# Statics of a Particle

• Sep 15th 2008, 05:00 PM
Statics of a Particle
Could someone help me with this problem
A weight of 20 kg is suspended from two strings of
length 10 cm and 12 cm, the ends of the strings being
attached to two points in a horizontal line, 15 cm apart.
Find the tension in each string.
Attachment 7791
The answer in the textbook is T1=14.99kg wt and T2=12.10kg wt, but i'm not sure how to get it. i tried finding the angles and then moving the vectors around but it wasn't right.
• Sep 15th 2008, 05:09 PM
arbolis
I won't do it, but the book is wrong. A tension is a force and its unit is commonly in newtons ($\displaystyle N$), not $\displaystyle kg$.
Put the equations :
The system is in equilibrium, so the sum of the vertical components of the tensions are equal in magnitude to the weight. Take note that the weight is a force and its unit is not in $\displaystyle kg$ but in $\displaystyle N$. The object has a mass of $\displaystyle 20 kg$ which means $\displaystyle 20kg\cdot \frac{9.8m}{s^2}=196N$.
• Sep 15th 2008, 08:15 PM
what are the vertical components of the tensions and how would you figure them out? would finding them help to solve the problem? would it be possible to use kilogram weight as the force instead of newtons, or does it have to be N? sorry but i still dont know how to work it out
• Sep 15th 2008, 08:54 PM
arbolis
Quote:

For example you know that the horizontal components of the tensions $\displaystyle T_1$ and $\displaystyle T_2$ are equal in magnitude and opposite in sense because the system is at equilibrium. If they are not equal in magnitude, then the system would be moving with an acceleration which is not the case.
On the x-axis you have that the horizontal component of $\displaystyle T_1$ times i unit vector $\displaystyle +$ horizontal component of $\displaystyle T_2$ times i unit vector is equal to $\displaystyle 0$.
I call $\displaystyle T_{1ver}$ the vertical component of $\displaystyle T_1$ and $\displaystyle T_{2ver}$ the vertical component of $\displaystyle T_2$. For the y-axis you have that $\displaystyle T_{1ver}+T_{2ver}=\text{weight of the mass}$.
And I'm sorry, your book is probably right about the weight. The weight is a force and the SI unit for it is the newton, $\displaystyle N$ which is worth $\displaystyle \frac{1kgm}{s^2}$, but I misread what your book says, it says $\displaystyle kg\cdot wt$. I never saw that before but it is probably a possible unit for a force. So excuse me here.