# A particle of mass 10kg is suspended by two strings that make an angle of 30° and 45° respectively with the horizontal. Find tensions in the strings?

Feb 9, 2018

So,this is the diagram of the described situation,the strings are attached to a rigid support at their one end and the other end is connected to the particle.

Let the string making an angle of $45$ with the horizontal has a tension of ${T}_{1}$ and the one making an angle of $30$ has a tension of ${T}_{2}$.

Now,what we have done is that we have taken the vertical and horizontal components of the tension in the two strings.

Now,considering the system to be in equilibrium,

see the diagram,the vertical component of the strings i.e (${T}_{1} \sin 45$ & ${T}_{2} \sin 30$) together will act to balance the weight of the particle($m g$)

So,we can say, ${T}_{1} \sin 45 + {T}_{2} \sin 30 = m g$...1

From horizontal equilibrium,we can say, ${T}_{1} \cos 45$ must balance ${T}_{2} \cos 30$ ,as no other force is present in horizontal direction.

So,${T}_{1} \cos 45 = {T}_{2} \cos 30$...2

Given, $m g = 10 \cdot 10 = 100 N$

So,solving equation 1 & 2 we get,

${T}_{1} = 89.65 N$ and, ${T}_{2} = 73.20 N$