Chapter 4, Problem 28P

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 0.15 m to one side? (b) What will be the tension in the wire?

To determine

Part (a) To Determine:

Horizontal force necessary to displace a 27-kg chandelier position 0.15 m to one side.

Answer to Problem 28P

Solution:

Horizontal force is equal to 9.6 N.

Explanation of Solution

Given:

A 27-kg chandelier hangs from a ceiling on a vertical 4.0-m- long wire, its position0.15 m.

Formula used:

$\theta ={\tan }^{-1}\left(\frac{0.15}{4.0}\right)$

Calculation:

Given that m = 27-kg

Length of wire = 4.0 m

Horizontal position 0.15 m.

$\theta ={\tan }^{-1}\left(\frac{0.15}{4.0}\right)$.

  $={2.09}^{\text{o}}$.

$Horizontalforce=mg\sin \theta$.

                                $=27\times 9.8\times \text{sin}\left(2.09\right)$.

                                $=9.6\text{N}$

To determine

Part (b) To Determine:

The tension in the wire.

Answer to Problem 28P

Solution:

$Tension\left(T\right)=264.77\text{ N}$

Explanation of Solution

Given:

A 27-kg chandelier hangs from a ceiling on a vertical 4.0-m- long wire, its position0.15m.

Formula used:

$\theta ={\tan }^{-1}\left(\frac{0.15}{4.1}\right)$,

$Horizontalforce=mg\sin \theta$,

$Tension\left(T\right)=\sqrt{{\left(9.6\right)}^2+{\left(27\times 9.8\right)}^2}$.

Calculation:

Given that m=27-kg,

Length of wire = 4.0 m.

Horizontal position 0.15m.

Hence, $\theta ={\tan }^{-1}\left(\frac{0.15}{4.1}\right)$

                $={2.09}^{\text{o}}$.

$Horizontalforce=mg\sin \theta$.

$=27\times 9.8\times \text{sin}\left(2.09\right)$.

$=9.6\text{N}$.

Tension is calculated as square root of square of horizontal component and vertical component.

$Tension\left(T\right)=\sqrt{{\left(9.6\right)}^2+{\left(27\times 9.8\right)}^2}$

$Tension\left(T\right)=264.77\text{ N}$.