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 the force \(F=24.5 \mathrm{~N}\) is applied to \(m_{1} ?\)

Short answer

Answer: The maximum applied force for which block 1 will not slide off block 2 is 11.226 N. The accelerations of blocks 1 and 2 when the applied force is 24.5 N are 3.978 m/s² and 2.652 m/s², respectively.

Step-by-step solution

Forces on both blocks when not sliding

First, we will analyze the forces acting on both blocks when they're not sliding with respect to each other. For m1 and m2 respectively, we have the following forces: - Gravitational forces acting downwards: \(m_1g\) and \(m_2g\). - Normal force from m2 on m1: \(N_{12}\) acting upwards. - Static friction force between both blocks: \(f_s\) acting on m1 to the right and on m2 to the left. Since m1 does not move vertically, we have \(N_{12}=m_1g\).

Determine maximum static friction force

To find the maximum applied force before m1 slides off m2, we need to consider the maximum static friction force given by: \(f_{s_{max}}=\mu_s N_{12}\) Plugging in the given coefficients of static friction and normal forces, we have \(f_{s_{max}}=(0.456)(m_1g)\)

Find the maximum applied force

Now, we can find the maximum applied force \(F_{max}\) using Newton's second law. Since m1 is not sliding on m2, the total horizontal force acting on the m1 is \(f_{s_{max}}\). Therefore: \(F_{max}=m_{1}a_{1}=f_{s_{max}}\) From step 2, we know the maximum static friction force \(f_{s_{max}}=(0.456)(m_1g)\). Plugging this in, \(F_{max}=(0.456)(m_1g)\) Now, we can plug in the mass of m1 and an acceleration due to gravity \(g=9.81\,\text{m/s}^2\): \(F_{max}=(0.456)(2.50\,\text{kg}\times 9.81\,\text{m/s}^2)=11.226\,\text{N}\)

Analyze forces with the given applied force

Now, let's analyze the forces acting on both blocks when \(F=24.5\,\text{N}\). Since \(F>F_{max}\), we know that m1 will slide across m2. So, we will have kinetic friction acting between the blocks, which is given by: \(f_k=\mu_k N_{12}\) Plugging in the given coefficient of kinetic friction and normal forces, we have \(f_k=(0.380)(m_1g)\)

Find the accelerations of both blocks

We'll use Newton's second law to determine the accelerations of m1 and m2. For m1, we have: \(F - f_k = m_1a_1\) For m2, we have: \(f_k = m_2a_2\) Using the values for \(F=24.5\,\text{N}\), \(f_k=(0.380)(m_1g)\), and mass of each block, we can solve these two equations for the accelerations: \(a_1 = \frac{F-f_k}{m_1}=\frac{24.5-(0.380)(2.50\,\text{kg}\times 9.81\,\text{m/s}^2)}{2.50\,\text{kg}}=3.978\,\text{m/s}^2\) \(a_2 = \frac{f_k}{m_2}=\frac{(0.380)(2.50\,\text{kg}\times 9.81\,\text{m/s}^2)}{3.75\,\text{kg}}=2.652\,\text{m/s}^2\)

Answers

So, the maximum applied force for which m1 will not slide off m2 is \(11.226\,\text{N}\), and the accelerations of m1 and m2 when the applied force is \(24.5\,\text{N}\) are \(3.978\,\text{m/s}^2\) and \(2.652\,\text{m/s}^2\), respectively.

Key concepts

Newton's Second Law

Understanding how forces affect the motion of objects is at the heart of physics, and Newton's second law is central to this understanding. This law states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. The law is commonly formulated as:

\[F = ma\]

where \(F\) represents the net force applied to the object, \(m\) is the mass of the object, and \(a\) is the acceleration produced. In our exercise, when a force \(F\) is applied to block 1, it will only start to slide if the applied force exceeds the maximum static frictional force between block 1 and block 2. At this point, the system transitions from a static to a dynamic state, and Newton's second law comes into play to calculate the subsequent acceleration of each block.

Kinetic Friction

Kinetic friction occurs when two objects are sliding against each other. It acts in the opposite direction of the movement, causing objects to naturally slow down over time. The kinetic frictional force can be calculated with the equation:

\[f_k = \mu_k N\]

where \(f_k\) is the force of kinetic friction, \(\mu_k\) is the coefficient of kinetic friction, and \(N\) is the normal force—the force perpendicular to the surface that's in contact with the object. In this exercise, when block 1 begins to slide over block 2, the force of kinetic friction \(f_k\) replaces the static frictional force and will influence the acceleration calculations for both blocks.

Normal Force

The normal force is the support force exerted upon an object that is in contact with another stable object. For example, if a book is resting upon a surface, the surface is exerting an upward force equal to the weight of the book; this force is called the normal force. It is typically represented with \(N\) and is always perpendicular to the contact surface. In our exercise, the normal force \(N_{12}\) exerted by block 2 on block 1 is equal to the weight of block 1 due to gravity. When operators calculate any type of frictional force, the normal force is a crucial component, as it defines the frictional force's magnitude based on the friction coefficient.