Two 1kg blocks rest on opposite sides of the double inclined plane. One inclined planes makes an angle of 30 degrees and the other an angle of 60 degrees with horizontal. The masses are connected by a light cord passing over a friction less pulley. What is the largest value of the coefficient of the static friction (assumed to be the same for each block) at which slipping can occur?
To find the largest value of the coefficient of static friction at which slipping can occur, we need to analyze the forces acting on the system.
Let's start by drawing a diagram to visualize the situation:
```
|\ /|
| \ / |
|| \\ 30° // ||
|| \\ // ||
|a \\ // b |
| \\_________________// |
|__--☆_______________☆--__|
```
Here:
- Mass 'a' rests on a double inclined plane inclined at 30 degrees.
- Mass 'b' rests on a double inclined plane inclined at 60 degrees.
- There is a light cord passing over a frictionless pulley, connecting the two masses.
Let's consider the forces acting on each mass:
For mass 'a':
1. The gravitational force (weight) acting straight down.
2. The normal force acting perpendicular to the inclined plane.
3. The force of static friction acting up the inclined plane.
For mass 'b':
1. The gravitational force (weight) acting straight down.
2. The normal force acting perpendicular to the inclined plane.
3. The force of static friction acting up the inclined plane.
Now, we can set up the equations of motion for each mass:
For mass 'a':
1. ΣFy = N - m_a * g * cos(30°) = 0 (since it is not accelerating in the y-direction)
2. ΣFx = m_a * g * sin(30°) - f_s = m_a * a (where 'a' is the acceleration of mass 'a')
For mass 'b':
1. ΣFy = N - m_b * g * cos(60°) = 0 (since it is not accelerating in the y-direction)
2. ΣFx = m_b * g * sin(60°) + f_s = m_b * a (where 'a' is the acceleration of mass 'b')
Now, let's calculate the mass and acceleration in terms of unknowns:
1. Mass 'a':
- Weight of mass 'a' = m_a * g = 1 kg * 9.8 m/s^2 = 9.8 N
- Normal force = m_a * g * cos(30°) = 1 kg * 9.8 m/s^2 * cos(30°) = 8.484 N
- Acceleration of mass 'a' = (m_a * g * sin(30°) - f_s) / m_a = (9.8 N * sin(30°) - f_s) / 1 kg
2. Mass 'b':
- Weight of mass 'b' = m_b * g = 1 kg * 9.8 m/s^2 = 9.8 N
- Normal force = m_b * g * cos(60°) = 1 kg * 9.8 m/s^2 * cos(60°) = 4.9 N
- Acceleration of mass 'b' = (m_b * g * sin(60°) + f_s) / m_b = (9.8 N * sin(60°) + f_s) / 1 kg
Since the masses are connected by a light cord passing over a frictionless pulley, the magnitudes of their accelerations should be equal.
Therefore, we can equate the accelerations:
(9.8 N * sin(30°) - f_s) / 1 kg = (9.8 N * sin(60°) + f_s) / 1 kg
Simplifying this equation, we have:
9.8 N * sin(30°) - f_s = 9.8 N * sin(60°) + f_s
Rearranging the equation, we get:
2 * f_s = 9.8 N * sin(60°) - 9.8 N * sin(30°)
Simplifying further:
f_s = (9.8 N * sin(60°) - 9.8 N * sin(30°)) / 2
Calculating this equation, we find:
f_s ≈ 0.85 N
Finally, we can calculate the maximum coefficient of static friction at which slipping can occur:
μ_s = f_s / N
Where N is the normal force.
Substituting the values:
μ_s = 0.85 N / (1 kg * 9.8 m/s^2 * cos(30°))
Calculating this equation, we get:
μ_s ≈ 0.185
Therefore, the largest value of the coefficient of static friction at which slipping can occur is approximately 0.185.
To find the largest value of the coefficient of static friction at which slipping can occur, we need to consider the forces acting on each block and the conditions for equilibrium.
Let's start by analyzing the forces acting on each block individually:
1. Block on the 30-degree inclined plane:
- Weight (mg): The weight of the block acts vertically downward and can be resolved into two components, one along the inclined plane and the other perpendicular to it.
- Normal force (N): The surface of the inclined plane exerts a normal force perpendicular to the plane.
- Static friction (f1): The force of static friction acts parallel to the inclined plane and opposes motion.
2. Block on the 60-degree inclined plane:
- Weight (mg): Similar to the first block, the weight can be resolved into two components along and perpendicular to the inclined plane.
- Normal force (N): The inclined plane exerts a normal force perpendicular to the surface.
- Static friction (f2): The force of static friction acts parallel to the inclined plane and opposes motion.
Since the masses are connected by a light cord passing over a frictionless pulley, the tension in the cord is the same for both blocks.
Now, let's set up the equations for equilibrium for each block:
1. Block on the 30-degree inclined plane:
- Along the inclined plane: f1 - m1 * g * sin(30°) = 0
- Perpendicular to the inclined plane: N - m1 * g * cos(30°) = 0
2. Block on the 60-degree inclined plane:
- Along the inclined plane: m2 * g * sin(60°) - f2 = 0
- Perpendicular to the inclined plane: N - m2 * g * cos(60°) = 0
Next, we need to relate the static friction to the normal force for each block:
For block 1:
- Since the block is at the point of slipping, the static friction force reaches its maximum value: f1_max = μ * N (where μ is the coefficient of static friction).
- We can substitute the normal force N from the equation N - m1 * g * cos(30°) = 0 into f1_max = μ * N to get f1_max in terms of m1, g, and μ.
For block 2:
- Using the same reasoning, f2_max = μ * N, where N = m2 * g * cos(60°).
Because the masses are connected by a light cord passing over the frictionless pulley, the tension in the cord is the same for both blocks. This tension is equal to f1_max (since both blocks are connected by the same cord).
Finally, we can equate f1_max and f2_max to get the largest value of the coefficient of static friction at which slipping can occur:
μ * N = μ * (m2 * g * cos(60°)) = Tension = f1_max = μ * (m1 * g * cos(30°))
By canceling out the common terms μ and g and rearranging the equation, we get:
m2 * cos(60°) = m1 * cos(30°)
Substituting the given masses, m1 = m2 = 1 kg, we have:
1 * cos(60°) = 1 * cos(30°)
0.5 = 0.866
Since the equation is not balanced (0.5 ≠ 0.866), there is no value of μ that satisfies the conditions for equilibrium. Therefore, the largest value of the coefficient of static friction at which slipping can occur is 0.