Question

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a student presses a book between his hands, as the drawing indicates. the forces that he exerts on the front and back covers of the book are perpendicular to the book and are horizontal. the book weighs 27.4 n. the coefficient of static friction between his hands and the book is 0.463. to keep the book from falling, what is the magnitude of the minimum pressing force that each hand must exert?

Explanation:

Step1: Analyze the forces acting on the book

The weight of the book $W = 27.4\ N$ acts down - ward. The frictional forces $f_1$ and $f_2$ from the two hands act upward. The normal forces $N$ from each hand are equal. The maximum static - friction force on each side is $f=\mu_sN$, and the total upward frictional force $F_f = 2f = 2\mu_sN$.

Step2: Set up the equilibrium equation

For the book not to fall, the total upward frictional force must balance the weight of the book. So, $F_f=W$. Substituting $F_f = 2\mu_sN$ into the equilibrium equation, we get $2\mu_sN = W$.

Step3: Solve for the normal force $N$

We want to find the normal force $N$ (the pressing force of each hand). Rearranging the equation $2\mu_sN = W$ for $N$, we have $N=\frac{W}{2\mu_s}$.
Given $W = 27.4\ N$ and $\mu_s=0.463$, then $N=\frac{27.4}{2\times0.463}$.
$N=\frac{27.4}{0.926}=29.6\ N$.

Answer:

$29.6\ N$