Four balls are suspended by cords. The longer, top cord loops over a frictionless pulley and pulls with a force of magnitude 95 N on the wall to which it is attached. The tensions in the shorter cords are T1 = 56.0 N (between ball A & B), T2 = 46.7 N (between ball B & C), and T3 = 9.3 N (between ball C & D. What are the masses of each ball in kilograms? Im not sure what type of equation to start

Question

Four balls are suspended by cords. The longer, top cord loops over a frictionless pulley and pulls with a force of magnitude 95 N on the wall to which it is attached. The tensions in the shorter cords are T1 = 56.0 N (between ball A & B), T2 = 46.7 N (between ball B & C), and T3 = 9.3 N (between ball C & D. What are the masses of each ball in kilograms? 

Im not sure what type of equation to start off with this..

would acceleration = 0? .. since the tensions between a ball are in equilibrium?

If the cords are fixed to the wall, acceleraltion is zero. 

Start with the tension at the wall, pulling toward the wall. It is supporting four balls. 
95=A+B+C+D 
Then, the next tension 
T1= B+C+D and so on.

Emily · Answer

The acceleration of the balls is zero... They are fixed in space... They aren't falling are they?... They are suspended in air.

To start, start from the bottom up, it's a whole lot easier. First you would start out using newton's second law, 
F=ma= T-M(of ball d)g(gravity constant)
0= T3-Mdg
-T= -Mdg
T=Mdg
T/g= Md
9.3/9.81=Md=.9480122324

That's the setup for finding the mass of ball d. 
The setup for finding Ball c it
F=ma=0
F=T2-(Md+Mc)g
0=T2-(Md+Mc)g
-T2=-(Md+Mc)g
T2=(Md+Mc)g
T2/g = (Md+Mc)

So follow that approach and you should find the right answers hope it helps.
(t2/g)-Md =Mc

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