Question

What is fs,max for sliding a 0.65 kg glass amulet on a glass display case?

Answer 1

What is fs, max for sliding a 0.65 kg glass amulet on a glass display case?

**Solution.**

$$m = 0.65kg, g = 9.8 \frac{m}{s^2};$$ $$F_f - ?$$

The force of friction is:

$$F_f = \mu F_n;$$

$\mu$ - the coefficient of friction;

$F_n$ - the normal force.

$$F_n = mg.$$

The coefficient of friction max for sliding a glass amulet on a glass display case is $\mu = 0.9$.

$$F_f = \mu mg.$$

The force of friction is:

$$F_f = 0.9 \cdot 0.65kg \cdot 9.8 \frac{m}{s^2} = 5.7N.$$

**Answer:** The force of friction is $F_f = 5.7N$.