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

Step-by-Step Solution

Answer

(a) \(\mu \geq 0.53\), (b) \(a = 2.61\) m/s² (table), \(a = 2.59\) m/s² (12.0-kg block), (d) Force needed: 83.2 N (a), 44.2 N (b).

Step 1: Understanding the forces on the 4.0-kg block

To prevent the 4.0-kg block from sliding off, the frictional force between the blocks must be equal to or greater than the force required to accelerate the 4.0-kg block at 5.2 m/s². This frictional force can be expressed as \( f = m_1 \cdot a \), where \( m_1 = 4.0 \) kg and \( a = 5.2 \) m/s².

Step 2: Calculating the minimum coefficient of friction \(\mu\)

The maximum frictional force available is given by \( f_{max} = \mu \cdot N \), where \( N \) is the normal force. Here, \( N = m_1 \cdot g \), where \( g = 9.8 \) m/s². So we have \( \mu \cdot m_1 \cdot g \geq m_1 \cdot a \). Simplifying gives \( \mu \geq \frac{a}{g} \), thus \( \mu \geq \frac{5.2}{9.8} \approx 0.53 \).

Step 3: Determine the acceleration with half the minimum \(\mu\)

When \( \mu \) is halved, \( \mu' = 0.53 / 2 = 0.265 \). The maximum frictional force becomes \( f'_{max} = \mu' \cdot m_1 \cdot g \). Calculating the friction gives \( f'_{max} = 0.265 \cdot 4.0 \cdot 9.8 = 10.372 \) N. The 4.0-kg block's acceleration with this friction is \( a' = \frac{f'_{max}}{m_1} \approx 2.59 \) m/s².

Step 4: Calculate acceleration of 4.0-kg block relative to the table

Relative to the table, the 4.0-kg block's acceleration is the remaining acceleration after accounting for friction. Since the system's desired acceleration is 5.2 m/s², the block moves as \( a_{table} = 5.2 - 2.59 = 2.61 \) m/s².

Step 5: Determine acceleration of 4.0-kg block relative to 12.0-kg block

For the acceleration relative to the 12.0-kg block, it is simply the frictional acceleration towards the right. This means the acceleration is the value \( a' = 2.59 \) m/s².

Step 6: Calculate applied force for case (a) and (b)

For case (a), the total force required is \( F = (m_1 + m_2) \cdot a \), so \( F = (4.0 + 12.0) \cdot 5.2 = 83.2 \) N. For case (b), the acceleration is reduced for the 4.0-kg block, needing \( F = 17.2 \cdot 2.61 = 44.2 \) N additional for counteraction of reduced friction.

Key Concepts

Coefficient of Friction

The coefficient of friction, represented as \( \mu \), is a crucial concept in physics, describing the force of friction between two surfaces. It is a dimensionless quantity that indicates how easily one object will slide over another. There are two primary types: static friction coefficient (\( \mu_s \)) and kinetic friction coefficient (\( \mu_k \)). Static friction comes into play when objects are at rest, while kinetic friction acts when objects are in motion. Both are vital for understanding everyday interactions where surfaces meet.

The value of \( \mu \) essentially tells us how "sticky" a surface is. For example, ice has a low \( \mu \), meaning it's slippery; sandpaper has a high \( \mu \), indicating it's rough. In the given exercise, understanding \( \mu \) is crucial to determining when the 4.0-kg block will start sliding off the 12.0-kg block as they accelerate together. Knowing \( \mu \) helps to predict whether the block will maintain its position or not as frictional forces counteract the movement.

Newton's Laws of Motion

Newton's laws of motion form the foundation for mechanics. These laws describe the relationship between the motion of an object and the forces acting on it.

1. **First Law (Inertia):** An object will remain at rest or in uniform motion unless acted upon by an external force. This law highlights the importance of friction in resisting the motion, as seen in the exercise, where friction keeps the 4.0-kg block from sliding.
2. **Second Law (F=ma):** The force acting on an object is equal to the mass of that object multiplied by its acceleration. It explains the acceleration of the block due to the force applied through friction. In the exercise, this law is used to determine the minimum friction necessary to keep the 4.0-kg block stationary relative to the accelerating 12.0-kg block.
3. **Third Law (Action-Reaction):** For every action, there's an equal and opposite reaction. In the context of friction, when the block tries to slide, the table or another surface pushes back with an equal force to resist the movement. This is crucial in maintaining the 4.0-kg block's relative position.

Kinetic and Static Friction

Kinetic and static friction are two types of frictional forces that resist motion between surfaces.

- **Static Friction (\( f_s \))**: This is the friction that prevents an object from starting to move. It needs to be overcome by an external force to set the object in motion. In our exercise, the static friction prevents the 4.0-kg block from sliding off as the larger block starts to accelerate.
- **Kinetic Friction (\( f_k \))**: Once an object is in motion, kinetic friction acts on it. This force is usually slightly lower than static friction, meaning less force is needed to keep an object moving than to start it moving.

In practical scenarios, engineers need to know both types to calculate the forces necessary for controlling object movement, like the acceleration in the exercise given. By knowing \( \mu_k \) and \( \mu_s \), they can design systems that operate efficiently under various conditions without losing control due to unexpected slips or jerks.