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Here is a picture of the masses.

"Three masses are connected by wires and hung vertically. Draw a Force Body Diagram for each mass and determine the tensions in the three wires."

I know there are forces only in the y-direction, so I started by trying to calculate the force of gravity on the last one.

Fnetx = 0

Fnety = Ft3 - Fg

I thought maybe Fnety could be 0 but then that doesn't make much sense since that would mean all the tensions cancel each other out but I don't think they do. This means I have two unknown variables and therefore can't solve.

Going to the middle mass, if you were to draw the force body diagram you would have two forces acting in the y-direction: Ft2 and Ft3. So: Fnety = Ft2 - Ft3

In this problem I still don't have any numbers to work with. Same thing happens when I try doing the top mass: Fnety = Ft1 - Ft2

In my book it says the answers are: Top Wire: 3.4 x 10^3 N Middle Wire: 2.0 x 10^2 N Bottom Wire: 1.3 x 10^2 N

I don't know how to get to this and I have tried. Thank you for any suggestions you may have at tackling this problem.

Liz
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    mass is the one going down, while tension is up, there will be 3 tension and 3 mass. So it should have 3 arrow up in the line, down in the masses. – wuiyang Apr 23 '15 at 07:00
  • physics.stackexchange.com is probably better suited for these questions. (I cant seem to get links to work in the comments...) – Asvin Apr 23 '15 at 22:47
  • @Asvin Okay, thank you! I will try it. – Liz Apr 23 '15 at 22:52

2 Answers2

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name from top to bottom, $m_1$, $m_2$, $m_3$ for each masses.

and each have tension, $T_1$, $T_2$, $T_3$ for each masses.

we calculate tension one by one.

First, the bottom one, $m_3$, because it is at stationary (it doesn't move), then $T_3=W_3$, using $W=mg$, where $W$ is weight, $m$ is mass, $g$ is gravity acceleration.

$$T_3 = W_3 =(13)(9.8)=127.4N$$

Next, move to $m_2$, because it weight was added, it need more tension. Formula is $T_2 = W_2+T_3$

$$T_2 = W_2 + 130N=(7)(9.8)+127.4N = 68.6N + 127.4N = 196N$$

Finally, we move to $m_1$, using above formula, which is $T_1 = W_1 + T_2$

$$T_1 = W_1 + 200N = (15)(9.8) + 196N = 147N + 196N = 343N$$

So, each line of the tension is $T_1=343N$, $T_2=196N$, $T_3=127.4N$

wuiyang
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Free Body Diagrams are drawn on a single body with parts connected in a certain way with several forces acting at specified joints/nodes/vertices.The purpose is to get resultant physical body and component forces body and moments.

In this case FBD is simply three vector diagram ( not to scale) with forces pointing down end to end when each wire tension supports summed weight of all masses below it as shown and a resultant total force pointing upwards as shown at point of fixity.

Hanging Masses

Narasimham
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