Calculating Tension in a Two Pulley Atwood Machine

[rossel]

Homework Statement

Calculate the force of tension in the string of an Atwood machine with masses (1 - on right) 69.95 grams, (2 - middle) 80.11 grams, and (3 - on right) 60.28 grams. When this value is determined, verify that the net force in the atwood machine is equal to zero....\...|.../
...44º.\.46º.|.60º./.30º
----------------------------- (all angles are measured from this line, x components)
...|...|...|
...|...|...|
...m1...m2...m3

Homework Equations

Fg = mg
Fnet = Fg + Ft (Fnet should equal approx. 0)

The Attempt at a Solution

First, calculating the force of gravity on each of the masses:

Fg1 = (0.06995)(9.8)
= 0.68 N

Fg2 = (0.08011)(9.8)
= 0.79 N

Fg3 = (0.060280)(9.8)
= 0.59 NSecondly, using the angles measured and the forces of gravity determined, solve for the x and y components. y1 = 0.68 N
y2 = 0.59Nx1/sin 46 = 0.68/sin 44
x1 = 0.70 N

x2/sin 60 = 0.59/sin 30
x2 = 1.02 NWriting Fnet statements for both x and y components produces:

Fnet(x) = 0.70 N + 1.02 N
= 1.72 N

Fnet(y) = 0.68 N + 0.79 N + 0.59
= 2.06 NFinally:
Fnet = Fx + Fy
Fnet = 1.72 - 2.06
Fnet = 0.34 NThe sum of each of these Fnet statements is supposed to be approximately zero. I understand why this should be so (as it is a static equilibrium), however I do not not know how to achieve that answer, or what I am doing wrong. I do not believe that 0.34 N is a relatively close answer.

These angles were calculating in a lab, so is it possible that I have done everything right and simply measured the angles wrong?

I understand some of the values are to be negative, but I am unaware of which (I think it is gravity?). My teacher is no help, so I would really appreciate someone explaining how to fix this. Thank you.

 

[haruspex]

Since there are two pulleys, I assume that these support the two outside weights and that the strings over them are tied together where a third string supports the central weight.

There are several errors in post #1.

First, the text and the diagram disagree regarding which is mass 1 and which is mass 3.

Secondly, although the data given are to four significant figures the weight calculations are truncated at 2. That seems reasonable since the g value is only being taken to two, but the value of g is irrelevant to checking for net zero force at the knot, so it can be factored out. This is an example of the many merits of working entirely symbolically as long as possible.

But most egregiously, the way the forces are combined makes nonsense. The horizontal forces are taken as positive and added although they act in opposite directions, and likewise the vertical forces; then the vertical force is subtracted from the horizontal force!