/*! 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 56 An ice cart of mass \(60 \mathrm... [FREE SOLUTION] | 91Ó°ÊÓ

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An ice cart of mass \(60 \mathrm{~kg}\) rests on a horizontal snow pateh with coefficient of statie friction \(1 / 3\) Assuming that there is no vertical acceleration, find the magnitude of the maximum horizontal force required to move the ice cart. \(\left(g=9.8 \mathrm{~ms}^{-2}\right)\) [BVP Engg, 2007] (a) \(100 \mathrm{~N}\) (b) \(110 \mathrm{~N}\) (c) \(209 \mathrm{~N}\) (d) \(196 \mathrm{~N}\)

Short Answer

Expert verified
The maximum horizontal force required is 196 N.

Step by step solution

01

Understanding Static Friction

Static friction is the force that keeps the object at rest. It is quantified by the equation \( f_s = \mu_s N \), where \( \mu_s \) is the coefficient of static friction and \( N \) is the normal force.
02

Calculate the Normal Force

For an object resting on a horizontal surface with no vertical acceleration, the normal force \( N \) is equal to the gravitational force \( mg \). Given the mass \( m = 60 \) kg and the acceleration due to gravity \( g = 9.8 \text{ m/s}^2 \), calculate \( N \):\[ N = mg = 60 \times 9.8 = 588 \text{ N} \]
03

Utilize the Coefficient of Static Friction

The coefficient of static friction is \( \mu_s = \frac{1}{3} \). We use this to calculate the maximum static friction force:\[ f_s = \mu_s N = \frac{1}{3} \times 588 = 196 \text{ N} \]
04

Determine the Maximum Horizontal Force

The maximum horizontal force required to just overcome static friction and move the ice cart is equal to the maximum static friction force, which is \( 196 \text{ N} \).

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Key Concepts

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

Normal Force
When an object, like the ice cart in this exercise, is placed on a horizontal surface, the force that acts perpendicularly to this surface is called the normal force. It is denoted by the symbol \( N \). For objects that are at rest and have no vertical acceleration, the normal force equals the gravitational force acting on the object.

To find the normal force, you consider the mass \( m \) of the object and multiply it by the acceleration due to gravity \( g \), giving you the equation:
  • \( N = mg \)
For our ice cart of mass \( 60 \text{ kg} \) on a snow patch, this results in a normal force of \[ N = 60 \times 9.8 = 588 \text{ N} \]
This means the cart is exerting a force of 588 Newtons on the snow patch because of its weight.
Coefficient of Friction
The coefficient of friction is a measure that defines the frictional force between two surfaces. In this exercise, we are concerned with static friction, which prevents the object from moving. The coefficient of static friction is represented by \( \mu_s \).

It usually ranges between 0 and 1, where higher values indicate stronger friction between surfaces. In our case, the snow and the ice cart have a static friction coefficient \( \mu_s = \frac{1}{3} \).

This value is crucial for calculating the maximum static friction force. It acts as a multiplier of the normal force, helping us understand how resistant the object is to starting movement. The smaller the coefficient, the easier it is for the object to move.
Maximum Static Friction Force
The maximum static friction force is the force threshold that must be exceeded to make an object move. Before this threshold is reached, static friction acts to keep the object at rest.

The equation to calculate maximum static friction force is:
  • \( f_s = \mu_s N \)
For the ice cart, with \( \mu_s = \frac{1}{3} \) and \( N = 588 \text{ N} \), the maximum static friction force calculates as:\[f_s = \frac{1}{3} \times 588 = 196 \text{ N}\]This result tells us that a horizontal force greater than 196 Newtons is needed to overcome static friction and move the ice cart. It's essentially the friction barrier that must be surpassed.

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Most popular questions from this chapter

Two masses of \(3 \mathrm{~kg}\) and \(5 \mathrm{~kg}\) are suspended from the ends of an unstreatchable massless cord passing over a frictionless pulley. When the masses are released, the pressure on the pulley is (a) \(2 \mathrm{~kg} \mathrm{f}\) (b) \(7.5 \mathrm{~kg}\) f (c) \(8 \mathrm{~kg} \mathrm{f}\) (d) \(15 \mathrm{kgf}\)

A chain lies on a rough horizontal table. It starts sliding when one-fourth of its length hangs over the edge of the table. The coefficient of static friction between the chain and the surface of the table is (a) \(\frac{1}{2}\) (b) \(\frac{1}{3}\) (c) \(\frac{1}{4}\) (d) \(\frac{1}{5}\)

A 5 kg stationary bomb is exploded in three parts having mass \(1: 1: 3\) respectively. Parts having same mass move in perpendicular directions with velocity \(39 \mathrm{~ms}^{-1}\), then the velocity of bigger part will be (a) \(10 \sqrt{2} \mathrm{~ms}^{-1}\) (b) \(\frac{10}{\sqrt{2}} \mathrm{~ms}^{-1}\) (c) \(13 \sqrt{2} \mathrm{~ms}^{-1}\) (d) \(\frac{15}{\sqrt{2}} \mathrm{~ms}^{-1}\)

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A ball is travelling with uniform translatory motion. This means that (a) it is at rest (b) the path can be a straight line or circular and the ball travels with uniform speed (c) all parts of the ball have the same velocity [magnitude and direction] and the velocity is constant (d) the centre of the ball moves with constant velocity and the ball spins about its centre unifomly

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