Chapter 5, Problem 77PQ

A heavy chandelier with mass 125 kg is hung by chains in equilibrium from the ceiling of a concert hall as shown in Figure P5.77, with θ₁ = 37.0° and θ ₂ = 64.0°. Assuming the chains are massless, what are the tensions F_T1, F_T2, and F_T3 in the three chains?

FIGURE P5.77

To determine

Find the values of tensions $F_{T1}$, $F_{T2}$ and $F_{T3}$.

Answer to Problem 77PQ

The values of tensions $F_{T1}$ is $549\text{ N}$, $F_{T2}$ is $\text{1}\text{.00}\times {\text{10}}^3\text{ N}$ and $F_{T3}$ is $\text{1}\text{.23}\times {\text{10}}^3\text{ N}$.

Explanation of Solution

The free body diagram is given below.

      

Applying Newton’s laws.

    $\Sigma F_x=F_{T2}\cos \left({\theta }_2\right)-F_{T1}\cos \left({\theta }_1\right)=0$                                                                (I)

Here, $F_x$ is the force, $F_{T1}$ is the tension force, $F_{T2}$ is the tension force and ${\theta }_1$ & ${\theta }_2$ are the angles.

    $\Sigma F_y=F_{T1}\sin \left({\theta }_1\right)+F_{T2}\sin \left({\theta }_2\right)-F_g=0$                                                        (II)

Here, $F_g$ is the gravitational force.

Rewrite the equation I to find $F_{T2}$.

    $\begin{array}{l}{F}_{T2}\cos \left({\theta }_2\right)-F_{T1}\cos \left({\theta }_1\right)=0\\ F_{T2}=\frac{F_{T1}\cos \left({\theta }_1\right)}{\cos \left({\theta }_2\right)}\end{array}$                                                                         (III)

Since the chandelier is at equilibrium then the gravitational force is the tension force.

    $F_{T3}=F_g=mg$                                                                          (IV)

Here, $F_{T3}$ is tension force, $F_g$ is the gravitational force, $m$ is the mass and $g$ is the acceleration due to gravity.

Substitute equation III in equation II.

    $F_{T1}\sin \left({\theta }_1\right)+\left(\frac{F_{T1}\cos \left({\theta }_1\right)}{\cos \left({\theta }_2\right)}\right)\sin \left({\theta }_2\right)-F_g=0$

  $\begin{array}{c}{F}_{T1}=\frac{F_g\cos \left({\theta }_2\right)}{\sin \left({\theta }_1\right)\cos \left({\theta }_2\right)+\sin \left({\theta }_2\right)\cos \left({\theta }_1\right)}\\ =\frac{F_g\cos \left({\theta }_2\right)}{\sin \left({\theta }_1+{\theta }_2\right)}\end{array}$                                                            (V)

Conclusion:

Substitute $9.81{\text{ m/s}}^2$ for $g$ and $125\text{ kg}$ for $m$ in equation IV.

    $\begin{array}{c}{F}_{T3}=\left(125\text{ kg}\right)\left(9.81{\text{ m/s}}^2\right)\\ =1.23\times {10}^3\text{ N}\end{array}$

Substitute $1.23\times {10}^3\text{ N}$ for $F_{T3}$, $37°$ for ${\theta }_1$ and $64°$ for ${\theta }_2$ in equation V.

    $\begin{array}{c}{F}_{T1}=\frac{\left(1.23\times {10}^3\text{ N}\right)\cos \left(37°\right)}{\sin \left(37°+64°\right)}\\ =549\text{ N}\end{array}$

Substitute $549\text{ N}$ for $F_{T1}$, $37°$ for ${\theta }_1$ and $64°$ for ${\theta }_2$ in equation III.

    $\begin{array}{c}{F}_{T2}=\frac{\left(549\text{ N}\right)\cos \left(37°\right)}{\cos \left(64°\right)}\\ =1.00\times {10}^3\text{ N}\end{array}$

Therefore, the values of tensions $F_{T1}$ is $549\text{ N}$, $F_{T2}$ is $\text{1}\text{.00}\times {\text{10}}^3\text{ N}$ and $F_{T3}$ is $\text{1}\text{.23}\times {\text{10}}^3\text{ N}$.