A sphere of mass $\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}\mathbf{}}\mathbf{kg}$is suspended from a cord. A steady horizontal breeze pushes the sphere so that the cord makes a constant angle role="math" localid="1657172689218" ${\mathbf{37}}^{\mathbf{o}}$ofwith the vertical. (a) Find the push magnitude and (b) Find the tension in the cord.

• The mass of the sphere is,$m=3.0\times {10}^{-4}kg.$
• The angle made by the cord with the vertical is,$\theta ={37}^o.$

Force acting on the object is equal to the mass multiplied by the acceleration of the object. It is called Newton’s second law of motion. The free-body diagram shows different forces acting on the object along with their directions.

Using the free-body diagram, Newton’s second law and vector resolution concept, the pushing force on the sphere and tension in the cord can be found.

Formulae:

$\overrightarrow{F}net=m\overrightarrow{a}$

Here,$\overrightarrow{F}net$is the net force acting on the object, $m$is the mass of the object, and$\overrightarrow{a}$is the acceleration of the object.

By applying Newton’s second law along x axis as,

$\overrightarrow{F}net=m\overrightarrow{a}$

Due to pushing force the sphere moves in upward direction as shown in the figure. Consider the sign convention according to the direction of forces acting on the sphere as shown in the figure.

By applying Newton’s second law along x axis as

localid="1657176304348" $$\begin{gathered} \overrightarrow{F}net=m\overrightarrow{a} \\ F.\cos \theta -mg.\sin \theta =0 \\ F.\cos \theta =mg.\sin \theta \\ F=\frac{3.0\times {10}^{-4}kg\times 9.8m/s^2\times \sin \left({37}^o\right)}{\cos \left({37}^o\right)} \\ =2.2\times {10}^{-3}N \end{gathered}$$

Hence, the magnitude of pushing force is $2.2\times {10}^{-3}N$.

The tension in the cord is acting in the upward direction, hence applying Newton’s second law in the vertical direction. The vertical acceleration is zero.

$$\begin{gathered} T-mg.\cos \theta -F.\sin \theta =0 \\ T=mg.\cos \theta +F.\sin \theta \\ =\left[(3.0\times {10}^{-4})kg\times 9.8m/s^2\times \cos \left({37}^o\right)\right]+(2.2\times {10}^{-3}N\times \sin ({37}^o\left)\right) \\ =3.7\times {10}^{-3}N \end{gathered}$$

Hence, the tension in the chord is $3.7\times {10}^{-3}N$.