Question

A hanging mass, \(M_{1}=0.400 \mathrm{~kg}\), is attached by a light string that runs over a frictionless pulley to a mass \(M_{2}=1.20 \mathrm{~kg}\) that is initially at rest on a frictionless ramp. The ramp is at an angle of \(\theta=30.0^{\circ}\) above the horizontal, and the pulley is at the top of the ramp. Find the magnitude and direction of the acceleration, \(a_{2}\), of \(M_{2}\).

Short answer

Answer: The magnitude of the acceleration, \(a_2\), is \(0.842\) m/s^2, and the direction is downward along the ramp.

Step-by-step solution

Identify the forces acting on the masses

We will first identify all the forces acting on each mass: - Gravitational force, \(W_1\), acting on \(M_1\) downwards. - Gravitational force, \(W_2\), acting on \(M_2\) vertically downwards, which can be broken into two components: \(W_{2\parallel}\) parallel to the ramp and \(W_{2\perp}\) perpendicular to the ramp. - Tension force, \(T\), in the string connecting the masses, acts upwards on \(M_1\) and parallel to the ramp in the direction opposite to that of the acceleration for \(M_2\).

Write Newton's second law equations for the masses

Using Newton's second law, we can write the equations for the forces acting on both masses (\(M_1\) and \(M_2\)): 1) For \(M_1\): \(T - W_1 = M_1a_1\) 2) For \(M_2\), only considering the forces parallel to the ramp: \(W_{2\parallel} - T = M_2a_2\)

Determine the value of the gravitational forces

To determine the values of the gravitational forces, let's use the following formulas: \(W_1 = M_1g\), where g is the acceleration due to gravity (9.81 m/s^2) \(W_2 = M_2g\) Next, we need to find the component of \(W_2\) parallel to the ramp (\(W_{2\parallel}\)) \(W_{2\parallel} = W_2 \sin{\theta}\)

Express the forces and accelerations

Now, let's express the known forces and accelerations in terms of the given values: \(W_1 = 0.400 \cdot 9.81\) N \(W_2 = 1.20 \cdot 9.81\) N \(W_{2\parallel} = 1.20 \cdot 9.81 \cdot \sin{30^{\circ}}\) N

Relate the accelerations of the two masses

The accelerations of \(M_1\) and \(M_2\) are related by the fact that the string connecting the masses has a constant length. Therefore, if one mass accelerates by a certain amount, the other mass must also have the same magnitude of acceleration but in the opposite direction: \(a_2 = -a_1\)

Solve for the acceleration

We can now substitute the values of forces and acceleration relationship into the Newton's second law equations to solve for \(a_2\): 1) For \(M_1\): \(T - (0.400 * 9.81) = 0.400 (-a_2)\) 2) For \(M_2\): \((1.20 * 9.81 * \sin{30^{\circ}}) - T = 1.20a_2\) Now, we can solve these two equations simultaneously to find the value of \(a_2\). Solving equation (1) for tension force T and then substituting it into equation (2): \((1.20 * 9.81 * \sin{30^{\circ}}) -(0.400 * 9.81 + 0.400(-a_2)) = 1.20a_2\) Solving this equation for \(a_2\), we find: \(a_2 = -0.842\) m/s^2 The negative sign indicates that the acceleration is in the opposite direction to the ramp's angle (\(\theta\)) as mentioned in step 5. Thus, the magnitude of the acceleration, \(a_2\), is \(0.842\) m/s^2, and the direction is downward along the ramp.

Key concepts

Gravitational Forces

Gravitational forces are the invisible hands that pull objects toward the Earth. This force acts downward and is proportional to the mass of an object. For any mass, the gravitational force can be calculated using the formula:

• \( W = mg \)

where \( m \) is the mass and \( g \) is the acceleration due to gravity, typically \( 9.81 \, \text{m/s}^2 \).
In the exercise, both objects, \( M_1 \) and \( M_2 \), experience gravitational forces. For \( M_1 \), this force is straightforward, pulling directly downward. For \( M_2 \), its gravitational force needs to be split into components due to the ramp's angle.
The component parallel to the ramp, \( W_{2\parallel} \), is what actually pulls \( M_2 \) down the slope, and it can be found by \( W_2 \sin \theta \). This separation helps us analyze how these forces play a role in setting the blocks in motion.

Acceleration Calculations

Acceleration is the change in velocity over time, and Newton's Second Law relates force and acceleration. When forces act on the masses, they cause accelerations that need to be calculated.
The core idea here is Newton's Second Law, which can be expressed mathematically as:

• \( F = ma \)

where \( F \) is the net force, \( m \) is the mass, and \( a \) is acceleration.
In the problem, both masses, \( M_1 \) and \( M_2 \), have their forces calculated and net forces applied. These forces include tension in the string and components of gravitational force.
Using these forces, we set up equations for each mass based on Newton's Second Law: one for each mass in the relevant direction, accounting for how they are affected by the forces, as shown in the solution steps. By solving these equations, we find that the accelerations are related. If one moves a certain way, the other does too, due to the constant tension and length of the string. This understanding leads to calculating the exact magnitude and direction of acceleration for \( M_2 \).

Tension in Strings

Tension in strings acts as a connector between the two masses, ensuring consistent force transmission. It’s an internal force exerted along the string, pulling equally and oppositely on both masses.
In the original exercise, the tension \( T \) is crucial. It acts upwards for \( M_1 \) and along the ramp for \( M_2 \). Because the string and pulley are frictionless, the tension remains constant.
Newton's law equations for both masses involve tension. We adjust these to find how much of the gravitational force is counteracted by tension, or contributes to the acceleration, through:

• For \( M_1 \), \( T - W_1 = M_1 a_1 \)
• For \( M_2 \), \( W_{2\parallel} - T = M_2 a_2 \)

Solving these equations helps us find both the tension in the string and acceleration. Ultimately, these calculations reveal how the tension mediated through the string influences the dynamics of this physical system.