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Chapter 5: Problem 42

A \(25.0-\mathrm{kg}\) box of textbooks rests on a loading ramp that makes an angle \(\alpha\) with the horizontal. The coefficient of kinetic friction is \(0.25,\) and the coefficient of static friction is \(0.35 .\) (a) As the angle \(\alpha\) is increased, find the minimum angle at which the box starts to slip. (b) At this angle, find the acceleration once the box has begun to move. (c) At this angle, how fast will the box be moving after it has slid 5.0 \(\mathrm{m}\) along the loading ramp?

Short Answer

Expert verified
(a) \( \alpha = \tan^{-1}(0.35) \); (b) \( a = g (\sin \alpha - 0.25 \cos \alpha) \); (c) \( v = \sqrt{2as} \).

Step by step solution

01

Identify Forces Involved

Let's first identify the forces acting on the box. The gravitational force acts downwards, and it has two components: one perpendicular to the ramp (normal force, \( F_n \)) and one parallel to the ramp (force of gravity along the ramp, \( F_{g, \parallel} \)). Additionally, there's the static frictional force acting against the box's tendency to slide.
02

Set Up Equations for Static Equilibrium

For the box to just start slipping, the static frictional force must equal the component of gravitational force parallel to the ramp. The static frictional force \( F_f \) is given by \( F_f = \mu_s F_n \), where \( \mu_s = 0.35 \). The normal force \( F_n = mg \cos \alpha \) and \( F_{g, \parallel} = mg \sin \alpha \). Equating these, we have:\[ \mu_s mg \cos \alpha = mg \sin \alpha \]
03

Solve for Minimum Angle \( \alpha \)

Solve the equation \( \mu_s mg \cos \alpha = mg \sin \alpha \) for \( \alpha \) by canceling \( mg \) from both sides:\[ \mu_s \cos \alpha = \sin \alpha \]Dividing both sides by \( \cos \alpha \), we get:\[ \mu_s = \tan \alpha \]Hence, \( \alpha = \tan^{-1}(\mu_s) = \tan^{-1}(0.35) \). Calculate \( \alpha \).
04

Calculate the Box's Acceleration Once It Moves

Once the box starts sliding, the kinetic frictional force \( F_{k} = \mu_k F_n \), where \( \mu_k = 0.25 \). Using Newton's second law parallel to the ramp:\[ F_{net} = ma = mg \sin \alpha - \mu_k mg \cos \alpha \]Solve for \( a \):\[ a = g (\sin \alpha - \mu_k \cos \alpha) \]
05

Calculate Speed After Sliding 5.0 m

Use the kinematic equation to find the final speed \( v \) after sliding 5.0 m:\[ v^2 = u^2 + 2as \]where \( u = 0 \) (initial velocity), \( a \) is the acceleration found in Step 4, and \( s = 5.0 \, \text{m} \). Substitute and solve for \( v \):\[ v = \sqrt{2as} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Friction
Friction is a force that opposes the relative motion or tendency of such motion of two surfaces in contact. There are two main types of friction we need to consider: static and kinetic friction.

- **Static Friction**: It acts on objects that are not moving. In this problem, it prevents the box from sliding down the inclined plane as we gradually increase the angle. The static frictional force can be calculated using the formula: \[ F_f = \mu_s F_n \] where \( \mu_s \) is the coefficient of static friction and \( F_n \) is the normal force. Static friction must be overcome for the object to start moving.
- **Kinetic Friction**: Once the box begins to slide, kinetic friction takes over. It is usually less than static friction, which is why objects start to move more easily after overcoming static friction. The kinetic frictional force is given by: \[ F_k = \mu_k F_n \] where \( \mu_k \) is the coefficient of kinetic friction. This force continues to act as long as the box is in motion.
Inclined Plane
An inclined plane is a flat surface tilted at an angle, often used to simplify lifting heavy objects. In our scenario, the loading ramp is an inclined plane where books are placed. Understanding the forces acting on an inclined plane is crucial for solving problems involving motion along a slope.

- **Normal Force**: The normal force \( F_n \) is perpendicular to the plane and counteracts the component of gravitational force pressing the box against the surface. It is calculated by: \[ F_n = mg \cos \alpha \] where \( m \) is mass, \( g \) is the gravitational acceleration, and \( \alpha \) is the angle of inclination.
- **Gravitational Force Parallel to the Plane**: The component of the gravitational force that causes the box to slide down the incline is given by: \[ F_{g, \parallel} = mg \sin \alpha \] Balancing these forces involves considering both static and kinetic friction.
Kinematics
Kinematics deals with the motion of objects without considering the causes of motion. In this problem, once the box starts sliding, we use kinematics to determine how fast it will be moving after sliding a certain distance.

- **Acceleration**: First, find the box鈥檚 acceleration using Newton's second law, taking into account all forces acting along the plane. The formula is: \[ a = g (\sin \alpha - \mu_k \cos \alpha) \] which combines the gravitational pull along the plane and the opposing kinetic friction.
- **Kinematic Equation**: To determine the velocity after sliding a distance, use the equation: \[ v^2 = u^2 + 2as \] Here, \( u \) is the initial velocity (zero in this case), \( a \) is the acceleration, and \( s \) is the distance traveled. Solve for \( v \) to find the final speed after sliding.
Gravitational Force
Gravitational force is the attractive force exerted by the Earth on objects, pulling them towards its center. It plays a crucial role in understanding the motion of objects on inclined planes. In this exercise, the gravitational force can be split into two components.

- **Perpendicular Component**: This component acts perpendicular to the plane and contributes to the normal force felt by the object. It is calculated as: \[ F_{g, \perp} = mg \cos \alpha \] serving as the counterbalance to the weight of the box pressing into the ramp.
- **Parallel Component**: This drives the box down the ramp. Calculated as: \[ F_{g, \parallel} = mg \sin \alpha \] which needs to be countered by static friction for the box to stay still. When the force exceeds static friction, it results in motion due to the unbalanced force, illustrating Newton's Laws of Motion in action.

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Most popular questions from this chapter

A small button placed on a horizontal rotating platform with diameter 0.320 \(\mathrm{m}\) will revolve with the platform when it is brought up to a speed of 40.0 rev/min, provided the button is no more than 0.150 \(\mathrm{m}\) from the axis. (a) What is the coefficient of station between the button and the platform? (b) How far from the axis can the button be placed, without slipping, if the platform rotates at 60.0 rev/min?

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