Question

the maximum force exerted onto a 1.65 kg book at rest is 11.3 n due to static friction.

what is the coefficient of static friction?

\mu = ?

Explanation:

🆕 New Concept Discovered: Coefficient of Static Friction
How surfaces grip each other before sliding starts

Step 1: Identify the normal force

The normal force \( F_N \) acting on a flat surface is equal to the weight of the book:

$$ F_N = m \cdot g $$

Given:

• Mass \( m = 1.65 \text{ kg} \)
• Acceleration due to gravity \( g \approx 9.8 \text{ m/s}^2 \) (or \( 9.81 \text{ m/s}^2 \))

$$ F_N = 1.65 \text{ kg} \times 9.8 \text{ m/s}^2 = 16.17 \text{ N} $$

Step 2: Use the static friction formula

The maximum static friction force \( F_{s,\text{max}} \) is related to the normal force by:

$$ F_{s,\text{max}} = \mu_s \cdot F_N $$

Given:

• Maximum static friction force \( F_{s,\text{max}} = 11.3 \text{ N} \)

Rearrange the formula to solve for the coefficient of static friction \( \mu_s \):

$$ \mu_s = \frac{F_{s,\text{max}}}{F_N} $$

Step 3: Calculate the coefficient

$$ \mu_s = \frac{11.3 \text{ N}}{16.17 \text{ N}} \approx 0.699 $$

Using \( g = 9.8 \text{ m/s}^2 \), the value is approximately \( 0.70 \).
(If using \( g = 9.81 \text{ m/s}^2 \), \( F_N = 16.1865 \text{ N} \), then \( \mu_s = \frac{11.3}{16.1865} \approx 0.698 \).)

Rounding to three significant figures gives \( 0.700 \) (or \( 0.698 \)).

Answer:

0.70