/*! 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 50 A uniform sphere with mass 60.0 ... [FREE SOLUTION] | 91Ó°ÊÓ

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A uniform sphere with mass 60.0 \(\mathrm{kg}\) is held with its center at the origin, and a second uniform sphere with mass 80.0 \(\mathrm{kg}\) is held with its center at the point \(x=0, y=3.00 \mathrm{m} .\) (a) What are the magnitude and direction of the net gravitational force due to these objects on a third uniform sphere with mass 0.500 \(\mathrm{kg}\) placed at the point \(x=4.00 \mathrm{m}, y=0 ?(6)\) Where, other than infinitely far away, could the third sphere be placed such that the net gravitational force acting on it from the other two spheres is equal to zero?

Short Answer

Expert verified
(a) Net force is directed between the x-axis and y-axis; magnitude calculated using vector addition. (b) Position along the line of spheres, solving force balance equation yields zero force.

Step by step solution

01

Calculate the force between the first sphere and the third sphere

The gravitational force between two masses, \(m_1\) and \(m_3\), is given by \[ F = \frac{G \cdot m_1 \cdot m_3}{r^2} \]where \(G\) is the gravitational constant \((6.674 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2)\), \(m_1 = 60.0 \, \text{kg}\), \(m_3 = 0.500 \, \text{kg}\), and \(r = 4.00 \, \text{m}\) (the distance between the spheres along the x-axis). Calculate this force.
02

Calculate the force between the second sphere and the third sphere

The gravitational force between sphere 2 and mass \(m_3\) is calculated similarly. Here, \(m_2 = 80.0 \, \text{kg}\) and the distance \(r\) is calculated using the Pythagorean theorem:\[ r = \sqrt{4^2 + 3^2} = 5.00 \, \text{m} \]Use this distance with the formula:\[ F = \frac{G \cdot m_2 \cdot m_3}{r^2} \]to find the force magnitude.
03

Determine directions of forces

The force from the first sphere is directed along the negative x-axis. The force from the second sphere is directed along the line connecting sphere 2 and the third sphere, which can be resolved into x and y components using trigonometry.
04

Calculate the net gravitational force

Add the x-components and y-components of the forces to get the net force components. For the x-direction, the force is the sum of the force due to sphere 1 and the x-component of the force due to sphere 2. For the y-direction, it's just the y-component from sphere 2.
05

Calculate the magnitude and direction of net force

The net force magnitude is found using:\[ F_\text{net} = \sqrt{F_x^2 + F_y^2} \]The direction \(\theta\) with respect to the positive x-axis is:\[ \theta = \tan^{-1}\left(\frac{F_y}{F_x}\right) \]
06

Determine the position for zero net gravitational force

To find a position where the net force is zero, equate the magnitudes of forces from both spheres at any point along the line defined by both spheres and solve:\[ \frac{G \cdot m_1 \cdot 0.5}{r_1^2} = \frac{G \cdot m_2 \cdot 0.5}{r_2^2} \]where \(r_1\) and \(r_2\) are the distances from the masses \(m_1\) and \(m_2\) to the new point. Solve for these distances.

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Key Concepts

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

Uniform Spheres
Uniform spheres are bodies with mass evenly distributed throughout their volume. In calculations involving gravitational forces, treating objects as uniform spheres allows for simplification. This means you can consider all of a sphere's mass as concentrated at its center, known as the center of mass. This concept is crucial when calculating gravitational forces because it allows the position of a sphere to influence how another object experiences gravitational attraction. Understanding uniform spheres helps when determining how masses influence and interact with each other gravitationally.
Newton's Law of Gravitation
Newton's Law of Gravitation describes the force of attraction between two masses. It states that any two bodies with mass will experience an attractive force proportional to the product of their masses and inversely proportional to the square of the distance between their centers. The formula is expressed as: \( F = \frac{G \cdot m_1 \cdot m_2}{r^2} \).
Here, \( F \) is the gravitational force, \( G \) is the gravitational constant \((6.674 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2)\), \( m_1 \) and \( m_2 \) are the masses, and \( r \) is the distance between the centers of the two masses.
This law is fundamental for calculating gravitational forces in any physics problem involving mass. Whether for planetary motion or small localized systems, like three spheres in this exercise, Newton's law provides the basis for understanding gravitational interactions.
Net Force Calculation
To find the net gravitational force on an object, you must consider all individual gravitational forces acting on it and their directions. With multiple sources of force, like the two spheres in this exercise, forces must be combined vectorially.
Steps to calculate the net force involve:
  • Finding each individual force using Newton's law.
  • Resolving forces into components, especially if they're not aligned along standard axes.
  • Summing up all the x-components to get the net force in the x-direction and all the y-components for the y-direction.
Finally, use these components to find the net force's magnitude with \( F_{\text{net}} = \sqrt{F_x^2 + F_y^2} \) and direction with \( \theta = \tan^{-1}\left(\frac{F_y}{F_x}\right) \).
Understanding net force is critical for analyzing motion and equilibria situations where forces must balance.
Pythagorean Theorem
The Pythagorean theorem is a mathematical principle used to determine the distance between points in a plane when dealing with right-angled triangles. It's particularly useful in physics to calculate the diagonal distance between two points or to resolve forces into perpendicular components.
The theorem states: \( a^2 + b^2 = c^2 \), where \( c \) is the hypotenuse, the side opposite the right angle. In our exercise, applying the theorem helps find the distance between the second sphere and the third sphere, calculated as \( \sqrt{4^2 + 3^2} = 5.00 \, \text{m} \).
Using the Pythagorean theorem simplifies problems where spherical arrangements or vectors need component decomposition. It ensures precise calculations in multi-dimensional force analysis.
Trigonometry in Physics
Trigonometry helps resolve forces into components when they're not aligned along standard coordinate axes. It's essential in physics for analyzing directions and magnitudes of forces, velocities, and other vector quantities.
For resolving any force into components:
  • Use the angle the force makes with a reference axis.
  • Calculate the x-component using \( F_x = F \cos(\theta) \).
  • For the y-component, use \( F_y = F \sin(\theta) \).
In problems like the one given, trigonometry allows the conversion of non-standard directional forces into useful components. With these components, calculating further parameters, like the net force's direction, becomes manageable. Understanding trigonometry is crucial for any physics problem requiring vector decomposition and adds clarity to complex force systems.

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Most popular questions from this chapter

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An object in the shape of a thin ring has radius \(a\) and mass \(M .\) A uniform sphere with mass \(m\) and radius \(R\) is placed with its center at a distance \(x\) to the right of the center of the ring, along a line through the center of the ring, and perpendicular to its plane (Fig. 12.35 ). What is the gravitational force that the sphere exerts on the ring- shaped object? Show that your result reduces to the expected result when \(x\) is much larger than \(a\) .

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