A mass of 5.00 kg hangs attached to three strings as shown in the figure (see image below). Find the tension in each string. Hint: Consider the equilibrium of the point where the strings join.

So finding $T_3$ was relative quick, knowing $W = mg$, $m = 5.00\text{ kg}$, $g = 10\ \mathrm{ms^{-2}}$

$$W = 50\text{ N} = T_3$$

It is the other two ($T_1$ and $T_2$) I'm not sure. In fact I'm at lost with what I learned in class and applying. I noticed the problem states a hint (equilibrium where strings join) but I'm not seeing how this information can be used.


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  • $\begingroup$ Hint: think about components of vectors. $\endgroup$ – David Z Jan 2 '12 at 1:44
  • $\begingroup$ You were right, thank you. the rest unfolded easily, I found the correct solution. $\endgroup$ – Cloud Jan 2 '12 at 2:05
  • 4
    $\begingroup$ Good to hear it. Perhaps you could write up the solution you found and post it as an answer. (Unfortunately the system requires you to wait 8 hours before answering your own question, but I'd encourage you to come back and do it tomorrow.) $\endgroup$ – David Z Jan 2 '12 at 2:13
  • $\begingroup$ (By the way, it's fantastic that such a small hint was enough to get you through the problem. I wish every homework question was like that.) $\endgroup$ – David Z Jan 2 '12 at 6:31
  • $\begingroup$ Anybody care to fix the image link? $\endgroup$ – Qmechanic Feb 2 '12 at 19:43

$\newcommand{\t}[1]{\frac{T_#1}{\sin\theta_#1}}$ The correct way is with vectors.

The shortcut way is to use Lami's theorem. Basically, if the tension in each string is $T_1$, and the angle opposite each string is $\theta_1$, then

$$\t 1=\t 2=\t 3$$

enter link description here
(source: wikimedia.org)

Referring to this diagram, $\frac{A}{\sin \alpha}=\frac{B}{\sin \beta}=\frac{C}{\sin \gamma}$, where $A,B,C$ are forces.


Here is the formal methodology for this type of problems.

When things are in equilibrium, then the sum of the forces acting in any direction equal to zero. So add up the vertical forces on the knot and make them equal to zero, as well as the horizontal forces equal to zero. From those two equations solve for the two unknowns $T_1$ and $T_2$.


Resolve the vector components. Then solve the individual tensions.

T3 = mg = 5*10 = 50

T2 sin 45 = T3 T2 sin 45 = 50 Therefore, T2 = 50√2

T1 = T2 cos 45 T1 = 50√2 cos 45 Therefore, T1 = 50

Hence, T1 = 50 T2 = 50√2 T3 = 50



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