# What is the gravitational force on the 52[kg] mass

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1. Mar 13, 2016

### heartshapedbox

1. The problem statement, all variables and given/known data
A particle of mass 52[kg] is at 5.0[m]ˆi a particle of mass 13[kg] is at 12[m]j, and a particle of mass 13[kg] is at −12[m]jˆ. What is the gravitational force on the 52[kg] mass due to the mass at 12[m]jˆ?

answer: −1.0 × 10−10[N]ˆı + 2.5 × 10−10[N]jˆ

2. Relevant equations
F2on1=Gm1m2/r^2

3. The attempt at a solution
Separated into components...

i component) f2on1=(6.67x10^-11)(52)(13)/5^2
= -1.0x10^-10 this is correct

j component) shouldn't this be solved in the same manner?
f2on1=(6.67x10^-11)(52)(13)/12^2
=3.13x10^-10

What am i missing from solving the y component?

2. Mar 13, 2016

### Kaura

I have little experience with gravity problems dealing with more than two bodies but in this case I think that it might work at least in my case to draw the particle positions on graph paper and proceed to solve for the magnitude of gravity between each body which could then be broken up into horizontal and vertical vector components

3. Mar 13, 2016

### heartshapedbox

This is what i have done :) The y component is giving me a hard time but I have found the x component. j stands for y and i stands for x.

4. Mar 13, 2016

### Staff: Mentor

The "r" in the gravitational force formula is the distance between the masses. You can't use individual component distances. Just think what would happen, for example, if two masses had the same $\hat{i}$ component but different $\hat{j}$ components. Would the gravitational force in the $\hat{i}$ direction be infinite?

One way to approach the problem is to first calculate the distance between the masses and find the magnitude of the force using the gravitational force formula, then split that into components using the geometry of the situation.

(I see that @Kaura beat me to the punch!)

5. Mar 13, 2016

### heartshapedbox

Great thanks so much guys, @Kaura and @gneill :)

I understand it easily now!

6. Mar 13, 2016

### Kaura

Glad to help I found this problem to be very interesting as well

7. Mar 13, 2016

### heartshapedbox

Awesome :) Sorry you had correctly told me what to do at the start, but I misunderstood. Thank you :)