# Calculate the gravitational force between these three masses

• hraghav
hraghav
Homework Statement
Three masses are located as follows:
Mass 1 with 𝑚1=5.25×106kg is at 𝑟⃗1=29.9m𝑖̂+17m𝑗̂. Mass 2 with 𝑚2=5.53×106kg is at 𝑟⃗2=40.4m𝑖̂+17m𝑗̂. Mass 3 with 𝑚3=3.2×106kg is at 𝑟⃗3=40.4m𝑖̂+35.4m𝑗̂. What angle does the net force on 𝑚3 make with the positive x-axis 𝑖̂? with the y-axis 𝑗̂?
Relevant Equations
I calculated the magnitude of the net force on 𝑚3 due to 𝑚1 and 𝑚2 to be 5.790N.
Fnet on3 = 1.2365i + 5.657j
for the angle with positive x axis I did: component of fnetx = (magnitude of fnet)*costheta
1.23651 = 5.790*costheta
1.23651/5.790 = costheta
theta = 77.668 degrees

for the angle with positive y axis I did: component of fnety = (magnitude of fnet)*costheta
5.657 = 5.790*costheta
5.657/5.790 = costheta
theta = 12.30 degrees

But I am still getting both wrong. My magnitude of the net force is correct. Could someone please help me with this and let me know where am I making an error?
Thank you

hraghav said:
What angle does the net force on 𝑚3 make with the positive x-axis 𝑖̂?
Measured anticlockwise from the positive x axis.

haruspex said:
Measured anticlockwise from the positive x axis.
Sorry I am just confused what would the equation look like then? would it be -component of fnetx = (magnitude of fnet)*costheta?

The components that you calculated place the resultant in the first quadrant. They would be correct if the gravitational force were repulsive instead of attractive. Make a drawing to see what's going on.

hraghav said:
Sorry I am just confused what would the equation look like then? would it be -component of fnetx = (magnitude of fnet)*costheta?
No. As @kuruman indicates, you have the wrong sign on "Fnet on 3".
When you have corrected that, you still need to be careful deducing angles from their trig functions because of quadrant ambiguities.

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