A constant horizontal force of 5N in magnitude is applied tangentially to the top most of a ring of mass 10 kg and radius 10 cm kept on a rough horizontal surface with coefficient of friction between surface and ring being 0.2. The friction force on the ring would be: `(g = 10 m//s^(2))`

Options

A. 5 N direction opposte to the applied force
B. 10 N directed opposite to the applied force
C. 10 N in the direction of applied force.
D. None of these

Explanation

To solve the problem, we need to determine the friction force acting on the ring when a constant horizontal force is applied. Here’s a step-by-step solution:

### Step 1: Understand the Forces Acting on the Ring
- A constant horizontal force of 5 N is applied tangentially to the top of the ring.
- The ring has a mass of 10 kg and is on a rough surface with a coefficient of friction (μ) of 0.2.

### Step 2: Calculate the Normal Force (N)
- The normal force (N) acting on the ring is equal to the weight of the ring since there are no vertical forces acting on it other than its weight.
- Weight (W) = mass (m) × gravitational acceleration (g)
- Given: m = 10 kg, g = 10 m/s²
- Therefore, W = 10 kg × 10 m/s² = 100 N
- Thus, N = 100 N.

### Step 3: Calculate the Maximum Friction Force (F_friction)
- The maximum friction force can be calculated using the formula:
\[ F_{\text{friction}} = \mu \times N \]
- Given: μ = 0.2
- Therefore:
\[ F_{\text{friction}} = 0.2 \times 100 N = 20 N \]

### Step 4: Determine the Direction of the Friction Force
- The applied force of 5 N tends to move the ring in the positive x-direction.
- Friction always acts in the direction opposite to the applied force to oppose the motion.
- Therefore, the friction force will act in the negative x-direction.

### Step 5: Conclusion
- The friction force acting on the ring is 20 N in the direction opposite to the applied force.

### Final Answer
The friction force on the ring would be **20 N** acting in the direction opposite to the applied force.

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