# Homework Help: Determine the direction and acceleration of the bead

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1. Aug 7, 2017

### jfnn

1. The problem statement, all variables and given/known data

Three lead spheres, of mass 10.0 kg each, are located at three corners of a square of ice length 45.0 cm, as shown. A bead is released at the forth corner. By considering the gravitational forces among the four objects only, determine the magnitude and direction of the acceleration of the bead when released.

2. Relevant equations

F=Gm1m2/r^2

F=ma

3. The attempt at a solution

Each of the spheres applies a gravitation force on the particle in the y direction and/or x direction.

F = G m1m2/r^2 where r is the distance between the bead and lead sphere. The distance between the bead and the sphere directly above it is 45 cm or 0.45 m (this is r)

The sphere directly above applies a force only in the positive y direction. This force is...

F = G m1m2/r^2 where r = 0.45 m and G = 6.67408 * 10 ^-11 and m1 = mass of bead and m2 is mass of the given sphere

F = (6.67408 * 10^-11)(m1)(10)/0.45^2

F = 6.67408*10^-10 * m1 / 0.2025 --> However I do not know the mass of the bead, so I continued on anyways.

The sphere directly to the right applies a force only in the positive x direction. This force is...

F = (6.67408*10^-11)(m1)(10)/0.45^2

F = 6.67408*10^-10 m1 / 0.2025

The sphere diagonally across from the bead exerts a force. This force is...

F = (6.67408*10^-11)(m1)(10)/r^2 --> the distance between them is not 45. I use a right triangle to find the hypotenuse if both sides are 0.45 m.

Thus, the distance between them is: the square root of (0.45^2 + 0.45^2) = hypotenuse

Hypotenuse is 0.63639 m

Therefore the force is:

F = (6.67408*10^-11)(m1)(10)/ (0.63639^2)

F = (6.67408*10^-10)(m1)/0.405

After I get these three forces, one in the x direction, one in the y direction only, and one having both x and y direction, where do I go from here to find the direction of the acceleration of the bead and magnitude?

I know F=ma.

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2. Aug 7, 2017

### TSny

Find the x-component of the total force on the bead and find the y-component of the total force on the bead. Use these to find the x and y components of the acceleration of the bead.

3. Aug 7, 2017

### jfnn

Okay so I actually was able to cancel out the mass of the bead. This is how I did it:

F = Gm1m2/r^2

While acceleration = F=m1a

Thus,

m1*a = Gm1m2/r^2

Thus, a = Gm2/r^2

I used this new equation to get the acceleration in the diagonal part:

acceleration of diagonal = (G)(10)/r^2

where G is the gravitational constant and r^2 = 0.405 (I got this from a right triangle doing a^2 + b^2 = r^2)

From this, I plugged the values in to get acceleration of diagonal = 2.00 * 10^-9 m/s^2

After this I did the same method to get the acceleration of the horizontal and vertical components, which are the same.

ACCELERATION OF VERTICAL = acceleration of horizontal = G(10)/(0,450)^2 --> R is the distance form a bead to sphere (all equivalent)

Thus acceleration of horizontal = acceleration of vertical = 3.00 * 10 ^ -9 m/s^2

From these two accerlations, I found the acceleration of the resultant from these two...

Thus acceleration of resultant = square root of (acceleration vertical ^2 + accerlation horizontal ^2)

a of resultant = square root of ( (3.00*10^-9)^2 + (3.00*10^-9)^2)

I got a of resultant = 1.8*10^-17 m/s^2

So total acceleration is (1.8*10^-17) = (2 * 10 ^-9) which gives the acceleration of the bead to be: 7.00*10^-9 m/s?

Is this an acceptable way of doing it?

The direction would be the diagonal, thus 45 degrees above horizontal

Any input is appreciated!

Thank you in advance.

4. Aug 7, 2017

### TSny

Your method looks OK. But for some reason, in your calculations you appear to be rounding to only one significant figure, so your answer is not very accurate.

For example, for the acceleration caused by the diagonal bead, you are writing 2.00 x 10-9 m/s2. The 2.00 should be 1.65

5. Aug 7, 2017

### jfnn

Ahh! My calculator does not go that far down so cuts off and rounds. When I use a different calculator I get the same value. I will go through this and fix these values. Thank you for your assistance.

6. Aug 7, 2017

### jfnn

Okay so I now got (6.31*10^-9 m/s^2) at a direction of theta = 45.0 degrees above positive horizontal axis

7. Aug 7, 2017

### TSny

That looks good!