A horizontal board is oscillating with an amplitude of 3 m. If the frequency of oscillation is 15 per minute, find the minimum value of coefficient of friction in order that a body placed on the board will A not slide when the board oscillates ?

Explanation

To solve the problem, we need to find the minimum value of the coefficient of friction (μ) that prevents a body from sliding on a horizontally oscillating board. Here’s a step-by-step solution:

### Step 1: Understand the given parameters
- Amplitude (A) = 3 m
- Frequency (f) = 15 per minute = 15/60 Hz = 0.25 Hz

### Step 2: Convert frequency to angular frequency (ω)
The angular frequency (ω) is given by the formula:
\[ \omega = 2\pi f \]
Substituting the value of f:
\[ \omega = 2\pi \times 0.25 = \frac{\pi}{2} \text{ rad/s} \]

### Step 3: Calculate the maximum acceleration (a_max)
The maximum acceleration (a_max) of the oscillating board can be calculated using the formula:
\[ a_{max} = A \omega^2 \]
Substituting the values of A and ω:
\[ a_{max} = 3 \times \left(\frac{\pi}{2}\right)^2 = 3 \times \frac{\pi^2}{4} \]
Calculating this:
\[ a_{max} \approx 3 \times 2.467 = 7.401 \text{ m/s}^2 \]

### Step 4: Set up the equation for friction
For the body to not slide, the frictional force must be equal to or greater than the pseudo force acting on the body due to the maximum acceleration. The pseudo force is given by:
\[ F_{pseudo} = m \cdot a_{max} \]
The frictional force (F_friction) is given by:
\[ F_{friction} = \mu m g \]
Setting these two forces equal for the condition of no sliding:
\[ \mu m g = m a_{max} \]

### Step 5: Simplify the equation
We can cancel the mass (m) from both sides:
\[ \mu g = a_{max} \]
Thus, we can express μ as:
\[ \mu = \frac{a_{max}}{g} \]

### Step 6: Substitute the values to find μ
Using \( g \approx 10 \text{ m/s}^2 \):
\[ \mu = \frac{7.401}{10} \approx 0.7401 \]

### Step 7: Final answer
The minimum value of the coefficient of friction (μ) required to prevent the body from sliding is approximately:
\[ \mu \approx 0.74 \]