"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.
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 $T_1=343N$, $T_2=196N$, $T_3=127.4N$