Question

Two blocks are stacked on a frictionless table, and a horizontal force \(F\) is applied to the top block (block 1). Their masses are \(m_{1}=2.50 \mathrm{~kg}\) and \(m_{2}=3.75 \mathrm{~kg}\). The coefficients of static and kinetic friction between the blocks are 0.456 and 0.380 , respectively a) What is the maximum applied force \(F\) for which \(m_{1}\) will not slide off \(m_{2} ?\) b) What are the accelerations of \(m_{1}\) and \(m_{2}\) when \(F=24.5 \mathrm{~N}\) is applied to \(m_{1}\) ?

Short answer

To find the maximum force, use the formula: $$ F_{max} = f_{s,max} = μ_s m_{1}g. $$ Plug in the values for \(μ_s\), \(m_{1}\), and \(g\). b) What are the accelerations of the two blocks when a force of 24.5 N is applied to the top block? To find the accelerations of the two blocks, first find the kinetic friction force: $$ f_k = μ_k m_{1}g. $$ Then, find the net forces acting on each block: $$ F_{net,1} = F - f_k \quad \text{and} \quad F_{net,2} = f_k. $$ Finally, use Newton's second law to find the accelerations: $$ a_{1} = \frac{F_{net,1}}{m_{1}} \quad \text{and} \quad a_{2} = \frac{F_{net,2}}{m_{2}}. $$ Calculate the values for \(a_{1}\) and \(a_{2}\).

Step-by-step solution

1. Calculate the maximum static friction force between the blocks

The maximum static friction force \(f_{s,max}\) between the two blocks is given by the product of the coefficient of static friction \(μ_s\) and the normal force \(N\). Since there is no vertical acceleration, the normal force between the blocks must equal the weight of block 1: \(N = m_{1}g\). Thus, $$ f_{s,max} = μ_s m_{1}g, $$ where \(g = 9.81 \,\text m/s^2\) is the gravitational acceleration.

2. Find the maximum force F

For block 1 to not slide off block 2, the applied force should not exceed the maximum friction force between the blocks. Therefore, $$ F_{max} = f_{s,max} = μ_s m_{1}g. $$ Plug in the values for \(μ_s\), \(m_{1}\), and \(g\) to find the maximum applied force \(F_{max}\).

3. Calculate the kinetic friction force between the blocks

When a force of \(F=24.5\,\text N\) is applied to block 1, the kinetic friction force \(f_k\) between the blocks needs to be calculated. Since the force is less than the maximum force calculated in step 2, the blocks will move together. The kinetic friction force can be calculated by the product of the coefficient of kinetic friction \(μ_k\) and the normal force: $$ f_k = μ_k N = μ_k m_{1}g. $$

4. Calculate the total force acting on each block

The total net force acting on block 1, \(F_{net,1}\), will be the applied force minus the kinetic friction force: $$ F_{net,1} = F - f_k. $$ The total net force acting on block 2, \(F_{net,2}\), will be equal to the kinetic friction force: $$ F_{net,2} = f_k. $$

5. Calculate the accelerations of the two blocks

Using Newton's second law, the acceleration of each block can be found as follows: $$ a_{1} = \frac{F_{net,1}}{m_{1}} \quad \text{and} \quad a_{2} = \frac{F_{net,2}}{m_{2}}. $$ Plug in the values for \(F_{net,1}\), \(F_{net,2}\), \(m_{1}\), and \(m_{2}\) to find the accelerations \(a_{1}\) and \(a_{2}\) of the blocks.

Key concepts

Static Friction

Static friction is the force that keeps an object at rest when a force is applied, preventing it from moving. When two surfaces are in contact, static friction acts parallel to the surface of contact. It's a balancing act—it resists the initial motion, making it crucial to understanding how objects begin to move.
In the given exercise, the static friction is what keeps block 1 from sliding off block 2 even when a horizontal force is applied.

• The maximum static friction force (\(f_{s, max}\)) is calculated by multiplying the coefficient of static friction (\(μ_s\)) by the normal force (\(N\)).
• In this scenario, since there’s no vertical movement, the normal force is equal to the gravitational force on block 1: \(N = m_{1}g\).
• The equation to find the force is \(f_{s, max} = μ_s m_{1}g\).

Understanding static friction helps us recognize that as long as the applied force stays below this maximum frictional force, block 1 stays put on block 2.

Kinetic Friction

Kinetic friction comes into play when an object is already moving over a surface. It’s usually less than static friction, meaning it takes less force to keep an object moving than to start it moving. While static friction prevents motion, kinetic friction acts in the opposite direction of the movement, working to slow it down.
In the exercise, once block 1 overcomes static friction with the applied force, kinetic friction keeps influencing the motion between block 1 and block 2.

• Kinetic friction force (\(f_k\)) can be calculated by the formula: \(f_k = μ_k m_{1}g\).
• The coefficient of kinetic friction (\(μ_k\)) determines how much friction exists while the object is in motion.
• In the situation given, the applied force is less than the maximum static friction force, so both blocks will move together, with kinetic friction working between them.

Realizing this distinction between static and kinetic friction is crucial in solving problems involving sliding objects.

Acceleration

Acceleration is the rate at which an object's velocity changes over time. Newton's Second Law of Motion tells us that acceleration happens when a net force is applied to an object, expressed in the equation \(F = ma\), where \(F\) is the net force, \(m\) is the mass, and \(a\) is the acceleration.
In this particular problem, once the force overcomes static friction and is less than maximum static friction, both blocks accelerate together. Determining their acceleration involves:

• Calculating the net force acting on each block.
• For block 1, the net force (\(F_{net,1}\)) is the applied force minus the kinetic friction: \(F_{net,1} = F - f_k\).
• For block 2, the net force (\(F_{net,2}\)) is simply the kinetic friction force since it is propelled by the friction from block 1: \(F_{net,2} = f_k\).
• Using \(F = ma\), solve for acceleration: \(a_1 = \frac{F_{net,1}}{m_1}\) and \(a_2 = \frac{F_{net,2}}{m_2}\).

Grasping how acceleration works in this perfect interplay of forces and mass is vital to determine how systems of objects behave when subjected to forces.