/*! 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 31 . SSM Two blocks, of weights \(3... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

. SSM Two blocks, of weights \(3.6 \mathrm{~N}\) and \(7.2 \mathrm{~N}\), are connected by a massless string and slide down a \(30^{\circ}\) inclined plane. The coefficient of kinetic friction between the lighter block and the plane is \(0.10\), and the coefficient between the heavier block and the plane is 0.20. Assuming that the lighter block leads, find (a) the magnitude of the acceleration of the blocks and (b) the tension in the taut string.

Short Answer

Expert verified
Acceleration is approximately 3.56 m/s², and tension is about 0.467 N.

Step by step solution

01

Calculate Normal Forces

The normal force for each block can be found using the equation \( N = W \cos(\theta) \), where \( W \) is the weight of the block and \( \theta \) is the angle of the incline. For the lighter block, \( N_1 = 3.6 \cos(30^{\circ}) \approx 3.12 \mathrm{~N}\). For the heavier block, \( N_2 = 7.2 \cos(30^{\circ}) \approx 6.24 \mathrm{~N}\).
02

Calculate Friction Forces

The friction force \( f \) for each block is calculated using \( f = \mu N \). For the lighter block, \( f_1 = 0.10 \times 3.12 \approx 0.312 \mathrm{~N} \). For the heavier block, \( f_2 = 0.20 \times 6.24 \approx 1.248 \mathrm{~N} \).
03

Find Total Forces Parallel to the Incline

Calculate the gravitational components and friction forces parallel to the incline. The gravitational force for the lighter block is \( 3.6 \sin(30^{\circ}) = 1.8 \mathrm{~N} \), and for the heavier block is \( 7.2 \sin(30^{\circ}) = 3.6 \mathrm{~N} \). Total parallel force \( F \) is given by \( F = (1.8 - 0.312) + (3.6 - 1.248) = 3.84 \mathrm{~N} \).
04

Calculate Net Accelerations

Use Newton's Second Law, \( F = ma \). The mass \( m \) of the blocks can be deduced from their weights (\( W = mg \)). Total mass is \( 3.6/9.8 + 7.2/9.8 = 1.08 \mathrm{~kg} \). So, \( a = \frac{3.84}{1.08} \approx 3.56 \mathrm{~m/s^2} \).
05

Calculate Tension in the String

Consider the lighter block. The net force on it is \( 3.6 \sin(30^{\circ}) - T - f_1 = m_1a \). Solving for the tension \( T \), we have \( T = 1.8 - m_1a - f_1 = 1.8 - (0.367)a - 0.312 \approx 0.467 \mathrm{~N} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Kinetic Friction
Kinetic friction is a force that opposes the motion of two surfaces sliding past each other. It acts in the opposite direction to the movement. Consider two blocks sliding down an inclined plane; one with a coefficient of kinetic friction of 0.10 and the other 0.20. The coefficient denotes how strong the frictional force is in relation to the normal force, essentially how 'grippy' the surface is.

The force of kinetic friction can be calculated with the formula:
  • \( f = \mu N \)
Where:
  • \( f \) is the frictional force,
  • \( \mu \) is the coefficient of kinetic friction,
  • \( N \) is the normal force.
In our example, the lighter block has a friction force of approximately 0.312 N, while the heavier block experiences a friction force of approximately 1.248 N. This illustrates how different coefficients affect the friction experienced by each block. Understanding kinetic friction is crucial in dynamics, as it directly influences how easily objects can move.
Inclined Plane
An inclined plane is a flat surface tilted at an angle to the horizontal. It is a classic object in physics that helps study how forces act on an object on a slope. When an object is on an inclined plane, gravity acts to pull it down the slope. However, not all the gravitational force goes straight down; it can be broken into two components: parallel and perpendicular to the incline.

To calculate these components, you use trigonometric functions:
  • Parallel to the incline: \( W \sin(\theta) \)
  • Perpendicular (Normal force): \( W \cos(\theta) \)
In this example, both blocks slide down a 30-degree incline. The normal force for each block is affected by the angle, and it is this force that is used to calculate kinetic friction. Inclined planes help to better understand gravitational components and calculate the forces acting on objects in dynamic scenarios.
Tension in a String
Tension in a string is the force exerted through a string, rope, or cable when it is pulled tight by forces acting from opposite ends. In our example, two blocks are connected by a massless string, and the tension in this string must be considered when analyzing their movement down the incline.

The tension in the string serves to accelerate both blocks at the same rate. When calculating tension for the lighter block, we account for gravitational pull, frictional force, and the acceleration of the system. By using Newton's Second Law, we find:
  • Net Force on lighter block = Gravitational force - Tension - Frictional force
  • \( T = 1.8 - (0.367)a - 0.312 \)
Solving this gives an approximate tension of 0.467 N. Understanding tension in the string is vital for problems involving connected objects, as it can significantly alter not only the motion but also how each object interacts dynamically with others connected by the string.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

SSM A \(68 \mathrm{~kg}\) crate is dragged across a floor by pulling on a rope attached to the crate and inclined \(15^{\circ}\) above the horizontal. (a) If the coefficient of static friction is \(0.50\), what minimum force magnitude is required from the rope to start the crate moving? (b) If \(\mu_{k}=0.35\), what is the magnitude of the initial acceleration of the crate?

Calculate the magnitude of the drag force on a missile \(53 \mathrm{~cm}\) in diameter cruising at \(250 \mathrm{~m} / \mathrm{s}\) at low altitude, where the density of air is \(1.2 \mathrm{~kg} / \mathrm{m}^{3}\). Assume \(C=0.75\).

\({ }^{\circ} 6 \cdot \) ssin A \(1000 \mathrm{~kg}\) boat is traveling at \(90 \mathrm{~km} / \mathrm{h}\) when its engine is shut off. The magnitude of the frictional force \(\vec{f}_{k}\) between boat and water is proportional to the speed \(v\) of the boat: \(f_{k}=70 v\), where \(v\) is in meters per second and \(f_{k}\) is in newtons. Find the time required for the boat to slow to \(45 \mathrm{~km} / \mathrm{h}\).

Suppose the coefficient of static friction between the road and the tires on a car is \(0.60\) and the car has no negative lift. What speed will put the car on the verge of sliding as it rounds a level curve of \(30.5 \mathrm{~m}\) radius?

A circular curve of highway is designed for traffic moving at \(60 \mathrm{~km} / \mathrm{h}\). Assume the traffic consists of cars without negative lift. (a) If the radius of the curve is \(150 \mathrm{~m}\), what is the correct angle of banking of the road? (b) If the curve were not banked, what would be the minimum coefficient of friction between tires and road that would keep traffic from skidding out of the turn when traveling at \(60 \mathrm{~km} / \mathrm{h} ?\)

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.