/*! 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 35 A coordinate system (in meters) ... [FREE SOLUTION] | 91Ó°ÊÓ

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A coordinate system (in meters) is constructed on the surface of a pool table, and three objects are placed on the table as follows: a \(2.0-\mathrm{kg}\) object at the origin of the coordinate system, a \(3.0-\mathrm{kg}\) object at \((0,2.0)\), and a \(4.0-\mathrm{kg}\) object at \((4.0,0)\). Find the resultant gravitational force exerted by the other two objects on the object at the origin.

Short Answer

Expert verified
The resultant gravitational force on the \(2.0-\mathrm{kg}\) object at the origin, due to the other two objects can be found by the method of vector addition. The magnitude and angle of the resultant force can be calculated using the Pythagorean theorem and trigonometric functions.

Step by step solution

01

Compute each gravitational force individually

Using the formula for gravitational force, which is \( F = G \frac{m1 \cdot m2}{r^2} \), where \( G \) is the universal gravitational constant (\( 6.674 \times 10^{-11} \, \mathrm{Nm}^2/\mathrm{kg}^2 \)), \( m1 \) and \( m2 \) are the masses of the two objects, and \( r \) is the distance between the centers of the two objects, calculate the force exerted on the \(2.0-\mathrm{kg}\) object by the \(3.0-\mathrm{kg}\) object and the \(4.0-\mathrm{kg}\) object, separately.
02

Determine the direction of each gravitational force

By convention, attractive forces are negative. The force due to the \(3.0-\mathrm{kg}\) object will be vertically up (negative in the y-direction), and the force due to the \(4.0-\mathrm{kg}\) object will be horizontally to the right (positive in the x-direction). Therefore, the force on the \(2.0-\mathrm{kg}\) object due to the \(3.0-\mathrm{kg}\) object will be negative, while that due to the \(4.0-\mathrm{kg}\) object will be positive.
03

Compute the resultant force

The total gravitational force on the object at the origin is the vector sum of the individual forces calculated in Step 1. To add these forces, first, identify their components in the x and y direction, and then add them vectorially using the Pythagorean theorem for the magnitudes and trigonometric functions for the angle of the resultant vector.

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Key Concepts

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

Newton's law of universal gravitation
Newton's law of universal gravitation is a fundamental principle that describes the attractive force between two masses. According to this law, every point mass attracts every other point mass in the universe with a force that is directly proportional to the product of their masses, and inversely proportional to the square of the distance between their centers.
This relationship is captured by the formula: \[ F = G \frac{m1 \cdot m2}{r^2} \]where:
  • \( F \) is the gravitational force between the masses.
  • \( G \) is the gravitational constant, approximately \( 6.674 \times 10^{-11} \, \mathrm{Nm}^2/\mathrm{kg}^2 \).
  • \( m1 \) and \( m2 \) are the masses of the objects.
  • \( r \) is the distance between the centers of the two masses.
In our exercise, this principle is applied to calculate the forces exerted on the 2.0-kg object at the origin by the 3.0-kg and 4.0-kg objects, respectively. Each force is computed by determining the mass products and the distance squared, which then allows for solving the gravitational force using the above formula.
This calculation reveals how the forces act in different directions, guiding us to the next step: vector addition.
Coordinate system
The coordinate system is crucial for solving problems involving multiple objects in physics. It allows us to clearly define the positions of objects within a structured framework, using axes for reference. In this problem, a coordinate system on a pool table defines the locations of masses with point coordinates.
The origin (0,0) holds the 2.0-kg object, the 3.0-kg object is at (0,2.0), and the 4.0-kg object is found at (4.0,0). Using this system, we can easily calculate the distances between different objects necessary for applying Newton's law of universal gravitation.
For instance:
  • Distance between the 2.0-kg and 3.0-kg objects is 2 meters vertical (y-direction).
  • Distance between the 2.0-kg and 4.0-kg objects is 4 meters horizontal (x-direction).
These distances, observed from the coordinate points, are critical in computing the forces correctly. Having a coordinate system not only simplifies the problem but also illustrates the direction of forces, clarifying the geometry involved.
Vector addition
After calculating individual forces, understanding how to add them helps determine the net gravitational effect on an object. Forces are vector quantities, meaning they have both magnitude and direction. Simply summing them requires consideration of their directional components.
The gravitational forces on the 2.0-kg object have specific directions:
  • Force from the 3.0-kg is upward (negative y-direction).
  • Force from the 4.0-kg is rightward (positive x-direction).
To find the resultant force, each force is broken into its component form. The force's x-component is horizontal, while the y-component is vertical. We add like components together: x-components with x-components and y-components with y-components.
This piecewise vector addition allows us to find the net vector, combining the magnitudes and directions into a single force vector acting on the object at the origin. This is exemplified in how one might merge the gravitational pull from different sides, leading into using the Pythagorean theorem for final calculations.
Pythagorean theorem
The Pythagorean theorem comes into play when determining the magnitude of the resultant vector, particularly after combining component forces in vector addition. This theorem is best known from geometry and states that in a right-angled triangle, the square of the length of the hypotenuse is equal to the sum of squares of the lengths of the other two sides. It can be expressed as: \[ c^2 = a^2 + b^2 \]where:
  • \( c \) is the hypotenuse or resultant vector magnitude.
  • \( a \) and \( b \) are the x and y components of the forces.
In this problem, each gravitational force contributes to the resultant force acting on the 2.0-kg object. By calculating the vector's magnitude using the Pythagorean theorem, we understand how different vectors add geometrically.
This technique distills all our prior work into a single clear answer. It's a succinct method to assert not just how strongly the object at the origin is attracted, but also in which direction the total force is acting, providing a comprehensive understanding of the forces in play.

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Most popular questions from this chapter

A \(40.0\)-kg child takes a ride on a Ferris wheel that rotates four times each minute and has a diameter of \(18.0 \mathrm{~m}\). (a) What is the centripetal acceleration of the child? (b) What force (magnitude and direction) does the seat exert on the child at the lowest point of the ride? (c) What force does the seat exert on the child at the highest point of the ride? (d) What force does the seat exert on the child when the child is halfway between the top and bottom?

A car moves at speed \(v\) across a bridge made in the shape of a circular arc of radius \(r\). (a) Find an expression for the normal force acting on the car when it is at the top of the arc. (b) At what minimum speed will the normal force become zero (causing the occupants of the car to seem weightless) if \(r=30.0 \mathrm{~m}\) ?

A massless spring of constant \(k=78.4 \mathrm{~N} / \mathrm{m}\) is fixed on the left side of a level track. A block of mass \(m=\) \(0.50 \mathrm{~kg}\) is pressed against the spring and compresses it a distance \(d\), as in Figure P7.76. The block (initially at rest) is then released and travels toward a circular loop-the-loop of radius \(R=1.5 \mathrm{~m}\). The entire track and the loop-the-loop are frictionless, except for the section of track between points \(A\) and \(B\). Given that the coefficient of kinetic friction between the block and the track along \(A B\) is \(\mu_{k}=0.30\), and that the length of \(A B\) is \(2.5 \mathrm{~m}\), determine the minimum compression \(d\) of the spring that enables the block to just make it through the loop-the-loop at point C. Hint: The force exerted by the track on the block will be zero if the block barely makes it through the loop-the-loop.

Q|C A bicycle is turned upside down while its owner repairs a flat tire. A friend spins the other wheel and observes that drops of water fly off tan- gentially. She measures the heights reached by drops moving vertically (Fig. P7.8). A drop that breaks loose from the tire on one turn rises vertically \(54.0\) cm above the tangent point. A drop that breaks loose on the next turn rises \(51.0 \mathrm{~cm}\) above the tan- gent point. The radius of the wheel is \(0.381 \mathrm{~m}\). (a) Why does the first drop rise higher than the second drop? (b) Neglecting air friction and using only the observed heights and the radius of the wheel, find the wheel's angular acceleration (assuming it to be constant).

A sample of blood is placed in a centrifuge of radius \(15.0 \mathrm{~cm}\). The mass of a red blood cell is \(3.0 \times\) \(10^{-16} \mathrm{~kg}\), and the magnitude of the force acting on it as it settles out of the plasma is \(4.0 \times 10^{-11} \mathrm{~N}\). At how many revolutions per second should the centrifuge be operated?

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