/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 36 A body of mass \(10 \mathrm{~kg}... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 36

A body of mass \(10 \mathrm{~kg}\) is lying on a rough horizontal surface. The coefficient of friction between the body and horizontal surface is \(0.577\). When the horizontal surface is inclined gradually, the body just begins to slide at a certain angle \(\alpha .\) This is called angle of repose. When angle of inelination is increased further, the body slides down with some acceleration. The minimum force required just to move the body up the incline is (a) \(100 \mathrm{~N}\) (b) \(57.7 \mathrm{~N}\) (c) \(111.5 \mathrm{~N}\) (d) \(157.7 \mathrm{~N}\)

Short Answer

Expert verified
(a) 100 N

Step by step solution

01

Understand the problem

We are given the mass of the body, the coefficient of friction and the situation in which it starts sliding down. Our task is to calculate the minimum force required to move the body upwards along the inclined plane just when it is about to slide.
02

Find the angle of repose

The angle of repose \( \alpha \) is the angle at which the body just starts to slide down. This angle depends on the coefficient of static friction \( \mu_s \). It is given by the formula:\[ \tan(\alpha) = \mu_s \]Given \( \mu_s = 0.577 \), we have:\[ \tan(\alpha) = 0.577 \]
03

Calculate angle \( \alpha \)

Using the formula \( \tan^{-1}(0.577) \), we can calculate the angle of repose \( \alpha \):\[ \alpha = \tan^{-1}(0.577) \approx 30掳 \]
04

Calculate gravitational component parallel to the incline

The gravitational force acting down the incline is given by \( mg \sin(\alpha) \).Given \( m = 10 \mathrm{~kg} \) and \( g = 9.8 \mathrm{~m/s^2} \), we have:\[ F_{gravity} = 10 \times 9.8 \times \sin(30掳) = 49 \mathrm{~N} \]
05

Calculate frictional force

The frictional force \( F_{friction} \) resisting motion up the plane is given by \( \mu_s mg \cos(\alpha) \).Calculating:\[ F_{friction} = 0.577 \times 10 \times 9.8 \times \cos(30掳) \approx 49.98 \mathrm{~N} \]
06

Calculate minimum force to move the body up

To move up, the applied force must overcome both the friction and the gravitational pull. So, \( F_{applied} = F_{friction} + F_{gravity} \).Thus:\[ F_{applied} = 49.98 + 49 = 98.98 \approx 100 \mathrm{~N} \]
07

Conclusion

The minimum force required just to move the body up the incline is \( 100 \mathrm{~N} \). Hence, the answer is (a) \( 100 \mathrm{~N} \).

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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 is a measure of how resistant a surface is to sliding. It is a dimensionless number, often denoted by \( \mu \), and it varies depending on the materials in contact. In this exercise, we deal with static friction, which is the resistance to the start of motion. Here, \( \mu_s = 0.577 \), which means that 57.7% of the gravitational force on the 10 kg object is resisted by friction.
  • Static friction must be overcome to initiate movement.
  • It is used to calculate the angle of repose for the inclined plane.
  • A higher coefficient indicates a rougher surface.
The coefficient of friction is crucial because it directly affects how easily an object can begin sliding and determines the force required to start and maintain motion.
Gravitational Force
Gravitational force is the force exerted by the Earth on any object resting on or near its surface. It's what keeps the object resting on the inclined plane. In our scenario, it not only acts vertically downwards but, when on an incline, it is split into two components: one parallel to the plane and one perpendicular to it.
  • The parallel component, \( mg \sin(\alpha) \), pulls the object down the incline and directly affects sliding.
  • The perpendicular component, \( mg \cos(\alpha) \), influences the frictional force.
  • Gravity鈥檚 strength depends on the mass \( m \) and Earth's gravity \( g = 9.8 \; \text{m/s}^2 \).
Understanding these components is essential for solving problems related to inclined planes, as they determine the net force acting on the object.
Frictional Force
Frictional force is a resisting force that acts opposite to the direction of applied force or motion. It prevents or slows down an object moving across a surface. For our inclined plane problem, this force depends on both the coefficient of friction and the normal force acting on the body.
  • The formula for the frictional force is \( F_{friction} = \mu_s mg \cos(\alpha) \).
  • It must be overcome by any applied force attempting to move the object uphill.
  • It remains significant until the object starts moving, after which kinetic friction takes over.
The frictional force, together with gravitational pull, defines the total force required to move the object and keep it in motion once it starts sliding.
Inclined Plane Physics
Inclined plane physics is the study of how forces operate on a slope. An inclined plane is a flat surface at an angle to the horizontal. It simplifies the mechanics of motion by breaking forces into components relative to the slope.
  • The angle of repose is key as it signifies when motion starts without any additional external force.
  • Understanding the vector components of gravity helps to analyze motion.
  • Inclined planes are classical physics problems that demonstrate how friction and gravitational forces interact.
By learning about inclined planes, we gain insights into how forces influence objects under real-world conditions, enabling us to solve practical problems involving ramps, slides, and even hills.

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Most popular questions from this chapter

A wooden box of mass \(8 \mathrm{~kg}\) slides down an inclined plane of inclination \(30^{\circ}\) to the horizontal with a constant acceleration of \(0.4 \mathrm{~ms}^{-2}\). What is the force of friction between the box and inclined plane? \(\left(g=10 \mathrm{~ms}^{-2}\right)\) (a) \(36.8 \mathrm{~N}\) (b) \(76.8 \mathrm{~N}\) (c) \(65.6 \mathrm{~N}\) (d) \(97.8 \mathrm{~N}\)

A block is gently placed on a conveyor belt moving horizontally with constant speed. After 4 s the velocity of the block becomes equal to the velocity of belt. If the coefficient of friction between the block and the belt is 0.2, then velocity of the conveyor belt is (a) \(2 \mathrm{~ms}^{-1}\) (b) \(4 \mathrm{~ms}^{-1}\) (c) \(6 \mathrm{~ms}^{-1}\) (d) \(8 \mathrm{~ms}^{-1}\)

Two weights \(w_{1}\) and \(w_{2}\) are suspended from the ends of a light string over a smooth fixed pulley. If the pulley is pulled up with acceleration \(g\), the tension in the string will be (a) \(\frac{4 w w_{2}}{w_{1}+w_{2}}\) (b) \(\frac{2 w w_{2}}{w_{1}+w_{2}}\) (c) \(\frac{w_{1}-w_{2}}{w_{1}+w_{j}}\) (d) \(\frac{w_{1} w_{2}}{2\left(w_{1}+w_{2}\right)}\)

A shell is fired from a cannon with velocity \(v \mathrm{~ms}^{-1}\) at an angle \(\theta\) with the horizontal direction. At the highest point in its path it explodes into two pieces of equal mass, One of the pieces retraces its path to the cannon and the speed in \(\mathrm{m} / \mathrm{s}\) of the piece immediately after the explosion is (a) \(3 v \cos \theta\) (b) \(2 v \cos \theta\) (c) \(\frac{3 \mathrm{~V}}{2} \cos \theta\) (d) \(\frac{\sqrt{3} v \cos \theta}{2}\)

A stone of mass \(0.25 \mathrm{~kg}\) tied to the end of a string is whirled round in a cirele of radius \(1.5 \mathrm{~m}\) with speed 40 rev/min in a horizontal plane. What is the tension in the string? What is the maximum speed with which the stone can be whirled around if the string can withstand a maximum tension of \(200 \mathrm{N?}\) :NCFRT] (a) \(200 \mathrm{~N}\) (b) \(300 \mathrm{~N}\) (c) \(2000 \mathrm{~N}\) (d) \(250 \mathrm{~N}\)
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