Three masses are arranged in the (x,y) plane. The X and Y units are meters
4kg at (-1,5
3kg at (1,1)
5kg at (0,0)
What is the magnitude of theresulting force on the 5kg mass at the origin? The universal gravitational constant is 6.673e-11 N*m^2/kg^2. Answer in units of N.
It is rather simple. Figure the two forces, and add as vectors.
I DID THAT BUT THE ANSWER IS WRONG. IDON'T KNOW WHY
Let me show you a shortcut:
fx= G5(4/1^1)(-1)+G5(3/1)(+1) That is the resultant fx
Fy=G5(5/5^2)(+1)+G5(3/1)(+1) that is the resultant fy
fx=G5(-1)
fy=G5(1/5+3)=G5(15/5)=5G*3
magnitude= 5G*sqrt(1+9)=5Gsqrt10
If you don't understand that shortcut,don't do it.
To find the magnitude of the resulting force on the 5kg mass at the origin, we need to calculate the individual forces exerted on it by the other two masses and then combine them.
The force between two masses can be calculated using Newton's Law of Universal Gravitation:
F = (G * m1 * m2) / r^2
where F is the force between the masses, G is the universal gravitational constant, m1 and m2 are the masses, and r is the distance between the masses.
Let's calculate the forces exerted on the 5kg mass by the 4kg mass at (-1,5) and the 3kg mass at (1,1).
1. Force exerted by the 4kg mass:
m1 = 5kg
m2 = 4kg
r = distance between the two masses = sqrt( (0 - (-1))^2 + (0 - 5)^2 )
Using the formula, we get:
F1 = (6.673e-11 * 5 * 4) / r^2
2. Force exerted by the 3kg mass:
m1 = 5kg
m2 = 3kg
r = distance between the two masses = sqrt( (0 - 1)^2 + (0 - 1)^2 )
Using the formula, we get:
F2 = (6.673e-11 * 5 * 3) / r^2
Now, we can calculate the total force on the 5kg mass by adding the forces F1 and F2:
Total force = F1 + F2
Simply substitute the values calculated above into this formula and perform the addition to get the magnitude of the resulting force on the 5kg mass at the origin in units of Newtons (N).