Question

A makeshift swing is constructed by making a loop in one end of a rope and tying the other end to a tree limb. A child is sitting in the loop with the rope hanging vertically when the child’s father pulls on the child with a horizontal force and displaces the child to one side. Just before the child is released from rest, the rope makes an angle of $\mathbf{15}\mathbf{°}$with the vertical and the tension in the rope is $\mathbf{280}\mathbf{\text{\hspace{0.17em}N}}$ .

(a) How much does the child weigh?

(b) What is the magnitude of the (horizontal) force of the father on the child just before the child is released?

(c) If the maximum horizontal force the father can exert on the child is $\mathbf{93}\mathbf{}\mathbf{N}$ , what is the maximum angle with the vertical the rope can make while the father is pulling horizontally?

Short answer

1. The weight of the child,$\text{W}=270.46\text{\hspace{0.17em}N}$.
2. The magnitude of the (horizontal) force of the father on the child,$\text{F}=72.47\text{\hspace{0.17em}N}$.
3. The maximum angle with the vertical the rope can make, $\mathrm{\theta }=18.98°.$

Step-by-step solution

Understanding the given information

The tension in the cord is$280\text{\hspace{0.17em}N}$.

The angle made by the rope to the vertical,$\mathrm{\theta }=15°$.

The maximum horizontal force, ${\text{F}}_{\text{max}}=93\text{\hspace{0.17em}N}$.

Concept and formula used in the given question

Using the equation for the equilibrium of force, you can find out the weight of the child, the horizontal force the father applied to the child, and the maximum angle made by the rope. The equations are given below.

$\begin{array}{l}\mathbf{W}\mathbf{=}\mathbf{Tcos}\mathbf{}\mathbf{\left(}\mathbf{15}\mathbf{\text{°}}\mathbf{\right)}\mathbf{}\\ \mathbf{F}\mathbf{=}\mathbf{Tsin}\mathbf{}\mathbf{\left(}\mathbf{15}\mathbf{\text{°}}\mathbf{\right)}\mathbf{}\end{array}$

(a) Calculation of the much the child weighs

The force equation for the child can be written as,

$\mathrm{W}=\mathrm{Tcos}(15°)$

And

$\mathrm{F}=\mathrm{Tsin}(15°)$

We have, the force equation for weight as,

$\mathrm{W}=\mathrm{Tcos}(15°)$

By using the value of $\text{T}=280\text{\hspace{0.17em}N}$,

we get,

$\text{W}=270.46\text{\hspace{0.17em}N}$

(b) Calculation for the magnitude of the (horizontal) force of the father on the child just before the child is released

We have, the force equation for the horizontal force

$\mathrm{F}=\mathrm{Tsin}(15°)$

So, we get

$\text{F}=72.47\text{\hspace{0.17em}N}$

(c) Calculation for the maximum  angle with the vertical the rope can make while the father is pulling horizontally

We have the force equation for horizontal force as

$\mathrm{F}=\mathrm{Tsin\theta }$

In the vertical direction, we have, the weight of the child,

$\text{W}=270.46\text{\hspace{0.17em}N}$

By rearranging this equation for angle, we get

$\begin{array}{c}\mathrm{\theta }=\left(\frac{{\text{F}}_{\text{max}}}{\text{W}}\right)\\ \mathrm{\theta }=18.98°\end{array}$