Asked by Skyler
Four 1,000,000- kg masses are arranged in a square, 150 meters on a side. To the nearest ten thousandth of a Newton, what is the magnitude of the force the mass in the lower left hand corner?
Can someone please explain this to me in a detailed manner? I know this somehow relates to gravitation/circular motion, but just would like to know a formula/method of going about solving this. I read up somewhere that since the 4 masses are in a square the opposite corners cancel out but then the diagonal needs to be taken into consideration. I also know that the degree measure of the bottom left corner's force is 45 degrees. Additional help would be greatly appreciated. Thanks.
Can someone please explain this to me in a detailed manner? I know this somehow relates to gravitation/circular motion, but just would like to know a formula/method of going about solving this. I read up somewhere that since the 4 masses are in a square the opposite corners cancel out but then the diagonal needs to be taken into consideration. I also know that the degree measure of the bottom left corner's force is 45 degrees. Additional help would be greatly appreciated. Thanks.
Answers
Answered by
bobpursley
draw the figure. figure the horizontal distance.
So the net force is the sum of the lleft-uleft;lleft-uright; lleft-lright
So it is adding three vectors. If you use arguements of symettry, you will see the ll-ul will add with the ll-lr to form a vector along the diagonal , of magnitude GMM/150^2 * cos45*2 the two at the last is because you are adding both. Add that to the force along the diagonal, and you have it.
So the net force is the sum of the lleft-uleft;lleft-uright; lleft-lright
So it is adding three vectors. If you use arguements of symettry, you will see the ll-ul will add with the ll-lr to form a vector along the diagonal , of magnitude GMM/150^2 * cos45*2 the two at the last is because you are adding both. Add that to the force along the diagonal, and you have it.
Answered by
Skyler
the way you explained it does not make sense to me.
I need to see an equation. I have no idea what the "that" means in 'Add "that" to the force (what force?) along the diagonal...
I need to see an equation. I have no idea what the "that" means in 'Add "that" to the force (what force?) along the diagonal...
Answered by
bobpursley
The last thing you need to see is an equation. You are adding three vectors. Compute the vectors, and add them with vector math, the result will be along the direction of the diagonal. I suspect you are having breaking vectors into components, and need to understand that.
Answered by
Akshay Goel
So, first calculate the force between two boxes which is on the right hand side
so apply formula F= (GMm)r^2
we know the value of G = 6.67* 10 ^(-11)
value r is 150
put the value for both masses nd calculte for F, after that box which is on left side up on the vertical line, calculate same for it, it is also the same force as above we get
so now calculate the force for the box which is remain (diagonally) . This time remember , you have to find "r" by pythagorean theorem , than do all same thing as we did above . than add all of them , you will get the force...
so apply formula F= (GMm)r^2
we know the value of G = 6.67* 10 ^(-11)
value r is 150
put the value for both masses nd calculte for F, after that box which is on left side up on the vertical line, calculate same for it, it is also the same force as above we get
so now calculate the force for the box which is remain (diagonally) . This time remember , you have to find "r" by pythagorean theorem , than do all same thing as we did above . than add all of them , you will get the force...
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.