/*! 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 76 A \(5.0-\mathrm{kg}\) block rest... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 76

A \(5.0-\mathrm{kg}\) block rests on a \(30^{\circ}\) incline. The coefficient of static friction between the block and the incline is \(0.20\). How large a horizontal force must push on the block if the block is to be on the verge of sliding \((a)\) up the incline and \((b)\) down the incline?

Short Answer

Expert verified
To slide up: 17.1 N; to slide down: 9.4 N.

Step by step solution

01

Analyze Forces on the Incline

Draw a free body diagram to identify the forces acting on the block. The forces include the gravitational force (weight), the normal force, the frictional force, and the horizontal force. The weight of the block is \( mg \) and acts vertically downward, where \( m = 5.0 \, \text{kg} \) and \( g = 9.8 \, \text{m/s}^2 \).
02

Decompose the Gravitational Force

Decompose the gravitational force into components parallel and perpendicular to the incline. The parallel component is \( mg \sin \theta \) and the perpendicular component is \( mg \cos \theta \), where \( \theta = 30^\circ \).
03

Determine the Normal Force

The normal force \( N \) is balanced by the perpendicular component of the gravitational force. Therefore, \( N = mg \cos \theta \).
04

Calculate Maximum Static Friction

The maximum static friction force \( f_{s \text{ max}} \) is given by \( f_{s \text{ max}} = \mu_s N \), where \( \mu_s = 0.20 \). Calculate this using the value of \( N \) found in the previous step.
05

Set Up Equilibrium Conditions for (a)

For the block to be on the verge of sliding up the incline, the horizontal force must overcome both the static friction and the component of the gravitational force parallel to the incline. Set up the equation: \( F\cos\theta - f_{s \text{ max}} = mg\sin\theta \).
06

Solve for Horizontal Force (a)

Rearrange the equation from Step 5 to solve for the horizontal force \( F \) required to make the block slide up the incline: \[ F = \frac{mg (\sin\theta + \mu_s \cos\theta)}{\cos\theta - \mu_s \sin\theta} \]. Insert the given values and compute \( F \).
07

Set Up Equilibrium Conditions for (b)

For the block to be on the verge of sliding down the incline, the horizontal force must counteract the force of friction. The equation is: \( mg\sin\theta - f_{s \text{ max}} = F\cos\theta \).
08

Solve for Horizontal Force (b)

Rearrange the equation from Step 7 to solve for the horizontal force \( F \) required to make the block slide down the incline: \[ F = \frac{mg (\sin\theta - \mu_s \cos\theta)}{\cos\theta + \mu_s \sin\theta} \]. Insert the given values and compute \( F \).

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Key Concepts

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

Free Body Diagram
A free body diagram is a visual representation that helps us analyze the forces acting on a particular object. It is like creating a map of forces.
To solve the problem involving a block on an inclined plane, we first need to identify all the forces at play.
  • The gravitational force, which acts downward towards the center of the Earth.
  • The normal force, which acts perpendicular to the surface of the incline.
  • The frictional force, resisting motion opposite to potential sliding.
  • An external horizontal force, applied to move the block.
Drawing each of these forces with accurate direction and relative magnitude is crucial.
This makes it easier to set up equations for equilibrium conditions and solve for unknowns later on.
Gravitational Force Components
When dealing with inclined planes, it is essential to break down the gravitational force into components.
These components help us derive equations pertinent to motion and equilibrium acts.
  • Parallel Component: It pulls the block down along the incline, calculated as \( mg \sin \theta \).
  • Perpendicular Component: It acts perpendicular to the incline, calculated as \( mg \cos \theta \).
In this case, we have a block with mass \( 5.0 \, \text{kg} \) on a \( 30^\circ \) slope.
The gravitational component parallel to the slope factors into exerting a force down the slope, challenging the static friction.
Equilibrium Conditions
Equilibrium conditions refer to the state where all forces balance each other out effectively, meaning the object remains at rest or moves at a constant velocity.
In this problem, we examine the block at the verge of motion up or down the incline.
  • To balance for upward motion, the horizontal force must overcome both friction and the parallel gravitational component.
  • The equation becomes: \( F\cos\theta - f_{s \text{ max}} = mg\sin\theta \).
  • For downward movement, it must equalize the friction and parallel force, leading to: \( mg\sin\theta - f_{s \text{ max}} = F\cos\theta \).
Establishing these conditions is integral for solving the required horizontal force.
Inclined Plane Physics
The inclined plane is a classic physics problem because it combines several fundamental principles. Understanding it leads to insights about how forces work on tilted surfaces.
Key factors include:
  • The angle of incline \( \theta \), which affects how the gravitational force is split into its components.
  • Static friction that prevents the block from moving until enough force is applied.
  • The coefficient of static friction, given as \( 0.20 \), which influences frictional resistance.
  • Using trigonometry to manipulate and calculate force values accurately.
By examining these aspects and breaking problems into simpler parts, you can tackle the full breadth of inclined plane queries with step-by-step integrity.

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Most popular questions from this chapter

A \(12-\mathrm{kg}\) box is released from the top of an incline that is \(5.0 \mathrm{~m}\) long and makes an angle of \(40^{\circ}\) to the horizontal. A 60-N friction force impedes the motion of the box. ( \(a\) ) What will be the acceleration of the box, and \((b)\) how long will it take to reach the bottom of the incline?

Having hauled it to the top of a tilted driveway, a child is holding a wagon from rolling back down. The driveway is inclined at \(20^{\circ}\) to the horizontal. If the wagon weighs \(150 \mathrm{~N}\), with what force must the child pull on the handle if the handle is parallel to and pointing up the incline?

A horizontal cable pulls a \(200-\mathrm{kg}\) cart along a horizontal track. The tension in the cable is \(500 \mathrm{~N}\). Starting from rest, (a) How long will it take the cart to reach a speed of \(8.0 \mathrm{~m} / \mathrm{s}\) ? (b) How far will it have gone?

The Moon, whose mass is \(7.35 \times 10^{22} \mathrm{~kg}\), orbits the Earth, whose mass is \(5.98 \times 10^{24} \mathrm{~kg}\), at a mean distance of \(3.85 \times 10^{8} \mathrm{~m}\). It is held in a nearly circular orbit by the Earth-Moon gravitational interaction. Determine the force of gravity due to the planet acting on the Moon. From the univesal law of gravitation $$ F_{G}=G \frac{m M}{R^{2}} $$we get $$ F_{G}=6.673 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg} \frac{\left(7.35 \times 10^{22} \mathrm{~kg}\right)\left(5.98 \times 10^{24}\right)}{\left(3.85 \times 10^{8} \mathrm{~m}\right)^{2}} $$ which yields $$ F_{G}=1.98 \times 10^{20} \mathrm{~N} $$ This is also the force on the Earth due to the Moon.

A 700-N man stands on a scale on the floor of an elevator. The scale records the force it exerts on whatever is on it. What is the scale reading if the elevator has an acceleration of ( \(a\) ) \(1.8 \mathrm{~m} / \mathrm{s}^{2}\) up? (b) \(1.8 \mathrm{~m} / \mathrm{s}^{2}\) down? (c) \(9.8 \mathrm{~m} / \mathrm{s}^{2}\) down?
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