Solution to Question 20
Let $N$ be the number of sheets (leaves) in each book. The problem states that each book has $n = 400$ pages. Since one sheet of paper consists of two pages (front and back), the number of sheets per book is:
$$ N = \frac{n}{2} = \frac{400}{2} = 200 $$When the two books are intermeshed, the sheets overlap. Let’s count the number of contact surfaces where friction acts. If we have $N$ sheets from one book interleaved with $N$ sheets from the other, the total number of overlapping interfaces is $2N – 1$.
Let $m$ be the mass of the overlapped portion of each sheet. The normal force on each sheet depends on the weight of all the sheets above it.
- 1st Interface (topmost): Supported by 1 sheet. Normal force $N_1 = mg$.
- 2nd Interface: Supported by 2 sheets. Normal force $N_2 = 2mg$.
- k-th Interface: Normal force $N_k = k mg$.
The total frictional force $f_{total}$ required to separate the books is the sum of the limiting friction on all active interfaces. The coefficient of static friction is $\mu$.
$$ F_{pull} = \sum_{k=1}^{2N-1} f_k = \sum_{k=1}^{2N-1} \mu N_k $$ $$ F_{pull} = \mu \sum_{k=1}^{2N-1} (k mg) = \mu mg \sum_{k=1}^{2N-1} k $$Using the formula for the sum of the first $K$ integers, $\sum_{i=1}^{K} i = \frac{K(K+1)}{2}$, where $K = 2N – 1$:
$$ F_{pull} = \mu mg \frac{(2N-1)(2N-1+1)}{2} = \mu mg \frac{(2N-1)(2N)}{2} $$ $$ F_{pull} = N(2N-1) \mu mg $$Now, we substitute the numerical values:
- $N = 200$
- $\mu = 0.1$
- $m = 0.5 \text{ g} = 0.5 \times 10^{-3} \text{ kg}$
- $g = 10 \text{ m/s}^2$