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A uniform sphere with mass 50.0 kg is held with its center at the origin, and a second uniform sphere with mass 80.0 kg is held with its center at the point x= 0, y= 3.00 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 kg placed at the point x= 4.00 m, y= 0? (b) 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) Find net force by calculating forces between Sphere 3 and each of Spheres 1 and 2, then summing components. (b) For zero force, place Sphere 3 along the line between Spheres 1 and 2, solving distances via gravitational balance.

Step by step solution

01

Identify Given Information and Convert to Coordinates

We have three spheres: Sphere 1 with mass m1=50.0 kg at (0,0), Sphere 2 with m2=80.0 kg at (0,3.00 m), and Sphere 3 with m3=0.500 kg at (4.00 m,0). We need to find the gravitational force on Sphere 3.
02

Apply Newton's Law of Gravitation

Use the formula for gravitational force: F=Gm1m2r2where G is the gravitational constant 6.674×1011 N(m2/kg2), m1 and m2 are the masses of two objects, and r is the distance between their centers.
03

Calculate Force Between Sphere 1 and Sphere 3

The distance r13 between Sphere 1 and Sphere 3 is 4.00 m. The force is directed along the x-axis.F13=G×50.0×0.5004.002Calculate F13.
04

Calculate Force Between Sphere 2 and Sphere 3

The distance r23 between Sphere 2 and Sphere 3 is calculated using the Pythagorean theorem:r23=(4.00)2+(3.00)2=5.00 mThen calculate the force:F23=G×80.0×0.5005.002This force has both x and y components, with the angle θ defined by tan1(3/4).
05

Resolve F23 into Components

The force F23 can be broken into components using the angle:- F23x=F23cos(θ)- F23y=F23sin(θ)
06

Calculate Net Gravitational Force on Sphere 3

Add the x-components and y-components:Fnet,x=F13F23xFnet,y=F23yThen calculate the magnitude and direction:Fnet=(Fnet,x)2+(Fnet,y)2θnet=tan1(Fnet,yFnet,x)
07

Find Equilibrium Point for Zero Gravitational Force

For Sphere 3 to experience zero net force, the forces from Sphere 1 and Sphere 2 must be equal and opposite. Setequations:G×50.0×0.500r12=G×80.0×0.500r22Solve for the distances r1 and r2 such that the center of mass position causes forces to cancel. Typically this will place Sphere 3 on the line connecting Spheres 1 and 2.

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Key Concepts

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

Newton's Law of Gravitation
Newton's Law of Gravitation is a fundamental principle of physics that describes the attractive force between any two masses in the universe. This force is responsible for the orbits of planets, the fall of objects on Earth, and many other phenomena.
The formula is F=Gm1m2r2where:
  • F is the gravitational force between two masses.
  • G is the gravitational constant, approximately 6.674×1011 N(m2/kg2).
  • m1 and m2 are the masses of the objects in kilograms.
  • r is the distance between the centers of the two masses, measured in meters.
Understanding this law enables us to calculate how much gravitational pull two objects exert on each other. It's essential to bear in mind that the force is mutual; both objects exert an equal force on one another, regardless of their size or mass.
Coordinate System
A coordinate system helps in specifying the location of an object in a spatial arrangement, like the positions of the spheres in this exercise. The system used here is a simple two-dimensional Cartesian plane with x and y axes.

  • Spheres are situated using ordered pairs like (x,y).
  • Sphere 1 is located at the origin (0,0), signifying that it's at the intersection of the two axes.
  • Sphere 2 is at (0,3.00), meaning it is 3 meters above the origin along the y-axis.
  • Sphere 3, the one we aim to analyze, is situated at (4.00,0), lying along the x-axis, 4 meters to the right of the origin.
Introducing a coordinate system allows us to calculate distances and directions between objects, create vectors, and determine angles—essentially helping in mathematical analysis of physical problems.
Vector Components
When dealing with forces in physics, it's beneficial to break them into vector components to simplify calculations of net force. The vector describes both the magnitude and direction of a force.

The gravitational force between objects can be resolved into:
  • **X-components:** These lie along the horizontal direction.
  • **Y-components:** These lie along the vertical direction.
For example, consider the force between Sphere 2 and Sphere 3. It is at an angle and has both x and y components. To find these components, we use trigonometric functions with the angle θ:
  • The x-component is calculated as F23x=F23cos(θ).
  • The y-component is F23y=F23sin(θ).
Breaking forces into components makes it easier to sum them and find resultant forces. This process is crucial for understanding the net effect on an object, especially if multiple forces act in different directions.
Equilibrium Point
An equilibrium point in the context of gravitational forces is a position where a body experiences no net force. It's as if the body is "floating," perfectly balanced by opposing forces.

To find such a point for Sphere 3 among Spheres 1 and 2, the forces exerted by the other two spheres should cancel out.Imagine placing Sphere 3 on an imaginary line between Spheres 1 and 2. By balancing the gravitational forces such that G×50.0×0.500r12=G×80.0×0.500r22we can determine at what point Sphere 3 will experience equal and opposite forces.

Such a balance ensures Sphere 3 remains stationary because the forces cancel out. Practically, this involves calculating precise locations where these conditions are true.

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Most popular questions from this chapter

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