Stacked Blocks: Videos & Practice Problems

In stacked block problems, two blocks are connected without ropes, with friction between them. When pulling the bottom block on a frictionless surface, the maximum acceleration before they slide apart is determined by static friction. The equation for maximum acceleration is $a=\mu g$, where μ is the coefficient of static friction and g is the acceleration due to gravity. Understanding these concepts is crucial for analyzing motion and forces in connected systems.

Stacked Blocks

In problems involving stacked blocks, understanding the dynamics of connected systems is crucial, especially when friction plays a role. Consider two blocks, A and B, where block A (10.5 kg) is on top of block B, and the surface beneath block B is frictionless. The friction between the two blocks allows them to move together when a force is applied to the bottom block (B). The goal is to determine the maximum acceleration that can be applied to block B without causing block A to slide off.

To analyze the situation, we start by drawing free body diagrams for both blocks. For block A, the forces acting on it include its weight (mg) and the normal force from block B. Since block A is not being pulled directly, there is no applied force acting on it. The frictional force between the two blocks, which is static friction, is what keeps block A moving with block B. This static friction acts in the same direction as the motion, which is a key distinction from previous problems where friction typically opposes motion.

For block B, the forces include its weight, the normal force from the ground, and the normal force exerted by block A. Additionally, there is a frictional force acting to the left on block B due to the static friction from block A. The static friction force is crucial because it prevents block A from sliding off as block B accelerates.

To find the maximum acceleration, we apply Newton's second law, \( F = ma \). For block A, the only horizontal force is the static friction force, which can be expressed as:

\[ F_{s, \text{max}} = \mu_s \cdot N_{AB} \]

Here, \( \mu_s \) is the coefficient of static friction, and \( N_{AB} \) is the normal force between the two blocks. Since block A is not accelerating vertically, the normal force \( N_{AB} \) equals the weight of block A, \( mg \). Thus, we can rewrite the equation as:

\[ F_{s, \text{max}} = \mu_s \cdot m_A \cdot g \]

Substituting this into the equation for block A gives:

\[ \mu_s \cdot m_A \cdot g = m_A \cdot a_{max} \]

Notably, the mass of block A cancels out, leading to:

\[ a_{max} = \mu_s \cdot g \]

For example, if the coefficient of static friction \( \mu_s \) is 0.7, the maximum acceleration can be calculated as:

\[ a_{max} = 0.7 \cdot 9.8 \, \text{m/s}^2 = 6.86 \, \text{m/s}^2 \]

This means that if the acceleration exceeds 6.86 m/s², block A will begin to slide off block B, and the friction will transition to kinetic friction. Understanding these dynamics is essential for solving stacked block problems effectively.

Stacked Block Tied to Wall

In this problem, we analyze a system of two stacked blocks, where block A rests on top of block B. Block B is being pulled to the right with a force of 45 newtons, while block A is tethered to a wall by a rope. Our goal is to determine the tension force acting on block A.

To begin, we create free body diagrams for both blocks. For block A, the forces include its weight (mg), the normal force from block B, and the tension force (T) from the rope. Additionally, since block A is on top of block B, there is a frictional force acting to the right, opposing the tendency of block A to slip off as block B is pulled. This frictional force is crucial in maintaining the equilibrium of block A.

Next, we analyze block B. The forces acting on it include its weight (m_Bg), the normal force from the ground, and the normal force exerted by block A. The applied force of 45 newtons pulls block B to the right, while the frictional force from block A acts to the left. Furthermore, since block B is moving, we consider the kinetic friction between block B and the ground, which also acts to the left.

Given that all surfaces are in motion, we identify the friction forces as kinetic. The coefficient of kinetic friction between the surfaces is 0.2. To find the tension force, we apply Newton's second law (F = ma) to block A. Since block A does not accelerate (it remains stationary relative to the wall), the net force acting on it is zero. This leads us to the conclusion that the frictional force equals the tension force:

F_friction = T

The frictional force can be expressed as:

F_friction = μ_k * N_A

Where μ_k is the coefficient of kinetic friction and N_A is the normal force on block A. The normal force is equal to the weight of block A, which is given by:

N_A = m_A * g

Substituting the values, we have:

F_friction = μ_k * (m_A * g) = 0.2 * (5 kg * 9.8 m/s²)

Calculating this gives us:

F_friction = 0.2 * 49 N = 9.8 N

Thus, the tension force (T) holding block A to the wall is 9.8 newtons. This analysis illustrates the balance of forces in a system where friction plays a critical role in maintaining equilibrium, even when external forces are applied.

A 4kg block sits on top of a 6kg block which is on a frictionless surface. The coefficients of friction between the two blocks are μs=0.5 and μk=0.3. Calculate the maximum force you can pull on the bottom block with so that the objects move together.

• A. 49 N
• B. 19.6 N
• C. 3.27 N
• D. 4.9 N

Answer: (A)

Here’s what students ask on this topic:

What is the maximum acceleration in stacked block problems?

The maximum acceleration in stacked block problems is determined by the coefficient of static friction (μ_s) between the two blocks and the acceleration due to gravity (g). The formula for maximum acceleration (a_max) is given by:

$a=\mu$_sg

Here, μ_s is the coefficient of static friction, and g is the acceleration due to gravity, approximately 9.8 m/s². This equation helps determine the point at which the blocks will start to slide relative to each other.

How do you draw free body diagrams for stacked block problems?

To draw free body diagrams for stacked block problems, follow these steps:

1. Identify all forces acting on each block. For the top block (A), consider its weight (m_Ag), the normal force from the bottom block (N_BA), and the static friction force (f_s).
2. For the bottom block (B), include its weight (m_Bg), the normal force from the floor (N_B), the normal force from the top block (N_AB), the applied force (F), and the friction force (f_s).
3. Ensure that action-reaction pairs are correctly represented, such as N_BA and N_AB being equal and opposite.

These diagrams help visualize the forces and solve for acceleration and other quantities.

What role does static friction play in stacked block problems?

In stacked block problems, static friction plays a crucial role in keeping the blocks moving together without sliding relative to each other. Static friction acts at the interface between the two blocks and opposes the relative motion. The maximum static friction force (f_s,max) is given by:

$f$_s,max=μ_sN

Here, μ_s is the coefficient of static friction, and N is the normal force between the blocks. When the applied force exceeds this maximum static friction, the blocks start to slide, and kinetic friction takes over.

How do you calculate the normal force between two stacked blocks?

The normal force between two stacked blocks is determined by the weight of the top block. If block A is on top of block B, the normal force (N_BA) exerted by block B on block A is equal to the weight of block A:

$N$_BA=m_Ag

Here, m_A is the mass of the top block, and g is the acceleration due to gravity (approximately 9.8 m/s²). This normal force is crucial for calculating the static friction force between the blocks.

What happens when the applied force exceeds the maximum static friction in stacked block problems?

When the applied force exceeds the maximum static friction in stacked block problems, the blocks begin to slide relative to each other. At this point, the friction between the blocks transitions from static to kinetic. The kinetic friction force (f_k) is generally less than the maximum static friction force and is given by:

$f$_k=μ_kN

Here, μ_k is the coefficient of kinetic friction, and N is the normal force between the blocks. The blocks will then move with different accelerations, and the analysis must account for kinetic friction.