To calculate the frictional force acting on a ring placed on a horizontal surface, we need to consider the following:

Key Factors:

1. Weight of the Ring (W): The weight of the ring is given by: $W = m \cdot g$ where:

   • $m$ = mass of the ring
   • $g$ = acceleration due to gravity ($9.8 \, \text{m/s}^2$).

2. Normal Force ($N$): On a horizontal surface, the normal force is equal to the weight: $N = W = m \cdot g$

3. Coefficient of Friction ($\mu$):

   • Static Friction ($f_s$): If the ring is at rest, the maximum static frictional force is: $f_s = \mu_s \cdot N$ Here, $\mu_s$ is the coefficient of static friction.
   • Kinetic Friction ($f_k$): If the ring is sliding, the kinetic frictional force is: $f_k = \mu_k \cdot N$ Here, $\mu_k$ is the coefficient of kinetic friction.

Frictional Force Value:

• If the ring is not moving: $f = \text{Force applied, if } f \leq f_s$ Otherwise, $f = f_s$.

• If the ring is moving: $f = f_k$

Example Calculation:

Suppose:

• $m = 5 \, \text{kg}$,
• $\mu_s = 0.4$,
• $\mu_k = 0.3$.

1. Weight: $W = 5 \cdot 9.8 = 49 \, \text{N}.$

2. Normal Force: $N = 49 \, \text{N}.$

3. Maximum Static Friction: $f_s = \mu_s \cdot N = 0.4 \cdot 49 = 19.6 \, \text{N}.$

4. Kinetic Friction: $f_k = \mu_k \cdot N = 0.3 \cdot 49 = 14.7 \, \text{N}.$

If an applied force is below $19.6 \, \text{N}$, the ring will not move. If it exceeds this value, the frictional force becomes $14.7 \, \text{N}$ as it slides.

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Do you need any clarifications or further examples?

Related Questions:

1. What is the effect of surface inclination on frictional force?
2. How does friction differ between smooth and rough surfaces?
3. What happens to friction if the ring's weight increases?
4. How do we calculate friction if additional forces are acting on the ring?
5. Can friction be negative, and under what conditions?

Tip:

Always remember that static friction adjusts up to its maximum limit ($f_s$) to counter external forces before motion begins.