Solution to Question 36
Analysis:
Let the velocity of load A be $\vec{v}$. The strings are pulling the load. The core principle here is that the component of the load’s velocity along the direction of each string must equal the speed at which that string is being pulled.
Let the two string segments connected to A be $S_1$ and $S_2$.
Let $\vec{n}_1$ and $\vec{n}_2$ be unit vectors along the strings pointing away from A.
We are given:
$$ \vec{v} \cdot \vec{n}_1 = u_1 $$
$$ \vec{v} \cdot \vec{n}_2 = u_2 $$
Geometric Solution:
Let $\alpha$ be the angle between $\vec{v}$ and the first string. Since the total angle between the strings is $\theta$, the angle between $\vec{v}$ and the second string is $(\theta – \alpha)$.
From the projections:
- $v \cos\alpha = u_1$ …(i)
- $v \cos(\theta – \alpha) = u_2$ …(ii)
Expand equation (ii):
$$v (\cos\theta \cos\alpha + \sin\theta \sin\alpha) = u_2$$Substitute $v \cos\alpha = u_1$ into this expansion:
$$u_1 \cos\theta + v \sin\alpha \sin\theta = u_2$$ $$v \sin\alpha = \frac{u_2 – u_1 \cos\theta}{\sin\theta}$$Now we have expressions for the two orthogonal components of $v$ relative to the first string ($v \cos\alpha$ and $v \sin\alpha$). We square and add them to find the magnitude of $v$:
$$v^2 = (v \cos\alpha)^2 + (v \sin\alpha)^2$$ $$v^2 = u_1^2 + \left( \frac{u_2 – u_1 \cos\theta}{\sin\theta} \right)^2$$Simplifying the algebra:
$$v^2 = \frac{u_1^2 \sin^2\theta + (u_2 – u_1 \cos\theta)^2}{\sin^2\theta}$$ $$v^2 = \frac{u_1^2 \sin^2\theta + u_2^2 + u_1^2 \cos^2\theta – 2 u_1 u_2 \cos\theta}{\sin^2\theta}$$ $$v^2 = \frac{u_1^2 (\sin^2\theta + \cos^2\theta) + u_2^2 – 2 u_1 u_2 \cos\theta}{\sin^2\theta}$$ $$v^2 = \frac{u_1^2 + u_2^2 – 2 u_1 u_2 \cos\theta}{\sin^2\theta}$$Final Expression:
$$v = \frac{\sqrt{u_1^2 + u_2^2 – 2 u_1 u_2 \cos\theta}}{\sin\theta}$$