Design a decorative “mobile” to consist of a low-mass rod of length $\mathbf{0}\mathbf{.}\mathbf{49}\mathbf{}\mathbf{m}$suspended from a string so that the rod is horizontal, with two balls hanging from the ends of the rod. At the left end of the rod hangs a ball with mass $\mathbf{0}\mathbf{.}\mathbf{484}\mathbf{}\mathbf{kg}$.At the right end of the rod hangs a ball with mass $\mathbf{0}\mathbf{.}\mathbf{273}\mathbf{}\mathbf{kg}$.You need to decide how far from the left end of the rod you should attach the string that will hold up the mobile, so that the mobile hangs motionless with the rod horizontal (“equilibrium”). You also need to determine the tension in the string supporting the mobile. (a) What is the tension in the string that supports the mobile? (b) How far from the left end of the rod should you attach the support string?

Tension is a force along the length of a medium, especially a force carried by a flexible medium, such as a rope or cable.

The tension$\left(T\right)$in the string that supports the mobile can be calculate as follows.

$T=Mg$ ….. (1)

Here,$M$is the mass and $g$is the acceleration due to gravity having a value of $9.8\mathrm{m}\backslash {\mathrm{s}}^2$.

Consider the given data as below.

Mass of ball at left end,$m_1=0.484kg$

Mass of the ball at right end,$m_2=0.273kg$

Acceleration due to gravity,$g=9.8m/s^2$

To hold up the mobile, you should string attach at center of mass of the system, at center of mass point total mass of the system is suppose to be concentrate.

Total mass of the system is given by

$\begin{array}{rcl}{M}_{total}& =& m_1+m_2\\ & =& 0.484\text{kg}+0.273\text{kg}\\ & =& 0.757\text{kg}\\ & & \end{array}$

The tension $\left(T\right)$in the string that supports the mobile can be calculate as follows.

$T=M_{total}g$

Here, $g$is the acceleration due to gravity having a value of$9.8\mathrm{m}/\mathrm{s}$.

Now, substitute known values in the above equation, and you have

$\begin{array}{rcl}T& =& \left(0.757\text{kg}\right)(9.8\mathrm{m}/{\mathrm{s}}^2)\\ & =& 7.42\mathrm{N}\\ & & \end{array}$

Hence, The tension in the string that supports the mobile is $7.42\mathrm{N}$.