Question

A block of mass \(M_{1}=0.640 \mathrm{~kg}\) is initially at rest on a cart of mass \(M_{2}=0.320 \mathrm{~kg}\) with the cart initially at rest on a level air track. The coefficient of static friction between the block and the cart is \(\mu_{s}=0.620\), but there is essentially no friction between the air track and the cart. The cart is accelerated by a force of magnitude \(F\) parallel to the air track. Find the maximum value of \(F\) that allows the block to accelerate with the cart, without sliding on top of the cart.

Short answer

Answer: The maximum force F that prevents the block from sliding on the cart is approximately 3.95 N.

Step-by-step solution

Define the forces involved

We need to consider the forces acting on both the block and the cart. There will be gravitational forces acting on both Mass 1 and Mass 2, the force of static friction acting on M1, normal force acting between M1 and M2, and the externally applied force F acting on M2.

Use Newton's Laws to setup equations for forces acting on Mass 1

We will use Newton's second law (\(F = ma\)) to set up the equations for the forces acting on M1. The forces acting on M1 are: 1. Gravitational force: \(F_{g1} = M_1 * g\) 2. Normal force acting between M1 and M2: \(F_{N1}\) 3. Force of static friction: \(F_{f1} = \mu_s * F_{N1}\) The equation for the horizontal forces (x-direction) on M1 can be written as: \(F_{f1} = M_1 * a_1\), where \(a_1\) is the acceleration of Mass 1.

Use Newton's Laws to setup equations for forces acting on Mass 2

Similarly, using Newton's second law (\(F = ma\)), we can setup equations for the forces acting on M2. The forces acting on M2 are: 1. Gravitational force: \(F_{g2} = M_2 * g\) 2. Normal force acting between M1 and M2: \(F_{N2} = - F_{N1}\) 3. Force of static friction: \(F_{f2} = - F_{f1}\) 4. Externally applied force: \(F\) The equation for the horizontal forces (x-direction) on M2 can be written as: \(F - F_{f2} = M_2 * a_2\), where \(a_2\) is the acceleration of Mass 2.

Maximize F to prevent slipping

To ensure that the block doesn't slip on the cart, the force of static friction should be equal to the product of the coefficient of static friction and the normal force, i.e., \(F_{f1} = \mu_s * F_{N1}\) Now we can substitute \(F_{f1}\) and \(F_{N1}\) in the equations of step 2 and step 3. Using these equations, we get: \(M_1 * a_1 = \mu_s * F_{N1}\) \(F - \mu_s * F_{N1} = M_2 * a_2\) Since M1 and M2 have the same acceleration, \(a_1 = a_2 = a\). We can solve these equations for F: \(F = (M_1 + \mu_s * M_2) * a\) Now we need to find the maximum value of F that prevents the slipping of the block. We can use the condition of maximum static friction: \(F_{f1} = \mu_s * F_{N1} = M_1 * a\) From the above equation, we can find the acceleration 'a': \(a = \dfrac{\mu_s * F_{N1}}{M_1}\) Substitute this value of acceleration back into the equation for F: \(F_{max} = (M_1 + \mu_s * M_2) * \dfrac{\mu_s * F_{N1}}{M_1}\) Now we need to find the normal force \(F_{N1}\). Since the system is in equilibrium vertically, we have: \(F_{N1} = M_1 * g\) Plug the values of \(M_1\), \(\mu_s\), \(M_2\), and \(g\) into the equation for \(F_{max}\): \(F_{max} = (0.640 + 0.620 * 0.320) * \dfrac{0.620 * (0.640 * 9.81)}{0.640}\)

Calculate the maximum force F

Now, we can calculate the maximum force F that prevents the sliding of the block: \(F_{max} = 3.95308 \, N\) The maximum value of F that allows the block to accelerate with the cart, without sliding on top of the cart, is approximately 3.95 N.

Key concepts

Newton's second law

Understanding Newton's second law is fundamental when dealing with motion-related problems. This principle is famously summarized by the equation \(F=ma\), which states that the force (\(F\)) applied to an object is equal to the mass (\(m\)) of the object multiplied by its acceleration (\(a\)).

In the case of the block and cart system, this law helps to predict how the block will behave when a force is applied to the cart. Both objects will accelerate together only if the force of static friction acting on the block is sufficient to prevent it from slipping. If we understand the second law, we can calculate the maximum acceleration the block can have before sliding, using the equation \( F_{max} = (M_1 + \frac{\bar{\mu}_s \bar{\cdot} M_2}{M_1}) \bar{\cdot} (\bar{\mu}_s \bar{\cdot} F_{N1}) \), and subsequently, we can determine the maximum force that can be applied.

Static Friction

The concept of static friction comes into play when an object is resting on a surface and a force is applied. It is the force that keeps the object stationary up to a certain point, despite applied external forces. Static friction kicks in to counteract the applied force until the threshold limit, known as the maximum static friction, is exceeded.

For the block on the cart, it's this very force that prevents it from slipping as the cart accelerates. The formula \(F_{f1} = \bar{\mu}_s \bar{\cdot} F_{N1}\) represents the maximum force of static friction where \(\bar{\mu}_s\) is the coefficient of static friction—one of the vital numbers in ensuring that our block doesn't turn into a bobsleigh. The correct identification of these values is crucial, which is why we calculate the normal force (\(F_{N1}\)) as the product of the block's mass and gravity, and then use it to quantify the maximum force of static friction.

Acceleration

The concept of acceleration describes how quickly the velocity of an object is changing. In physics exercises, it's usually represented by 'a' and measured in meters per second squared (m/s²). Acceleration can be the result of an increase or decrease in speed and/or a change in direction.

In the exercise involving the block and the cart, we work out that both objects should have the same acceleration if the block is not to slip. This shared acceleration underpins the synchronization of their motions, and coupled with the coefficient of static friction, it dictates the maximum force that can be applied to the cart. It's fundamental to understand that the higher the value of \(\bar{\mu}_s\), the higher the acceleration can be before the block begins to slide. Through the equation \(a = \frac{\bar{\mu}_s \bar{\cdot} F_{N1}}{M_1}\), we find the maximum acceleration that the static friction can sustain before being overcome.