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Three uniform spheres are located at the corners of an equilateral triangle. Each side of the triangle has a length of \(1.20 \mathrm{~m}\). Two of the spheres have a mass of \(2.80 \mathrm{~kg}\) each. The third sphere (mass unknown) is released from rest. Considering only the gravitational forces that the spheres exert on each other, what is the magnitude of the initial acceleration of the third sphere?

Short Answer

Expert verified
The magnitude of the initial acceleration of the third sphere is approximately \(1.946 \times 10^{-10} \text{ m/s}^2\).

Step by step solution

01

Identify Forces on the Third Sphere

The two spheres with mass 2.80 kg exert gravitational forces on the third sphere. Since the triangle is equilateral, the forces are at angles of 60 degrees relative to the line between the centers. We need to calculate these forces.
02

Calculate Gravitational Force

The gravitational force between two masses is given by the formula: \[ F = \frac{G imes m_1 imes m_2}{r^2} \]where \( G \) is the gravitational constant \((6.674 \times 10^{-11} \text{m}^3 \text{kg}^{-1} \text{s}^{-2})\), \( m_1 \) and \( m_2 \) are the masses, and \( r = 1.20 \text{ m} \) is the separation between the masses.For mass \( m_1 = m_2 = 2.80 \text{ kg} \):\[ F = \frac{6.674 \times 10^{-11} \times 2.80 \times m}{(1.20)^2} \]
03

Determine the Net Force Components

The forces due to each of the 2.80 kg spheres have both x- and y-components due to symmetry. Since the angle between forces and the equilateral triangle line is 60 degrees, the x-components cancel each other, while the y-components add up. The y-component for each force is given by:\[ F_y = F \times \cos(30^\circ) = \left(\frac{6.674 \times 10^{-11} \times 2.80 \times m}{1.44}\right) \times \frac{\sqrt{3}}{2} \]
04

Calculate Total Acceleration

The net force is the sum of the y-components:\[ F_{net} = 2 \times F_y \]The acceleration \( a \) of the third sphere can be calculated using Newton's second law:\[ a = \frac{F_{net}}{m} \]Substitute the expression for \( F_y \) to solve for \( a \):\[ a = \frac{2 \times \left(\frac{6.674 \times 10^{-11} \times 2.80 \times m}{1.44}\right) \times \frac{\sqrt{3}}{2}}{m} = \left(\frac{6.674 \times 10^{-11} \times 2.80 \times \sqrt{3}}{1.44}\right) \]
05

Simplify and Compute Numerical Solution

Notice that \( m \) cancels out in the equation, making the acceleration independent of the third sphere's mass. Substituting numerical values:\[ a = \frac{6.674 \times 10^{-11} \times 2.80 \times \sqrt{3}}{1.44} \approx 1.946 \times 10^{-10} \text{ m/s}^2 \]

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Key Concepts

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

Newton's Second Law
In physics, Newton's Second Law serves as a centerpiece for understanding motion and its causes. It is expressed by the equation \( F = ma \), where \( F \) represents the force applied, \( m \) denotes the mass of the object, and \( a \) is the acceleration produced by the force.

This law shows the relationship between an object's mass, the forces acting upon it, and the resulting acceleration. It's a straightforward concept that explains:
  • How more force is required to accelerate a heavier object at the same rate as a lighter one.
  • An increase in force results in an increase in acceleration, assuming mass remains constant.

In our exercise, Newton's second law helps us determine the acceleration of the third sphere. By calculating the net gravitational force acting on it and using the known mass, we can solve for acceleration through the rearranged formula \( a = \frac{F_{net}}{m} \).

Understanding this law highlights why objects with the same force act differently based on their mass, and it's crucial for solving problems involving dynamic systems.
Equilateral Triangle
The equilateral triangle is a fascinating geometric shape, marked by three equal sides and equal angles, each measuring 60 degrees. This symmetry brings about interesting properties helpful in simplifying problems in physics and engineering.

An equilateral triangle’s properties are often leveraged to:
  • Simplify symmetry analysis when dealing with forces or components.
  • Understand the direction and magnitude of forces acting upon an object at each vertex.

In our problem, the equilateral triangle forms the basis for calculating the gravitational forces between the spheres. Since each side of the triangle is the same length, and each of its internal angles is 60 degrees, it makes it easier to resolve force components.

With every vertex subject to equal gravitational pull, the forces due to mass are symmetrically distributed. Solving such problems often involves using trigonometric identities related to the equilateral triangle, like realizing force components at 60 and 30-degree angles are repeatedly useful.

This symmetrical property enables straightforward calculations needed to resolve the net force acting on the sphere.
Acceleration Calculation
Calculating acceleration in physics is key to understanding how quickly an object's velocity changes. It ties directly to Newton’s Second Law, reflecting the effects of applied forces relative to mass.

To calculate acceleration, you need to determine the net force acting on the object and apply the formula \( a = \frac{F_{net}}{m} \).

Here’s how to tackle this kind of problem:
  • Identify all forces acting on the object.
  • Combine these forces to determine the net force (consider both magnitude and direction).
  • Use the object's mass to find acceleration through the relationship \( a = \frac{F_{net}}{m} \).

In the given exercise, gravitational forces from two 2.80 kg spheres influenced the unknown mass sphere. These forces affected both the x and y components. Due to the equilateral triangle’s symmetry, the x-components canceled, making it simpler to calculate the net force via the y-components.

This shows that in real-world applications, understanding how forces interact is essential to accurately calculating acceleration, a derived quantity crucial for resolving many physics scenarios.

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Most popular questions from this chapter

Refer to Multiple-Concept Example 10 for help in solving problems like this one. An ice skater is gliding horizontally across the ice with an initial velocity of \(+6.3 \mathrm{~m} / \mathrm{s}\). The coefficient of kinetic friction between the ice and the skate blades is \(0.081,\) and air resistance is negligible. How much time elapses before her velocity is reduced to \(+2.8 \mathrm{~m} /\) s?

A car is driving up a hill. (a) Is the magnitude of the normal force exerted on the car equal to the magnitude of its weight? Why or why not? (b) If the car drives up a steeper hill, does the normal force increase, decrease, or remain the same? Justify your answer. (c) Does the magnitude of the normal force depend on whether the car is traveling up the hill or down the hill? Give your reasoning. Problem A car is traveling up a hill that is inclined at an angle of \(\theta\) above the horizontal. Determine the ratio of the magnitude of the normal force to the weight of the car when (a) \(\theta=15^{\circ}\) and \((\mathrm{b}) \theta=35^{\circ} .\) Check to see that your answers are consistent with your answers to the Concept Questions.

An arrow, starting from rest, leaves the bow with a speed of \(25.0 \mathrm{~m} / \mathrm{s}\). If the average force exerted on the arrow by the bow were doubled, all else remaining the same, with what speed would the arrow leave the bow?

A bowling ball (mass \(=7.2 \mathrm{~kg},\) radius \(=0.11 \mathrm{~m}\) ) and a billiard ball (mass \(=0.38 \mathrm{~kg}\), radius \(=0.028 \mathrm{~m}\) ) may each be treated as uniform spheres. What is the magnitude of the maximum gravitational force that each can exert on the other?

A block whose weight is \(45.0 \mathrm{~N}\) rests on a horizontal table. A horizontal force of \(36.0 \mathrm{~N}\) is applied to the block. The coefficients of static and kinetic friction are 0.650 and 0.420 , respectively. Will the block move under the influence of the force, and, if so, what will be the block's acceleration? Explain your reasoning.

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