Mass suspended by 3 ropes (midterm practice)

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To determine the tension in the rope attached at -4m\hat{x}, start by drawing a free-body diagram of the system. The mass of 100kg generates a gravitational force of 980N acting downward. Analyze the forces in the x, y, and z directions to set up equilibrium equations. By applying the conditions for static equilibrium, solve for the tension in the specified rope. The expected answer is 810N, indicating the calculations align with the problem's requirements.
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A 100kg mass is supended by 3 ropes. one rope is attached at a point 1m\hat{x} + 1m\hat{y}, one is attached at 1m\hat{x} - 1m\hat{y} and one is attached at -4m\hat{x}. The three ropes all connect at -1m\hat{z}, at which point the mass is attached. What is the tension T in the rope attached at -4m\hat{x}?

I found this problem in my midterm study sheet, but I have no idea how to solve it.
I only know the answer is supposed to be 810N but I don't understand how to get there. Thanks
 
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First draw a picture, then construct the free-body diagram.

ehild
 
Thread 'Variable mass system : water sprayed into a moving container'

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}...
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