimum force she must apply with each of her hands (F_min) to keep the book from falling, we need to consider the forces acting on the book. She needs to apply a force greater than or equal to the maximum static frictional force to prevent the book from sliding out of her hands.

The maximum static frictional force (F_friction) can be calculated using the following equation:

F_friction = µs * F_normal

Where:

µs = Coefficient of static friction between her hand and the book = 0.53

F_normal = Normal force

The normal force (F_normal) is equal to the weight of the book (F_gb) since there is no vertical acceleration (the book is not moving vertically):

F_normal = F_gb = 10.78 N

Now, calculate the maximum static frictional force:

F_friction = 0.53 * 10.78 N ≈ 5.72 N

Therefore, the minimum force she must apply with each of her hands (F_min) to keep the book from falling is approximately 5.72 Newtons.