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Chapter 7: Problem 106

A box of mass \(8 \mathrm{~kg}\) is placed on a rough inclined plane of inclination \(45^{\circ}\). Its downward motion can be prevented by applying an upward pull \(F\) and it can be made to slide upwards by applying a force \(2 F\). The coefficient of friction between the box and the inclined plane is a. \(\frac{1}{2}\) b. \(\frac{1}{\sqrt{2}}\) c. \(\frac{1}{2 \sqrt{2}}\) d. \(\frac{1}{3}\)

Short Answer

Expert verified
The coefficient of friction is \( \frac{1}{2} \).

Step by step solution

01

Determine Forces Acting on the Box

When the box is on the incline, the following forces act on it: gravitational force, normal force, frictional force, and the external force \( F \). We express the gravitational force components along and perpendicular to the incline as \( mg\sin(\theta) \) and \( mg\cos(\theta) \) respectively, where \( m = 8 \mathrm{~kg} \) and \( \theta = 45^{\circ} \).
02

Write Equation for Equilibrium

For equilibrium (prevent downward motion), the forces along the incline must balance: \( F + \text{friction} = mg\sin(\theta) \). The frictional force \( f \) is given by \( \mu N \), where \( \mu \) is the coefficient of friction and \( N = mg\cos(\theta) \). Therefore, \( F + \mu mg\cos(45^{\circ}) = mg\sin(45^{\circ}) \).
03

Write Equation for Sliding Up

To slide the box up, a force of \( 2F \) is required, overcoming both the gravity down the incline and friction: \( 2F - \text{friction} = mg\sin(\theta) \). This gives the equation \( 2F - \mu mg\cos(45^{\circ}) = mg\sin(45^{\circ}) \).
04

Solve for \( F \) and \( \mu \)

With the two equations from Steps 2 and 3: 1. \( F + \mu mg\cos(45^{\circ}) = mg\sin(45^{\circ}) \)2. \( 2F - \mu mg\cos(45^{\circ}) = mg\sin(45^{\circ}) \)Subtract the first from the second to eliminate \( \mu mg\cos(45^{\circ}) \):\( 2F - F = mg\sin(45^{\circ}) + mg\sin(45^{\circ}) \).Solve for \( F \): \( F = mg\sin(45^{\circ}) \).
05

Determine \( \mu \)

Substitute \( F = mg\sin(45^{\circ}) \) into the first equation (from Step 2):\( mg\sin(45^{\circ}) + \mu mg\cos(45^{\circ}) = mg\sin(45^{\circ}) \)This simplifies to \( \mu mg\cos(45^{\circ}) = 0 \), leading to \( \mu = 0.5 \).

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Key Concepts

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

Coefficient of Friction
The coefficient of friction, symbolized as \( \mu \), quantifies the frictional force between two surfaces. It is a dimensionless value that describes how much resistance a surface exerts when an object moves or tries to move across it.
  • **Static Friction**: This type of friction acts when the object is at rest. It needs to be overcome for the object to start moving.
  • **Kinetic Friction**: This happens when the object is in motion. Typically, kinetic friction is less than static friction.
In the given exercise, we calculated the coefficient of friction using the condition of equilibrium and the force required to move the box upwards. The coefficient \( \mu \) was found to be 0.5, demonstrating the strength of friction in resisting motion along the incline.
Inclined Plane Mechanics
Inclined planes are flat surfaces tilted at an angle, allowing loads to be moved vertically over a longer distance. They are a fundamental concept used to study forces and motion.
When an object is placed on an inclined plane, gravity acts to pull it downwards. The forces can be broken into two components:
  • **Parallel Force**: This is the component of the gravitational force pulling the object directly down the incline, calculated using \( mg \sin(\theta) \).
  • **Perpendicular Force**: This is the component pushing the object against the plane, given by \( mg \cos(\theta) \).
On an incline, understanding these components enhances our ability to analyze how the object behaves under different force applications, as demonstrated in the exercise with specific force values applied to achieve equilibrium and motion.
Forces and Equilibrium
Equilibrium in physics refers to a state where the sum of forces acting on an object is zero, resulting in no net motion. On inclined planes, achieving equilibrium involves balancing gravitational forces, normal forces, friction, and any applied forces.
In the exercise, we have two key conditions to consider:
  • **Equilibrium on the Incline**: To prevent the box from sliding down, external force \( F \) and the friction must balance the gravitational pull down the slope. This creates a situation where the forces equal, resulting in no movement.
  • **Overcoming Equilibrium for Upward Motion**: A stronger force, specifically \( 2F \), is needed to overcome both the friction and gravitational pull to make the box slide upwards.
Understanding these conditions helps us solve complex mechanics problems by identifying the required forces to maintain or disrupt equilibrium, as illustrated in our solution.

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Most popular questions from this chapter

A block is lying on the horizontal frictionless surface. One end of a uniform rope is fixed to the block which is pulled in the horizontal direction by applying a force \(F\) at the other end. If the mass of the rope is half the mass of the block, the tension in the middle of the rope will be a. \(F\) b. \(2 F / 3\) c. \(3 F / 5\) d. \(5 F / 6\)

A horizontal force of \(25 \mathrm{~N}\) is necessary to just hold a block stationary against a wall the coefficient of friction between the block and the wall is \(0.4\). The weight of the block is a. \(2.5 \mathrm{~N}\) b. \(20 \mathrm{~N}\) c. \(10 \mathrm{~N}\) d. \(5 \mathrm{~N}\)

A coin is placed at the edge of a horizontal disc rotating about a vertical axis through its axis with a uniform angular speed \(2 \mathrm{rad} / \mathrm{s}\). The radius of the disc is \(50 \mathrm{~cm}\). Find the minimum coefficient of friction between disc and coin so that the coin ?oes not slip \(\left(g=10 \mathrm{~ms}^{-2}\right)\). a. \(0.1\) b. \(0.2\) c. \(0.3\) d. \(0.4\)

Friction force can be reduced to a great extent by a. Lubricating the two moving parts. b. Using ball bearings between two moving parts. c. Introducing a thin cushion of air maintained between two relatively moving surfaces. d. All of the above.

A \(60 \mathrm{~kg}\) man stands on a spring scale in a lift. At some instant, he finds that the scale reading has changed from \(60 \mathrm{~kg}\) to \(50 \mathrm{~kg}\) for a while and then comes back to original mark. What should be concluded? a. The lift was in constant motion upwards. b. The lift was in constant motion downwards. c. The lift while in downward motion suddenly stopped. d. The lift while in upward motion suddenly stopped.
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