Newton's first law, forces, charges

In summary, the initial acceleration of a sphere with mass 0.01 kg released from rest at point P is 9.63x10^-11 N/m2.
  • #1
heartshapedbox
30
0

Homework Statement


Two uniform masses of .260kg are fixed at points A and B. Find the magnitude and direction of the initial acceleration of a uniform sphere with mass 0.01kg released from rest at point P that is acted on the gravitational forces of attraction from spheres A and B.

The diagram shows the spheres arranged in a triangle. If you were to divide the triangle in half, making two IDENTICAL right angles triangles, you would have a hypotenuse of 10cm, x value of 8cm, and y value of 6 cm.

Homework Equations


F=G Mm/r^2
G=6.67x10^-11
F=ma

The Attempt at a Solution



The x values cancel as vectors are equal in magnitude and opposite in direction.
This leaves the y values. Solve for F between mass of 0.26kg and 0.01kg.
F=(6.67x10^-11)(0.26)(0.01)/(0.06^2m)
=4.8x10^-11
Multiply this by two because there are two identical masses acting on the sphere at point P
=9.63x10^-11

Now, using F=ma, equate the result to this...
9.63x10^-11=ma
9.63x10^-11=(0.01)a
divide 9.63x10^-11 by .01
a=9.0x10^-9

This is wrong... the correct answer should be 2.1x10^-9

Thanks :)
 
Last edited:
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  • #2
heartshapedbox said:
F=(6.67x10^-11)(0.26)(0.01)/(0.06^2m)

The 0.26 kg masses are not 0.06 m away. You need to use the actual distance and then consider what the component in the relevant direction is.
 
  • #3
Orodruin said:
The 0.26 kg masses are not 0.06 m away. You need to use the actual distance and then consider what the component in the relevant direction is.
I understand that the two masses are not 0.06 m away. I am solving for the y component of distance between the 0.26mass and the 0.01 mass.
 
  • #4
heartshapedbox said:
I understand that the two masses are not 0.06 m away. I am solving for the y component of distance between the 0.26mass and the 0.01 mass.
But that is not the right way to find the force.
You need to find the direct force of attraction, then take the component of that in the y direction.
 
  • #5
heartshapedbox said:
I understand that the two masses are not 0.06 m away. I am solving for the y component of distance between the 0.26mass and the 0.01 mass.
You are computing the force as if it was 0.06 m away and not taking components. You cannot get the components by inserting the component of the separation instead of the separation in the force law. You need to first compute the total force (using the actual distance) and then split it into components.

To see this more clearly, if things worked the way you are using the force law, the components would be larger than the total force, which is clearly not the case.
 

1. What is Newton's first law?

Newton's first law, also known as the law of inertia, states that an object at rest will remain at rest and an object in motion will remain in motion at a constant velocity unless acted upon by an external force.

2. What are forces?

Forces are interactions between two objects that cause a change in motion. They can be either contact forces, such as pushing or pulling, or non-contact forces, such as gravity or magnetism.

3. How are forces related to Newton's first law?

In Newton's first law, an object will only change its state of motion if a force is applied to it. This means that forces are necessary for any changes in an object's velocity.

4. What is the difference between balanced and unbalanced forces?

Balanced forces are equal in magnitude and opposite in direction, resulting in a net force of zero and no change in an object's motion. Unbalanced forces, on the other hand, result in a net force and cause a change in an object's motion.

5. What are charges?

Charges are a fundamental property of matter that create electric fields and interact with each other through electromagnetic forces. They can be positive, negative, or neutral, and can be transferred from one object to another through various processes.

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