Problem 32 - Physics for Scientists and Engineers with Modern Physics

Question

(III) A 4.0-kg block is stacked on top of a \(12.0-\mathrm{kg}\) block, which is accelerating along a horizontal table at \(a=5.2 \mathrm{~m} / \mathrm{s}^{2}\) (Fig. 5-40). Let \(\mu_{\mathrm{k}}=\mu_{\mathrm{s}}=\mu .\)( \(a\) ) What minimum coefficient of friction \(\mu\) between the two blocks will prevent the \(4.0-\mathrm{kg}\) block from sliding off? \((b)\) If \(\mu\) is only half this minimum value, what is the acceleration of the \(4.0-\mathrm{kg}\) block with respect to the table, and \((c)\) with respect to the \(12.0-\mathrm{kg}\) block? (d) What is the force that must be applied to the \(12.0-\mathrm{kg}\) block in \((a)\) and in \((b)\), assuming that the table is frictionless?

Step-by-Step Solution

Answer

Minimum friction coefficient is 0.531. With reduced friction, acceleration is 2.6 m/s². Required forces are 83.2 N and 62.4 N.

Step 1: Understanding the Problem

The problem consists of two blocks where the 4.0 kg block is placed on top of the 12.0 kg block. We need to find the minimum static friction coefficient to prevent the top block from sliding and other values derived from it, considering friction and forces acting on these blocks.

Step 1: Calculating Minimum Coefficient of Friction

The force required to prevent the sliding of the top block is provided by static friction: \( f_s = m_1 \cdot a \), where \( m_1 \) is the mass of the top block.Since \( f_s = \mu_s \cdot N \) and \( N = m_1 \cdot g \) (the normal force), then \( \mu_s \cdot m_1 \cdot g = m_1 \cdot a \).Therefore, \( \mu_s = \frac{a}{g} \). Substitute \( a = 5.2 \ \text{m/s}^2 \) and \( g = 9.8 \ \text{m/s}^2 \):\[ \mu_s = \frac{5.2}{9.8} \approx 0.531 \].

Step 2: Finding Acceleration with Reduced Friction

Given \( \mu \) is half of the minimum value, \( \mu = 0.2655 \).To find the horizontal acceleration of the 4.0 kg block when it slides, use: \( f_k = \mu_k \cdot m_1 \cdot g = m_1 \cdot a_1 \).So, \( a_1 = \mu_k \cdot g \), where \( \mu_k = 0.2655 \):\[ a_1 = 0.2655 \times 9.8 \ \approx 2.6 \ \text{m/s}^2 \].

Step 3: Relative Acceleration to the 12 kg Block

The relative acceleration \( a_{rel} \) of the 4.0 kg block with respect to the 12.0 kg block is the difference between their accelerations: \( a_{rel} = a - a_1 \).Thus, \( a_{rel} = 5.2 \ \text{m/s}^2 - 2.6 \ \text{m/s}^2 = 2.6 \ \text{m/s}^2 \).

Step 4: Calculating Force Applied to 12 kg Block

For no sliding condition (part a):Force applied, \( F = (m_1 + m_2) \cdot a = (4.0 + 12.0) \cdot 5.2 \ \text{N} = 83.2 \ \text{N} \).For sliding condition (part b) where \( \mu \) is reduced:Force applied only to the 12 kg block considering its acceleration (as the 4 kg block now slides and has its own acceleration):\( F = m_2 \cdot a = 12.0 \cdot 5.2 \ \text{N} = 62.4 \ \text{N} \).

Step 6: Conclusion

Summarize the findings: The minimum coefficient of friction is approximately 0.531. With the reduced friction, the block accelerates at 2.6 m/s² relative to the table and to the lower block. The force required to keep it stationary when friction is enough is 83.2 N, and it is 62.4 N when friction is reduced.

Key Concepts

Friction Coefficient

To understand how the two blocks interact, we need to dive into the concept of the friction coefficient. The coefficient of friction is a number that represents the resistance to sliding along a surface. This force keeps objects from moving past each other too easily. It has two types: static (\(\mu_s\)) and kinetic friction (\(\mu_k\)). In this exercise, the same value is given for both static and kinetic friction.

Static friction supports an object until an external force is strong enough to initiate movement. Kinetic friction then acts when the object is in motion. The specific equation for static friction is:

• The force of static friction \(f_s = \mu_s \cdot N\).
• \(N\) is the normal force, which is the force exerted by a surface to support the weight of an object resting on it, equal to the gravitational force \(N = m \cdot g\).

In this problem, static friction must be greater or equal to the accelerating force to keep the 4.0-kg block from sliding off the 12.0-kg block.

Newton's Laws

Newton's Laws of Motion provide the foundation for solving physics problems about forces and motion. In this scenario, Newton's second law is especially important.

According to Newton's second law, acceleration (\(a\)) is produced when a force (\(F\)) acts on a mass (\(m\)). The relationship is defined by the equation:

• \(F = m \cdot a\).

Here, this principle helps us understand how much force is required to prevent motion versus allowing it.

It ties the force needed to move the 12.0-kg block with the 4.0-kg block on top, emphasizing that such force must be able to overcome the static friction to keep the smaller block from sliding. When motion is applied, Newton's laws help us calculate how friction impacts the blocks' acceleration.

Acceleration Calculation

Acceleration is a key concept when discussing motion. It is defined as a change in velocity over time. In this problem, you calculate acceleration when considering the force of friction and its impact.

When given half the minimum friction coefficient, we calculate the acceleration of the 4.0-kg block using kinetic friction principles:

• The kinetic friction equation: \(f_k = \mu_k \cdot N\), with \(N = m \cdot g\).
• \(a_1 = \mu_k \cdot g\), where \(\mu_k\) is reduced to half its initial value.

This approach adjusts the acceleration based on the coefficient of friction, allowing us to understand how its alteration impacts the motion directly on the surface where friction is reduced.

Forces and Motion

The relationship between forces and motion is central to this exercise. Forces come from interactions such as gravity and friction, impacting how objects move. In our problem, a 4.0-kg block sits atop a 12.0-kg block, and both are subject to various forces.

Key forces to consider are:

• Gravitational force, pulling both blocks downward.
• Frictional force, preventing the top block from sliding when sufficient.
• Applied force, needed to accelerate the 12.0-kg block with the 4.0-kg block on top.

When calculating the required force, we use:

• \(F = (m_1 + m_2) \cdot a\), summing up both block masses and using the target acceleration.

This calculation shows how changes in friction coefficients modify the interplay of these forces and how they result in different motion scenarios, demonstrating Newton's laws effectively.