A student presses a book between his hands, as the drawing indicates. The forces that he exerts on the front and back cover are perpendicular to the book and are horizontal. The book weighs 29.7kg. The coefficient of static friction between his hands and the book is 0.44. To keep the book from falling, what is the magnitude of the minimum pressing force that each hand must exert?
2F*0.44 = M g
Solve for F, which will be in Newtons
To determine the magnitude of the minimum pressing force that each hand must exert to keep the book from falling, we need to consider the equilibrium condition.
Let's break down the forces acting on the book. We have the weight of the book acting vertically downwards, and the forces exerted by the student's hands acting horizontally.
Since the book is in equilibrium (not moving), the sum of the forces in both the horizontal and vertical directions must be zero.
1. Vertical forces:
The weight of the book (W) is given as 29.7 kg. We can calculate the weight using the formula: W = mg, where g is the acceleration due to gravity (approximately 9.8 m/s²).
W = 29.7 kg × 9.8 m/s² = 290.46 N (rounded to two decimal places).
2. Horizontal forces:
The force exerted by each hand on the front and back covers of the book must balance out to prevent it from falling. Let's call this force "F."
Since the force of friction (Ff) opposes the motion (which in this case is zero), it can be calculated as Ff = μN, where μ is the coefficient of static friction and N is the normal force.
The normal force (N) is equal to the weight of the book (W) since the book is on a horizontal surface.
N = W = 290.46 N.
Now, we can calculate the force of friction:
Ff = μN,
Ff = 0.44 × 290.46 N = 127.68 N (rounded to two decimal places).
Since the force of friction is the force exerted by each hand, the magnitude of the minimum pressing force that each hand must exert is 127.68 N.