Question

Question: (II) A 27 kg chandelier hangs from a ceiling on a vertical 4.0 m long wire. (a) What horizontal force would be necessary to displace its position by 0.15 m to one side? (b) What will be the tension in the wire?

Short answer

(a) The horizontal force that would be needed to displace its position by 0.15 m to one side is $9.93\mathrm{N}$. (b) The tension in the wire is $264.8\mathrm{N}$.

Step-by-step solution

Step 1. Application of Newton’s second law

The wire makes an angle with the vertical when you displace the position of the hanging object that is attached to the vertical string. Resolve tension in the string into components along the horizontal and vertical.

Apply Newton's second law to the hanging object to calculate the applied force and tension in the string.

Step 2. Given information

Given data:

The mass of the chandelier is $m=27\mathrm{kg}$.

The length of the wire is $l=4.0\mathrm{m}$.

The displacement from its initial position is $b=0.15\mathrm{m}$.

Step 3. Calculate the required horizontal force

(a)

The free-body diagram of the chandelier hanging from a ceiling can be drawn as follows:

Here, $F$is the required horizontal force, and $T$is the tension in the wire.

The angle made by the string with the vertical can be calculated as

$\begin{array}{c}\sin \theta =\frac{b}{l}\\ \sin \theta =\frac{\left(0.15\mathrm{m}\right)}{\left(4.0\mathrm{m}\right)}\\ \theta =2.15°\end{array}$

Applying the equilibrium condition along the x-direction,

role="math" localid="1645434121199" $\begin{array}{c}\sum F_x=0\\ F-T\sin \theta =0\\ T\sin \theta =F\dots \left(i\right)\end{array}$

Applying the equilibrium condition along the y-direction,

role="math" localid="1645434157776" $\begin{array}{c}\sum F_y=0\\ T\cos \theta -mg=0\\ T\cos \theta =mg\dots \left(ii\right)\end{array}$

Dividing equation (i) by equation (ii),

$\begin{array}{c}\frac{T\sin \theta }{T\cos \theta }=\frac{F}{mg}\\ \tan \theta =\frac{F}{mg}\\ F=mg\tan \theta \end{array}$

Substituting the values in the above equation,

$\begin{array}{c}F=\left(27\mathrm{kg}\right)\left(9.8{\mathrm{ms}}^2\right)\tan \left(2.15°\right)\\ F=9.93\mathrm{N}\end{array}$

Thus, the required horizontal force is $9.93\mathrm{N}$.

Step 4. Calculate the tension in the wire

(b)

The tension in the wire can be calculated as

$\begin{array}{c}T=\frac{mg}{\cos \theta }\\ T=\frac{\left(27\mathrm{kg}\right)\left(9.8{\mathrm{ms}}^2\right)}{\cos \left(2.15°\right)}\\ T=264.8\mathrm{N}\end{array}$

Thus, the tension in the wire is $264.8\mathrm{N}$.