Calculate the gravitational force between these three masses

  • Thread starter Thread starter hraghav
  • Start date Start date
AI Thread Summary
The discussion focuses on calculating the gravitational force between three masses and determining the angles of the net force components. The user correctly calculated the magnitude of the net force but is confused about the angles, initially finding 77.668 degrees for the x-axis and 12.30 degrees for the y-axis. The error lies in the sign of the net force on mass 3, as gravitational forces are attractive, not repulsive. Additionally, there are warnings about quadrant ambiguities when deducing angles from trigonometric functions. A visual representation of the forces is recommended to clarify the situation.
hraghav
Messages
48
Reaction score
5
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
 
Physics news on Phys.org
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.
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top