Finding X and Y components of forces in equilibrium experiment

[aek]

im having trouble filling in a table similar to this:

FORCE X COMPONENT (N) Y COMPONENT (N)
F1
F2
F3
SUM

these are the results:
when angle is 30 degrees
f1 =0.5N
f2 = 0.5N
f3 = 1.0N
when angle is 50 degrees
f1=1.6N
f2=1.3N
f3=1.4N

I used approx 100 grams on each vector. there is a diagram also i placed for added detail, the rectangles are the spring balances

and finally how do i prove it is a equillibrium and the expected magnitude and direction of the next force?

Thanks in advance.

 

[xanthym]

In the drawing, F1 is shown perfectly vertical, and F2 is shown perfectly horizontal. How was this maintained (or obtained) during the experiment?? (Or did these angles change when force F3 was applied??)
Also, what and where were the "100 gram" objects??


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[aek]

well this experiment was conducted on table and the weights were laying the table connected to the spring balances. the 3 forces stood even throughout the experiment..xanthym? do you think you can help me please?

 

[xanthym]

To complete your "Force Equilibrium Table", you need to establish an X-Y coordinate system for your experimental setup. Referencing your drawing, a natural choice might be:
X-axis ---> horizontal line thru center, with (+) direction to the RIGHT
Y-axis ---> vertical line thru center, with (+) direction UP
Now you can resolve your forces into X & Y components, with components having the following signs:
Force component pulling RIGHT ---> (+)X
Force component pulling LEFT ---> (-)X
Force component pulling UP ---> (+)Y
Force component pulling DOWN ---> (-)Y

If all forces are in equilibrium, all the X-components will add to zero AND all the Y-components will add to 0.

For instance, for your first data set:
when angle is 30 degrees
f1 =0.5N
f2 = 0.5N
f3 = 1.0N
Force "f1" would be listed in the "Y Component" column with (-0.5 N) because it is pulling DOWN. Force "f2" would be listed in the "X Component" column with (-0.5 N) because it is pulling LEFT. Force f3 would have 2 listings because it is pulling at an angle and has both an X-component & Y-component: #1) (1.0N)*cos(30 deg)=(+0.866 N) in the "X Component" column because that component is pulling RIGHT, and #2) (1.0N)*sin(30 deg)=(+0.5 N) in the "Y-Component" column because that component is pulling UP. All the force components in the "X Component" column AND all the force components in the "Y Component" column should add to 0. (If they don't, there is some "experimental error" in your experiment.)

...X Component...Y Component
f1...0......-0.5
f2....-0.5......0
f3...+0.866......+0.5
SUM...+0.366......0.0 <--- Add each col. (This row should have 0's)

If you understand the above, you should be able to complete the Table for the 50 degree case. (A totally new Table is started for this next case.)
when angle is 50 degrees
f1=1.6N
f2=1.3N
f3=1.4N


Try it and see what results you get.


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[aek]

THANKS A MILLION, if there's anything i can do in return please don't hesitate. Thanks again.