# Simple gravitational attraction forces problem

• kirby27

#### kirby27

Each mass in the figure below is 2.00. Find the magnitude of the net gravitational force on mass A due to the other masses in figure b (the square).

Figure: http://i.imgur.com/fjSgU.jpg

attempt:
i used the formula F=G [(m1*m2)/r^2]. all of the forces are in the direction of D so i resolved the B and C using the above equation and SOH CAH TOA into their directions that point toward D. Then i used the above equation without SOH CAH TOA to get D and added the componenets of B and C to that of D. I got 8.88 * 10^-8 but this is wrong. where did i go wrong? thanks

ehild

before i write it all out is this the right method?

before i write it all out is this the right method?

The result is wrong. I can not help without seeing what you did. I am sorry.

Good luck!

ehild

i re-did it and got 6.67E-8 is this correct?

calculations:

D acting on A:
F=(6.673E-11)[4/.1414^2]. i got the .1414 from pythagorean using .1 as a and b.
F=1.3346E-8

B acting on A:
F=(6.673E-11)[4/.1^2]
F=2.669E-8 --> but this is not downward pull
using SOH CAH TOA i did
cos(45)=x/2.669E-8
x=1.89E-8 --> this needs to be multiplied by 2 for both B acting on A and C acting on A (they are the same).

so (1.89E-8)*2 + 1.3346E-8 = 5.11E-8

i somehow get a different answer everytime i do it

i somehow get a different answer everytime i do it

It is correct now but you need to add the unit.

There is less chance to make mistakes in calculations if you solve the problem symbolically and substitute the data at the end.

So you have the resultant force along the diagonal AD. The magnitude of the force is

F=G q^2/a^2(1/2 +2cos(45)) =6.673E-11*400(0.5+1.414)=5.11 N.

Do not forget the unit!

ehild

you said it is 5.11 N. isn't it 5.11E-8 N?

you said it is 5.11 N. isn't it 5.11E-8 N?

Oppps, I left out that E-8. You are right!

ehild