/*! 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 111 A horizontal force of \(25 \math... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 111

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. \(5 \mathrm{~N}\)

Short Answer

Expert verified
The weight of the block is 10 N (option c).

Step by step solution

01

Understand the Problem

We are given a horizontal force and a coefficient of friction, and we need to find the weight of a block that the force is holding against a wall. The problem involves the balance of forces including frictional force and gravitational force.
02

Identify the Forces Involved

The forces acting on the block are the force due to gravity (weight), the normal force exerted by the wall, and the frictional force that prevents the block from sliding down. The normal force here is equal to the horizontal force applied, which is given as 25 N.
03

Apply the Friction Formula

The frictional force can be calculated using the formula: \( f = \mu N \), where \(f\) is the frictional force, \(\mu\) is the coefficient of friction, and \(N\) is the normal force. Here, \(N = 25\, \mathrm{N}\) and \(\mu = 0.4\).
04

Calculate the Frictional Force

Substitute the given values into the friction formula: \( f = 0.4 \times 25 = 10 \mathrm{~N} \). This frictional force acts to balance the weight of the block.
05

Weight Equals Frictional Force

For the block to remain stationary, the weight of the block must equal the frictional force that prevents it from sliding. Therefore, the weight of the block is 10 N.

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Key Concepts

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

Normal Force
When dealing with objects resting against surfaces, the concept of normal force is crucial. The normal force is the perpendicular force that a surface exerts on an object in contact with it. In this scenario, the block is pressed against a wall by a horizontal force.
As a reaction, the wall exerts an equal and opposite normal force back on the block.
In our exercise, this normal force is equal to the horizontal force applied, which is given as 25 N. Hence, the normal force is essentially the amount of force that acts perpendicularly on the surface.
  • Normal force helps balance other forces like friction.
  • It's essential in the calculation of frictional force.
Understanding normal force assists in our broader understanding of how forces interact with objects in equilibrium.
Frictional Force
Frictional force is a resistive force that occurs when two surfaces are in contact. It opposes the motion between the surfaces. In this exercise, it is what prevents the block from sliding down the wall.
The frictional force can be calculated with the formula:
\[ f = \mu N \] where:
  • \( f \) is the frictional force.
  • \( \mu \) is the coefficient of friction.
  • \( N \) is the normal force.
In our context, the coefficient of friction \( \mu \) is 0.4 and the normal force \( N \) is 25 N.
Substituting these values gives us a frictional force of 10 N.
This friction is crucial because it determines the maximum load (weight) that can be supported without movement, holding the block stationary against gravity.
Balance of Forces
The balance of forces is a fundamental principle in physics that ensures an object remains in equilibrium. In this exercise, the block is stationary, implying the forces acting on it are balanced.

The forces involved include:
  • The gravitational force acting downward (the weight of the block).
  • The frictional force acting upward to oppose this weight.
  • The normal force exerted by the wall.
Each of these forces plays a critical role in maintaining the block's stationary position.
The key here is that the frictional force counteracts the weight of the block, keeping it from sliding down. This shows us the relationship and delicate balance between the forces:
  • The downward gravitational force is equal to the upward frictional force.
  • The normal force supports the frictional force needed.
This balance allows for the block to remain in place without moving.

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Most popular questions from this chapter

A professor holds an eraser against a vertical chalkboard by pushing horizontally on it. He pushes with a force that is much greater than it required to hold the eraser. The force of friction exerted by the board on the eraser increases if he a. pushes eraser with slightly greater force. b. pushes eraser with slightly less force. c. raises his elbow so that the force he exerts is slightly downward but has same magnitude. d. lowers his elbow so that the force he exerts is slightly upward but has the same magnitude.

A vehicle is moving with a velocity \(v\) ou a curved road of width \(b\) and radius of curvature \(R\). For counteracting the centrifugal force on the vehicle, the difference in elevation required in between the outer and inner edges of the rod is a. \(v^{2} b / R g\) b. \(v b / R g\) c. \(v b^{2} / R g\) d. \(v b / R^{2} g\)

The masses of the blocks \(A\) and \(B\) are \(m\) and \(M .\) Between \(A\) and \(B\) there is a constant frictional force \(F\), and \(B\) can slide frictionlessly on horizontal surface (see Fig. \(7.378\) ). \(A\) is set in motion with velocity while \(B\) is at rest. What is the distance moved by \(A\) relative to \(B\) before they move with the same velocity? a. \(\frac{m M v_{0}^{2}}{F(m-M)}\) b. \(\frac{m M v_{0}^{2}}{2 F(m-M)}\) c. \(\frac{m M v_{0}^{2}}{F(m+M)}\) d. \(\frac{m M v_{0}^{2}}{2 F(m+M)}\)

Two trolleys 1 and 2 are moving with accelerations \(a_{1}\) and \(a_{2}\), respectively, in the same direct. A bock of mass \(m\) on trolley 1 is in equilibrium from the frame of observer stationary with respect to trolley 2 . The magnitude of friction force on block due to trolley is (assume that no horizontal force other than friction force is acting on block) a. \(m\left(a_{1}-a_{2}\right)\) b. \(m a_{2}\) c. \(m a_{1}\) d. Data insufficient

A block of mass \(1 \mathrm{~kg}\) is at rest on a horizontal table. The coefficient of static friction between the block and the table is \(0.50\). If \(g=10 \mathrm{~ms}^{-2}\), then the magnitude of a force acting upwards at an angle of \(60^{\circ}\) from the horizontal that will just start the block moving is: a. \(5 \mathrm{~N}\) b. \(5.36 \mathrm{~N}\) c. \(74.6 \mathrm{~N}\) d. \(10 \mathrm{~N}\)
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