Solution for Question 45
System Analysis
Consider the equilibrium of the entire rope.
- Mass of rope $m = 2$ kg.
- Length $L = 4$ m.
- Linear mass density $\lambda = m/L = 0.5$ kg/m.
- Total Weight $W = mg = 20$ N.
Derivation of Vertical Height ($y$)
Consider a small element of length $dl$ on the rope. The vertical component of tension at any point balances the weight of the segment below it.
Let $T$ be the tension at an angle $\theta$. The vertical component is $T_y = T \sin \theta$. The horizontal component is constant everywhere: $T_x = T \cos \theta = T_0$.
From the free body diagram of an element $dl$ with mass $dm = \lambda dl$:
$$ dT_y = (dm)g = (\lambda dl)g $$Since $dy = dl \sin \theta$, we can relate the change in tension to the change in height. We know that the net tension vector magnitude $T$ is related to the vertical height. A more direct approach uses energy or forces on a finite segment.
Method using Forces on a Finite Segment:
Consider a segment of length $l$ from the bottom (horizontal tangent) to a point at height $y$.
The forces are:
1. Horizontal tension $T_0$ at the bottom.
2. Tension $T$ at the top (tangent to rope).
3. Weight of the segment $W_{seg} = \lambda l g$ acting downwards.
Resolving forces vertically: $T \sin \theta = \lambda l g$.
Resolving forces horizontally: $T \cos \theta = T_0$.
Now, consider the tension magnitude relationship: $T^2 = (T \sin \theta)^2 + (T \cos \theta)^2$. $$ T^2 = (\lambda l g)^2 + T_0^2 $$ Differentiating with respect to the vertical coordinate $y$ is complex involving the catenary shape $y = a \cosh(x/a)$. A simpler property relates the work done to lift the chain.
Alternative: Direct Force Integration
The vertical component of tension is $T_y = \lambda g s$ (where $s$ is arc length).
The tension $T$ is directed along the tangent: $dy/dx = \tan \theta = T_y/T_x = \lambda g s / T_0$.
However, we specifically need the relation between $T$, $T_0$, and $y$.
Consider the vertical equilibrium of a small element $dy$:
$$ dT_y = \lambda g dl $$
Since $T_y = T \sin \theta$ and $T_x = T \cos \theta = T_0$, we have $T = T_0 / \cos \theta = T_0 \sec \theta$.
Also $T_y = T_0 \tan \theta$. So $dT_y = T_0 \sec^2 \theta d\theta$.
Equating the two expressions for $dT_y$:
$$ T_0 \sec^2 \theta d\theta = \lambda g dl $$
Since $dy = dl \sin \theta$:
$$ dy = \frac{T_0 \sec^2 \theta \sin \theta}{\lambda g} d\theta = \frac{T_0}{\lambda g} \tan \theta \sec \theta d\theta $$
Integrating from bottom ($\theta=0, T=T_0$) to top ($\theta, T$):
$$ y = \int_0^{\theta} \frac{T_0}{\lambda g} \sec \theta \tan \theta d\theta $$
$$ y = \frac{T_0}{\lambda g} [\sec \theta]_0^{\theta} = \frac{T_0}{\lambda g} (\sec \theta – 1) $$
Rearranging: $\lambda g y = T_0 \sec \theta – T_0$.
Since $T = T_0 \sec \theta$ (from horizontal component definition):
$$ \lambda g y = T – T_0 $$
$$ y = \frac{1}{\lambda g}(T – T_0) $$
Calculations
Using the derived relation with $\lambda g = 5$ N/m and $y=2$ m:
$$ 2 = \frac{1}{5}(T – T_0) \implies T – T_0 = 10 \quad \dots(1) $$From the vector triangle of forces for the entire rope ($T_0$ horizontal, $W=20$ vertical, $T$ hypotenuse):
$$ T^2 = T_0^2 + W^2 $$ $$ T^2 – T_0^2 = 20^2 = 400 $$ $$ (T – T_0)(T + T_0) = 400 $$ Substituting eq(1) ($T – T_0 = 10$): $$ 10(T + T_0) = 400 \implies T + T_0 = 40 \quad \dots(2) $$Solving (1) and (2):
$$ 2T = 50 \implies T = 25 \text{ N} $$ $$ T_0 = 15 \text{ N} $$Find Angle
Using the vertical component: $T \sin \theta = mg = 20$
$$ 25 \sin \theta = 20 $$ $$ \sin \theta = \frac{20}{25} = \frac{4}{5} $$ $$ \theta = 53^\circ $$Answer: (b) $53^\circ$ above the horizontal