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  4. Problem 42
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Question: 

Four 7.5-kg spheres are located at the corners of a square of side 0.80 m. Calculate the magnitude and direction of the gravitational force exerted on one sphere by the other three.

Source: Giancoli, Douglas C., Physics: Principles with Applications, 7th Edition, 2014.

Quick Answer: 

$1.1\times 10^{-8}\textrm{ N}$

Giancoli 7th Edition, Chapter 5, Problem 42

(4:18)

Chapter 5, Problem 42 is solved.

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Transcript for this Giancoli solution

This is Giancoli Answers with Mr. Dychko. We have a square with a side length of 0.8 meters and a mass of 7.5 kilograms at each corner and we'll find the net force on one of the masses due to the other three. There's gonna be a force upwards; let's call it force one force due to mass one and we have numbered all the masses - 1, 2, 3 and 4. Force due to mass 1 and the force due to mass 3 will be the same size forces because their masses are equal distances away; they are just different directions. And we can combine force 1 and 3 using Pythagoras because they are at right angles; it's a square after all, and we get our result there of F 3 plus 1. And then the direction of that result of force one and three will be the same as the direction of force 4 and so we can add those without any trigonometry because they are collinear; this resultant of F 1 and 3 is along the same line as force 4 because this is square and force 4 is gonna be on the diagonal, at a 45 degree angle, to a side and same with the resultant of F 1 and F 3. So we are using some shortcuts here because of the symmetry of this situation and if this mass was a different size mass than this one then the resultant of F 1 and 3 would not be along the diagonal and then we would have to use trigonometry and get into more complicated work. The concept will be the same but it would be a lot of busy work involved in calculating components and so on. So let's first do F 1 plus F 3 and that's gonna be combining F 1 and 3 using Pythagoras. And so we have a square root of F 1 squared plus F 3 squared and then down here I wrote, F 1 in place of F 3 because they have the same size and so it's 2 times F 1 squared, in other words and that's square root 2 times F 1. And then the resultant final force on this mass number 2 is going to be the result from here - square root 2 times F 1 plus the force 4. And this size force is gonna be different than the size of F 1 and 3. because the distance along the diagonal is not equal to the side length, it's gonna be further away so it's gonna be a smaller force and it's gonna be G times the one mass times other mass but they are the same size so we'll just use m squared divided by the distance to mass 4 squared. And then plus root 2 times F 1; G mass squared times distance to mass one squared. And you can factor out the G and the m squared and you have 1 over r 4 squared plus root 2 over r 1 squared. r 4 squared is r 1 squared plus r 1 squared because it's a right triangle so the distance along the diagonal here is this is r 4, that distance squared, this being the hypotenuse here is gonna be r 1 squared plus r 1 squared which is 2r 1 squared and so that's what I have written here. And substituting that in for r 4 we get this and then it makes it look a little cleaner by having a common denominator. So I multiply this term by 2 over 2 and now with the common denominator, we can add the numerator. So we have 1 plus 2 root 2 over 2 times r 1 squared and then I'll plug it into the calculator and get the answer. So we have gravitational constant times 7.5 kilograms squared times 1 plus 2 times square root 2 all divided by 2 divided by 0.8 meters squared and we get 1.1 times 10 to the negative 8 newtons is the resultant force and that's on a 45 degree angle to a side.