Chapter 4, Problem 54P

(a)

To determine

The horizontal component of the force.

Explanation of Solution

Given:

An arch is grounded at the both ends and helium filled balloons are placed at equal interval on the arch with a mass less rope of length $L$ .

Introduction:

Under equilibrium condition net force acting on a static or a dynamic body sums up to zero. No net force acts on a body under the equilibrium condition.

The arch with the helium filled balloons is in equilibrium. The horizontal component of the force acting on the balloons is of same magnitude.

Write the expression for $i^{\text{th}}$ and ${\left(i-1\right)}^{\text{th}}$ component of the horizontal force acting on the balloons.

  $T_i\text{cos}{\theta }_i$ And $T_{\left(i-1\right)}\text{cos}{\theta }_{\left(i-1\right)}$

Here $T_i$ is the $i^{\text{th}}$ component of the horizontal force, $T_{\left(i-1\right)}$ is the ${\left(i-1\right)}^{\text{th}}$ component of the horizontal force and $\theta$ is the angle between the arch and the mass less rope.

Under equilibrium the net force is zero. The horizontal components of the forces are equal in magnitude.

Write the expression of force under equilibrium condition.

  $\begin{array}{l}{T}_i\text{cos}{\theta }_i=T_{\left(i-1\right)}\text{cos}{\theta }_{\left(i-1\right)}\\ T_i\text{cos}\theta -T_{\left(i-1\right)}\text{cos}{\theta }_{\left(i-1\right)}=0\end{array}$

Substitute $T_{\text{H}}$ for $T_i\text{cos}\theta$ in the above equation.

  $\begin{array}{l}{T}_{\text{H}}-T_{\left(i-1\right)}\text{cos}{\theta }_{\left(i-1\right)}=0\\ T_{\text{H}}=T_{\left(i-1\right)}\text{cos}{\theta }_{\left(i-1\right)}\end{array}$

Conclusion:

Thus, the horizontal component of the forces has equal magnitude.

(b)

To determine

The tension of the mass-less rope.

Explanation of Solution

Given:

An arch is grounded at the both ends and helium filled balloons are placed at equal interval on the arch with a mass less rope of length $L$ .

Introduction:

The arch with the helium filled balloons is in equilibrium. The horizontal component of the force acting on the balloons is of same magnitude.

Write the expression for Newton’s second law

  $F=ma$

Here, $F$ is the force, $m$ is the mass of the object and $a$ is the acceleration of the body.

Under equilibrium condition all the vertical component of the force balances each other.

  $\sum F_{\text{y}}=0$

Here, $\sum F_{\text{y}}$ is the summation of all the vertical component of the forces.

Write the expression for $i^{\text{th}}$ and ${\left(i-1\right)}^{\text{th}}$ component of the vertical force acting on the balloons.

  $T_i\text{sin}{\theta }_i$ And $T_{\left(i-1\right)}\text{sin}{\theta }_{\left(i-1\right)}$

Under equilibrium condition

  $\begin{array}{l}F+T_i\text{sin}{\theta }_i-T_{\left(i-1\right)}\text{sin}{\theta }_{\left(i-1\right)}=0\\ F=T_{\left(i-1\right)}\text{sin}{\theta }_{\left(i-1\right)}-T_i\text{sin}{\theta }_i\end{array}$

Conclusion:

Thus, under equilibrium condition the relation between forces of tension is $T_{\left(i-1\right)}\text{sin}{\theta }_{\left(i-1\right)}-T_i\text{sin}{\theta }_i$

(c)

To determine

The, trigonometric identity $\text{tan}{\theta }_0$ equals to a constant value.

Explanation of Solution

Given:

An arch is grounded at the both ends and helium filled balloons are placed at equal interval on the arch with a mass less rope of length $L$ .

Introduction:

There are two supports at the two ends of the arch. Each supports equally shares the vertical component of the force.

Write the expression for vertical component of the force shared by the two supports.

  $h=NF/2$

Here, $N$ is the number of balloons and $F$ is the upward lift force.

Write the expression for the trigonometric identity $\text{tan}{\theta }_0$

  $\text{tan}{\theta }_0=\frac{\text{height}}{\text{base}}$

Substitute $NF/2$ for height and $T_{\text{H}}$ for base in the above equation.

  $\begin{array}{c}\text{tan}{\theta }_0=\frac{NF/2}{T_{\text{H}}}\\ =NF/2T_{\text{H}}\end{array}$

Write the expression for symmetry of the angle $\theta$ .

  ${\theta }_{\left(N+1\right)}=-{\theta }_0$

Here $(-)$ sign indicates that $\text{tan}$ is a odd function.

Write the expression for $\text{tan}{\theta }_{\left(N+1\right)}$ .

  $\begin{array}{c}-\text{tan}{\theta }_{\left(N+1\right)}=\text{tan}{\theta }_0\\ =NF/2T_{\text{H}}\end{array}$

Conclusion:

Thus, under equilibrium condition the expression for $-\text{tan}{\theta }_{\left(N+1\right)}$ is $NF/2T_{\text{H}}$ .

(d)

To determine

The, trigonometric identity $tan{\theta }_i$ and the $y$ co-ordinate equals to a constant value.

Explanation of Solution

Given:

An arch is grounded at the both ends and helium filled balloons are placed at equal interval on the arch with a mass less rope of length $L$ .

Introduction:

Under equilibrium condition the components of force due to tension of the rope balances each other.

Write the expression for $T_{\text{H}}$

  $T_{\text{H}}=T_i\text{cos}{\theta }_i$

Substitute $T_{\left(i-1\right)}\text{cos}{\theta }_{\left(i-1\right)}$ for $T_i\text{cos}{\theta }_i$ in the above expression.

  $T_{\text{H}}=T_{\left(i-1\right)}\text{cos}{\theta }_{\left(i-1\right)}$

Write the expression for force.

  $F=T_{\left(i-1\right)}\text{sin}{\theta }_{\left(i-1\right)}-T_i\text{sin}{\theta }_i$

Divide both sides of the above expression by $T_{\text{H}}$ .

  $\begin{array}{c}\frac{F}{T_{\text{H}}}=\frac{\left(T_{\left(i-1\right)}\text{sin}{\theta }_{\left(i-1\right)}-T_i\text{sin}\theta \right)}{T_{\text{H}}}\\ =\text{tan}{\theta }_{\left(i-1\right)}-\text{tan}{\theta }_i\end{array}$

Rearrange the above equation.

  $\text{tan}{\theta }_i=\text{tan}{\theta }_{\left(i-1\right)}-F/T_{\text{H}}$

Substitute $1$ for $i$ in theabove expression.

  $\text{tan}{\theta }_1=\text{tan}{\theta }_0-F/T_{\text{H}}$

Substitute $NF/2T_{\text{H}}$ for $\text{tan}{\theta }_0$ in the above expression.

  $\begin{array}{c}\text{tan}{\theta }_1=\left(NF/2T_{\text{H}}\right)-\left(F/T_{\text{H}}\right)\\ =\left(N-2\right)\left(F/T_{\text{H}}\right)\end{array}$

Write the expression of length of rope between two consecutive balloons.

  $L_{\text{in}\text{-between}}=\frac{L}{N+1}$

Here $N$ is the number of balloons.

Write the expression for the horizontal coordinate of the $i^{\text{th}}$ balloons.

  $x_i=\left(L/N+1\right){\displaystyle \sum _{J=0}^{i-1}\text{cos}{\theta }_j}$

Write the expression for the vertical coordinate of the $i^{\text{th}}$ balloons.

  $y_i=\left(L/N+1\right){\displaystyle \sum _{J=0}^{i-1}\text{sin}{\theta }_j}$

Conclusion:

Thus, the horizontal coordinate of the $i^{\text{th}}$ balloon is $\left(L/N+1\right){\displaystyle \sum _{J=0}^{i-1}\text{cos}{\theta }_j}$ and the vertical coordinate is $\left(L/N+1\right){\displaystyle \sum _{J=0}^{i-1}\text{sin}{\theta }_j}$ .