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 = 1.7 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?

Answer 1

Answer:

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

Answer 2

The tendency of a force to rotate the body to which it is applied.  It depends on the axis of rotation. The force is defined as the product of mass and acceleration.

The minimum force required to hold the book is 4.3316N. The solution is as follows:-

F' = μsR

Where F' is the Friction force and μs is the Coefficient of static friction, R is the normal reaction.

while $R = mg$

Where m is the mass of the book and g is the acceleration due to gravity.

After solving both the equation.

F' = μsmg

All the data is given in the question that is m = 1.7 kg, μs = 0.52, g = 9.8 m/s²

After solving it.

$F' = 1.7*(9.8)*(0.52)F' = 8.6632 N$

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

The force applied on each hand is

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

For more information, refer to the link:-

https://brainly.com/question/25026730