Problem 71 - University Physics with Modern Physics

Question

A block with mass \(m_{1}\) is placed on an inclined plane with slope angle \(\alpha\) and is connected to a second hanging block with mass \(m_{2}\) by a cord passing over a small, frictionless pulley (Fig. P5.68). The coefficient of static friction is \(\mu_{\mathrm{s}}\) and the coefficient of kinetic friction is \(\mu_{\mathrm{k}}\) (a) Find the mass \(m_{2}\) for which block \(m_{1}\) moves up the plane at constant speed once it is set in motion. (b) Find the mass \(m_{2}\) for which block \(m_{1}\) moves down the plane at constant speed once it is set in motion. (c) For what range of values of \(m_{2}\) will the blocks remain at rest if they are released from rest?

Step-by-Step Solution

Answer

(a) \(m_2 = m_1 (\sin(\alpha) + \mu_k \cos(\alpha))\); (b) \(m_2 = m_1 (\sin(\alpha) - \mu_k \cos(\alpha))\); (c) \(m_1 (\sin(\alpha) - \mu_s \cos(\alpha)) \leq m_2 \leq m_1 (\sin(\alpha) + \mu_s \cos(\alpha))\).

Step 1: Analyze Forces and Motion for m1 and m2

Identify the forces acting on block \(m_1\) on the inclined plane. These include gravitational force \(m_1g\), normal force \(N\), friction force \(f_k\), and tension force \(T\). For block \(m_2\), the forces are gravitational force \(m_2g\) and tension force \(T\). This preliminary analysis helps in setting up the equations for the motion of the blocks.

Step 2a: Establish Equations for Constant Speed Up the Plane

When block \(m_1\) moves up at constant speed, the net force along the incline is zero. Thus, the equation along the incline is:\[ T - m_1 g \sin(\alpha) - f_k = 0 \]where the kinetic friction \(f_k = \mu_k m_1 g \cos(\alpha)\). For block m2, the equation is:\[ m_2 g - T = 0\]Thus, \(T = m_2 g\). By equating and solving, find the expression for \(m_2\).

Step 2b: Solving for m2 (Constant Speed Up)

Substitute \(T = m_2 g\) from the second equation into the first:\[ m_2 g = m_1 g \sin(\alpha) + \mu_k m_1 g \cos(\alpha) \]Rearrange to solve for \(m_2\):\[ m_2 = m_1 (\sin(\alpha) + \mu_k \cos(\alpha)) \]

Step 3a: Establish Equations for Constant Speed Down the Plane

When block \(m_1\) moves down at constant speed, the net force along the incline is zero, thus the equation along the incline becomes:\[ m_1 g \sin(\alpha) - T - f_k = 0 \]And for block \(m_2\), the equation is:\[ T - m_2 g = 0\]Thus, \(T = m_2 g\). Substitute and solve for \(m_2\).

Step 3b: Solving for m2 (Constant Speed Down)

Substitute \(T = m_2 g\) into the first equation:\[ m_1 g \sin(\alpha) - \mu_k m_1 g \cos(\alpha) = m_2 g \]Rearrange this to solve for \(m_2\):\[ m_2 = m_1 (\sin(\alpha) - \mu_k \cos(\alpha)) \]

Step 4: Determine the Range for Static Equilibrium

For the blocks to remain at rest, we need to consider static friction (\(f_s = \mu_s N\)). The friction must counter any potential motion up or down:For upward motion potential:\[ T \leq m_1 g \sin(\alpha) + \mu_s m_1 g \cos(\alpha) \]For downward motion potential:\[ T \geq m_1 g \sin(\alpha) - \mu_s m_1 g \cos(\alpha) \]Combining gives:\[ m_1 (\sin(\alpha) - \mu_s \cos(\alpha)) \leq m_2 \leq m_1 (\sin(\alpha) + \mu_s \cos(\alpha)) \]

Step 7: Conclusion

The solutions provide values of \(m_2\) under which the system behaves according to different conditions, either moving at constant speed in different directions or remaining at rest.

Key Concepts

Static and Kinetic Friction

In inclined plane physics problems, understanding friction is vital. Friction opposes movement and comes in two forms: static and kinetic. Static friction prevents a stationary object from starting to move. It is the frictional force experienced until the applied force exceeds a threshold. For an object on an inclined plane, this can be expressed as:\[ f_s = \mu_s N \]where \( \mu_s \) is the coefficient of static friction and \( N \) is the normal force. This friction is at play when objects are still.

On the other hand, kinetic friction comes into play when an object is already in motion. It acts to slow down moving objects. Kinetic friction is generally less than static friction and is given by:\[ f_k = \mu_k N \]where \( \mu_k \) is the kinetic friction coefficient. Understanding these forces allows us to solve for the motion of objects on inclined planes when movement occurs.

Newton's Laws of Motion

Newton's laws are crucial in physics to understand how objects move and interact. Specifically, the first law of motion, which is often summed up as "an object at rest stays at rest, and an object in motion stays in motion," is foundational when examining forces on inclined planes. In the exercise, this is applied when the block moves at constant speed, implying net forces balance out, resulting in zero acceleration.

Additionally, the second law, \( F = ma \), indicates that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. This law helps us set up equations based on the forces acting upon both masses in the problem, especially considering the tension and gravitational forces involved.

Understanding and applying these laws allow us to predict how forces interact, helping to determine the behavior of the blocks on the incline and the necessary conditions for movement.

Pulley Systems

The pulley system in this exercise is a classic way to navigate force transmission. A pulley enables alterations in the direction of forces applied to an object and can minimize the effort needed to lift or move a mass. Here, the pulley connects the two blocks. Block \( m_1 \) on the incline and block \( m_2 \) suspended in the air. The rope's tension collaborates with gravitational and frictional forces for both blocks.

For any pulley system, the sum of forces around the pulley must be understood. It is crucial to know that the tension is the same throughout a frictionless pulley, which simplifies calculations significantly. This equality allows us to set up equations correlating tension and weights, as seen in solving for \( m_2 \) when block \( m_1 \) is in motion. Encountering pulley systems often means balancing multiple forces across different objects to ensure equilibrium or achieve desired motion.

Mastering such systems involves visualizing the setup and understanding how each component influences the rest.

Forces on Inclined Planes

Solving inclined plane problems in physics often requires breaking forces into components. These components help in analyzing real-world problems where slopes add complexity to standard force equations.

An inclined plane is typically analyzed by looking at the forces parallel and perpendicular to the surface. The gravitational force \( m_1 g \) is split into two components: \( m_1 g \sin(\alpha) \) parallel and \( m_1 g \cos(\alpha) \) perpendicular to the plane. The normal force counteracts the perpendicular component and is crucial for calculating both static and kinetic friction.

In solving this problem, the balance of parallel forces ensures the block moves at constant speed or stays still. By equating these forces with tension and friction, you can determine the conditions required for block \( m_1 \) to move up or down the plane. Thus, understanding how to decompose and utilize these forces is key to mastering inclined plane physics.