/*! 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 1 A rectangular wooden box \(10 \m... [FREE SOLUTION] | 91Ó°ÊÓ

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A rectangular wooden box \(10 \mathrm{~cm} \times 20 \mathrm{~cm} \times 40 \mathrm{~cm} \mathrm{~m}\) A rectangunw horizontal surface with its face of largest area is kept on a hor on the surface. A minimum force of \(10 \mathrm{~N}\) applied parallel to the surface sets the box in sliding motion along the surface. If the box is now kept with its face of smaller area in contact with the surface, the minimum force applied parallel to the surface, to set the box in motion, is (1) less than \(10 \mathrm{~N}\) (2) may be greater or less than \(10 \mathrm{~N}\) (3) greater than \(10 \mathrm{~N}\) (4) equal to \(10 \mathrm{~N}\)

Short Answer

Expert verified
Greater force than 10 N is required.

Step by step solution

01

Understand the Problem

A wooden box is initially placed with its largest face on the surface, and a minimum force of 10 N is needed to slide it. We need to determine the force required when the box rests on its smallest face.
02

Identify Surface Areas

Calculate the area of each face of the box. The largest face area is 20 cm \( \times \) 40 cm = 800 cm². The smallest face area is 10 cm \( \times \) 20 cm = 200 cm².
03

Concept of Friction

The force required to slide an object depends on the frictional force, which is proportional to the contact area when the object weight remains the same. Larger contact areas typically lower the pressure and hence friction slightly.
04

Compare Contact Areas

Since the initial face in contact was the largest, switching to the smallest face increases the pressure on the surface. Increased pressure generally increases the frictional force.
05

Conclusion Based on Frictional Force

The frictional force increases when the box is placed on its smaller face, meaning a greater force than 10 N is needed to overcome this friction and set the box in motion.

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Key Concepts

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

Sliding Friction
Sliding friction is a force that opposes the motion of two surfaces sliding past one another. It plays a crucial role when you try to move objects like our wooden box from rest. Sliding friction depends not just on the nature of the materials in contact but also on how they interact. When you push an object on a horizontal surface, this type of friction is what you need to overcome to set the object in motion. The measure of this resistance is given by the frictional force. In our example, we have a box that requires a certain force (10 N) to initiate sliding when positioned with its largest surface area touching the ground. By rearranging it so that a smaller surface is in contact with the floor, the sliding friction becomes greater, thus affecting the required force to maintain motion.
Force Calculation
Calculating the force required to move an object like our wooden box involves understanding the frictional force at play. The frictional force (\( F_f \)) is calculated by multiplying the coefficient of friction (\( \, \mu \)) with the normal force (\( F_N \)), which is typically the object's weight if the surface is horizontal: - \[ F_f = \mu \times F_N \]Here, the coefficient of friction is a constant that depends on the materials in contact. The normal force is the force exerted by a surface perpendicular to the object. For our box, it's simply its weight. Thus, any increase in the normal force, controlled by the box's orientation, affects the magnitude of \( F_f \), ultimately requiring more or less applied force to move the box. By flipping the box to let its smaller face contact the ground, we increase the pressure on the contact surface, thereby affecting the force needed to set it into motion.
Surface Area and Pressure
The surface area in contact with another surface can dramatically alter pressure, and by extension, friction. In physics, pressure is defined as the force exerted per unit area. When the box rests on its larger face, the contact area is greater with lower pressure since the same weight is distributed over a broader space. When turned to rest on a smaller face, the contact area decreases, leading to increased pressure, because- \[ \text{Pressure} = \frac{\text{Force}}{\text{Area}} \]Higher pressure implies that molecules on contrasting surfaces press more fiercely upon each other, enhancing the frictional force. This is why switching the box's orientation to make a smaller contact area results in needing greater force to move it, as the higher pressure increases resistance between the box and ground. This concept explains why more force is needed to induce motion when a smaller face is in contact.

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Most popular questions from this chapter

A box is accelerating with acceleration \(=20 \mathrm{~m} / \mathrm{s}^{2}\). A block of mass \(10 \mathrm{~kg}\) placed inside the box and is in contact with the vertical wall as shown. The friction coefficient between the block and the wall is \(\mu=0.6\) and take \(g=10 \mathrm{~m} / \mathrm{s}^{2}\). Then (1) The acceleration of the block will be \(20 \mathrm{~m} / \mathrm{s}^{2}\) (2) The friction force acting on the block will be \(100 \mathrm{~N}\) (3) The contact force between the vertical wall and the block will be \(100 \sqrt{5} \mathrm{~N}\) (4) The force between the vertical wall and the block is only electromagnetic in nature

A horizontal force just sufficient to move a body 0 \(4 \mathrm{~kg}\) lying on a rough horizontal surface is applied on it. The coefficient of static and kinetic friction between the body and the surface are \(0.8\) and \(0.6\), respectively. If the force continues to act even after the block has started moving, the acceleration of the block in \(\mathrm{ms}^{-2}\) is \(\left(g=10 \mathrm{~ms}^{-2}\right)\) (1) \(1 / 4\) (2) \(1 / 2\) (3) 2 (4) 4

A block \(A\) of mass \(2 \mathrm{~kg}\) is placed over another block \(B\) of mass \(4 \mathrm{~kg}\), which is placed over a smooth horizontal floor. The cocfficient of friction between \(A\) and \(B\) is \(0.4\), When a honzontal force of magnitude \(10 \mathrm{~N}\) is applied on \(A\), the accelcration of blocks \(A\) and \(B\) are (1) \(\mid \mathrm{ms}^{2}\) and \(2 \mathrm{~ms}^{-3}\), respectively (2) \(5 \mathrm{~ms}^{2}\) and \(2.5 \mathrm{~ms}^{-2}\), respectively (3) Both the blocks will moves together with acceleration \(1 / 3 \mathrm{~ms}^{-2}\) (4) Both the blocks will move together with acceleration, \(5 / 3 \mathrm{~ms}^{2}\)

A triangular prism of mass \(M\) with a block of mass \(m\) placed on it is released from rest on a smooth inclined plane of inclination \(\theta\). The block does not slip on the prism. Then (1) The acceleration of the prism is \(g \cos \theta\). (2) The acceleration of the prism is \(g \tan \theta\). (3) The minimum coefficient of friction between the block and the prism is \(\mu_{\min }=\cot \theta\). (4) The minimum coefficient of friction between the block and the prism is \(\mu_{\min }=\tan \theta\).

A monkey of mass \(m\) clings to a rope slung over a fixed pulley. The opposite end of the rope is tried to a weight of mass \(M\) lying on a horizontal table. The coefficient of friction between the weight and the table is \(\mu\). Find the acceleration of weight and the tension of the rope for two cases. The monkey moves downwards with respect to the rope with an acceleration \(b\). The tension of rope is (1) \(\frac{M m(\mu g+g+b)}{M+m}\) (2) \(\frac{M m(\mu g-g+b)}{M+m}\) (3) \(\frac{M m(\mu g+g-b)}{M+m}\) (4) \(\frac{M m(\mu g-g-b)}{M+m}\)

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