One block of 2⋅0 kg mass is placed on top of another block of 3⋅0 kg mass. The coefficient of static friction between the two blocks is 0⋅2. The bottom block is pulled with a horizontal force F such that both the blocks move together without slipping. Taking acceleration due to gravity as 10 m/s2, the maximum value of the frictional force is :
4 N
Understanding Friction Between Stacked Blocks
This problem involves two blocks stacked one on top of the other, where the bottom block is pulled horizontally. We are asked to find the maximum value of the frictional force between the two blocks such that they move together without slipping.
Analyzing Forces and Friction
When the bottom block is pulled, a static friction force acts between the two blocks. This static friction is what causes the top block to move along with the bottom block. For the blocks to move together without slipping, the static friction force must provide the necessary acceleration for the top block. The maximum possible value of this static friction force is determined by the coefficient of static friction and the normal force between the two surfaces.
Calculating the Maximum Static Friction
First, let's identify the mass of the top block and the coefficient of static friction between the blocks:
- Mass of the top block ($m_1$) = 2.0 kg
- Coefficient of static friction ($\mu_s$) = 0.2
- Acceleration due to gravity ($g$) = 10 m/s²
The normal force ($N$) acting on the top block is equal to its weight because it is resting on a horizontal surface (the bottom block). The weight is calculated as mass × gravity:
\(N = m_1 g\)
Substituting the values:
\(N = (2.0 \, \text{kg})(10 \, \text{m/s}^2)\)
\(N = 20 \, \text{N}\)
The maximum static friction force ($f_{s,max}$) between the two blocks is given by the product of the coefficient of static friction and the normal force:
\(f_{s,max} = \mu_s N\)
Substituting the values:
\(f_{s,max} = (0.2)(20 \, \text{N})\)
\(f_{s,max} = 4 \, \text{N}\)
This value, 4 N, is the maximum possible static friction force that can exist between the two blocks. If the required static friction force to make the top block accelerate at the same rate as the bottom block exceeds 4 N, slipping will occur.
Maximum Frictional Force for No Slipping
The question asks for the maximum value of the frictional force when the blocks move together without slipping. This happens when the applied force F on the bottom block is such that the required acceleration of the system (both blocks together) makes the static friction force on the top block exactly equal to its maximum possible value, $f_{s,max}$. Beyond this point, any increase in the applied force F will cause the top block to slip relative to the bottom block.
Therefore, the maximum value of the frictional force between the two blocks when they are moving together without slipping is equal to the maximum static friction force calculated above.
\(f_{\text{friction, max}} = f_{s,max} = 4 \, \text{N}\)
figure class="table"| Quantity | Symbol | Value |
|---|---|---|
| Mass of top block | \(m_1\) | 2.0 kg |
| Mass of bottom block | \(m_2\) | 3.0 kg |
| Coefficient of static friction (between blocks) | \(\mu_s\) | 0.2 |
| Acceleration due to gravity | \(g\) | 10 m/s² |
| Normal force on top block | \(N\) | 20 N |
| Maximum static friction | \(f_{s,max}\) | 4 N |
The maximum value of the frictional force when the blocks move together without slipping is 4 N.
Revision Table: Key Concepts
figure class="table"| Concept | Description |
|---|---|
| Static Friction | Friction force that prevents relative motion between surfaces in contact. Its value varies from zero up to a maximum value. |
| Maximum Static Friction | The largest possible value of static friction, given by \(f_{s,max} = \mu_s N\), where \(\mu_s\) is the coefficient of static friction and \(N\) is the normal force. |
| Kinetic Friction | Friction force that opposes relative motion between surfaces that are sliding past each other. Its value is typically constant for given surfaces and normal force. |
| Normal Force | The force exerted by a surface perpendicular to the surface of contact. For an object on a horizontal surface, it often equals the weight. |
Additional Information: System Acceleration
While not explicitly asked, we can also find the maximum acceleration of the system (both blocks) when they move together without slipping. This acceleration ($a$) is determined by the maximum static friction acting on the top block:
\(f_{s,max} = m_1 a\)
\(4 \, \text{N} = (2.0 \, \text{kg}) a\)
\(a = \frac{4 \, \text{N}}{2.0 \, \text{kg}} = 2 \, \text{m/s}^2\)
So, the maximum acceleration the system can have while moving together is 2 m/s². The frictional force between the blocks when they move together at this acceleration is exactly 4 N.
The total force F applied to the bottom block to achieve this acceleration of the combined system ($m_1 + m_2$) would be:
\(F = (m_1 + m_2) a\)
\(F = (2.0 \, \text{kg} + 3.0 \, \text{kg})(2 \, \text{m/s}^2)\)
\(F = (5.0 \, \text{kg})(2 \, \text{m/s}^2) = 10 \, \text{N}\)
If the applied force F exceeds 10 N, the required acceleration would be greater than 2 m/s², which would require a static friction force greater than 4 N, leading to slipping.
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