Question

In the figure shown, find out the value of $\theta$ at this instant [assume string to be tight]

• A. $tan^{-1}\frac{3}{4}$
• B. $tan^{-1}\frac{4}{3}$
• C. $tan^{-1}\frac{3}{8}$
• D. none of these

Answer 1

Answer: (A)

$\theta = tan^{-1}\frac{3}{4}$

Answer 2

To find the value of $\theta$, we use the principle of constraint motion for an inextensible string. The total length of the string remains constant.

Let $L_A$ be the length of the string segment connecting block A to the pulley, and $L_B$ be the length of the string segment connecting block B to the pulley. Let $x_A$ be the horizontal distance of block A from the vertical line passing through the pulley, and $x_B$ be the horizontal distance of block B from the vertical line passing through the pulley. Let $h$ be the vertical height of the pulley from the ground.

The total length of the string $L$ can be written as: $L = L_A + L_B + \text{constant (length over pulley)}$ $L = \sqrt{x_A^2 + h^2} + \sqrt{x_B^2 + h^2} + \text{constant}$

Since the string is inextensible, its total length $L$ is constant, so its time derivative is zero: $\frac{dL}{dt} = 0$

Differentiating the expression for $L$ with respect to time $t$: $\frac{1}{2\sqrt{x_A^2 + h^2}} \left(2x_A \frac{dx_A}{dt} + 2h \frac{dh}{dt}\right) + \frac{1}{2\sqrt{x_B^2 + h^2}} \left(2x_B \frac{dx_B}{dt} + 2h \frac{dh}{dt}\right) = 0$ $\frac{x_A}{L_A} \frac{dx_A}{dt} + \frac{h}{L_A} \frac{dh}{dt} + \frac{x_B}{L_B} \frac{dx_B}{dt} + \frac{h}{L_B} \frac{dh}{dt} = 0$

From the figure, we can relate the geometric terms to the angles: For the left side (block A): $\cos\theta = \frac{x_A}{L_A}$ $\sin\theta = \frac{h}{L_A}$

For the right side (block B): $\cos 30^\circ = \frac{x_B}{L_B}$ $\sin 30^\circ = \frac{h}{L_B}$

Now, let's relate the rates of change of coordinates to the given velocities: Velocity of block A, $v_A = 3.25 \, \text{m/s}$, is to the right. Since $x_A$ is the horizontal distance from the pulley (assuming A is to the left), moving right means $x_A$ is decreasing. So, $\frac{dx_A}{dt} = -v_A$. Velocity of block B, $v_B = \sqrt{3} \, \text{m/s}$, is to the right. Since $x_B$ is the horizontal distance from the pulley (assuming B is to the right), moving right means $x_B$ is increasing. So, $\frac{dx_B}{dt} = v_B$. Velocity of the pulley, $v_P = 1 \, \text{m/s}$, is upwards. So, $\frac{dh}{dt} = v_P$.

Substitute these into the differentiated length equation: $(\cos\theta)(-v_A) + (\sin\theta)(v_P) + (\cos 30^\circ)(v_B) + (\sin 30^\circ)(v_P) = 0$ $-v_A \cos\theta + v_P \sin\theta + v_B \cos 30^\circ + v_P \sin 30^\circ = 0$

Rearranging the terms: $v_P (\sin\theta + \sin 30^\circ) = v_A \cos\theta - v_B \cos 30^\circ$

Now, substitute the given values: $v_A = 3.25 \, \text{m/s} = \frac{13}{4} \, \text{m/s}$ $v_B = \sqrt{3} \, \text{m/s}$ $v_P = 1 \, \text{m/s}$ $\sin 30^\circ = \frac{1}{2}$ $\cos 30^\circ = \frac{\sqrt{3}}{2}$

$1 \left(\sin\theta + \frac{1}{2}\right) = \frac{13}{4} \cos\theta - \sqrt{3} \left(\frac{\sqrt{3}}{2}\right)$ $\sin\theta + \frac{1}{2} = \frac{13}{4} \cos\theta - \frac{3}{2}$ $\sin\theta + 0.5 = 3.25 \cos\theta - 1.5$ $\sin\theta + 0.5 + 1.5 = 3.25 \cos\theta$ $\sin\theta + 2 = 3.25 \cos\theta$ Multiply by 4 to clear the decimal/fraction: $4\sin\theta + 8 = 13\cos\theta$

We need to find $\theta$. We can express this in terms of $\tan\theta$. Divide by $\cos\theta$ (assuming $\cos\theta \neq 0$): $4\tan\theta + \frac{8}{\cos\theta} = 13$ This form is not directly solvable for $\tan\theta$.

Let's try to rearrange $4\sin\theta + 8 = 13\cos\theta$ as $13\cos\theta - 4\sin\theta = 8$. We know that $\sin^2\theta + \cos^2\theta = 1$. From $4\sin\theta = 13\cos\theta - 8$, square both sides: $16\sin^2\theta = (13\cos\theta - 8)^2$ $16(1 - \cos^2\theta) = 169\cos^2\theta - 208\cos\theta + 64$ $16 - 16\cos^2\theta = 169\cos^2\theta - 208\cos\theta + 64$ $185\cos^2\theta - 208\cos\theta + 48 = 0$

This is a quadratic equation in $\cos\theta$. Let $x = \cos\theta$. $185x^2 - 208x + 48 = 0$ Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $x = \frac{208 \pm \sqrt{(-208)^2 - 4(185)(48)}}{2(185)}$ $x = \frac{208 \pm \sqrt{43264 - 35520}}{370}$ $x = \frac{208 \pm \sqrt{7744}}{370}$ To find $\sqrt{7744}$: $80^2 = 6400$, $90^2 = 8100$. Ends in 4, so unit digit is 2 or 8. Try $88^2 = (90-2)^2 = 8100 - 360 + 4 = 7744$. So, $\sqrt{7744} = 88$.

$x = \frac{208 \pm 88}{370}$ Two possible values for $\cos\theta$: $x_1 = \frac{208 + 88}{370} = \frac{296}{370} = \frac{148}{185} = \frac{4 \times 37}{5 \times 37} = \frac{4}{5}$ $x_2 = \frac{208 - 88}{370} = \frac{120}{370} = \frac{12}{37}$

If $\cos\theta = \frac{4}{5}$, then $\sin\theta = \sqrt{1 - (4/5)^2} = \sqrt{1 - 16/25} = \sqrt{9/25} = \frac{3}{5}$ (since $\theta$ is an acute angle in the context of the problem, $\sin\theta > 0$). Let's check this pair in the equation $4\sin\theta + 8 = 13\cos\theta$: $4\left(\frac{3}{5}\right) + 8 = 13\left(\frac{4}{5}\right)$ $\frac{12}{5} + 8 = \frac{52}{5}$ $\frac{12 + 40}{5} = \frac{52}{5}$ $\frac{52}{5} = \frac{52}{5}$ This solution is consistent.

For $\cos\theta = \frac{4}{5}$ and $\sin\theta = \frac{3}{5}$, we have $\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{3/5}{4/5} = \frac{3}{4}$. So, $\theta = \tan^{-1}\left(\frac{3}{4}\right)$.

Let's check the second solution for $\cos\theta = \frac{12}{37}$. If $\cos\theta = \frac{12}{37}$, then $\sin\theta = \sqrt{1 - (12/37)^2} = \sqrt{\frac{37^2 - 12^2}{37^2}} = \sqrt{\frac{(37-12)(37+12)}{37^2}} = \sqrt{\frac{25 \times 49}{37^2}} = \frac{5 \times 7}{37} = \frac{35}{37}$. Check this pair in the equation $4\sin\theta + 8 = 13\cos\theta$: $4\left(\frac{35}{37}\right) + 8 = 13\left(\frac{12}{37}\right)$ $\frac{140}{37} + \frac{8 \times 37}{37} = \frac{156}{37}$ $\frac{140 + 296}{37} = \frac{156}{37}$ $\frac{436}{37} = \frac{156}{37}$ This is false. So, $\cos\theta = \frac{12}{37}$ is an extraneous solution introduced by squaring.

Therefore, the only valid solution is $\theta = \tan^{-1}\left(\frac{3}{4}\right)$.

Core Solution:

1. Define the total length of the string $L = \sqrt{x_A^2 + h^2} + \sqrt{x_B^2 + h^2}$.
2. Differentiate $L$ with respect to time and set to zero: $\frac{dL}{dt} = 0$.
3. Substitute $\frac{dx_A}{dt} = -v_A$, $\frac{dx_B}{dt} = v_B$, $\frac{dh}{dt} = v_P$.
4. Use trigonometric relations: $\cos\theta = \frac{x_A}{\sqrt{x_A^2 + h^2}}$, $\sin\theta = \frac{h}{\sqrt{x_A^2 + h^2}}$, and similarly for $30^\circ$.
5. The resulting equation is $-v_A \cos\theta + v_P \sin\theta + v_B \cos 30^\circ + v_P \sin 30^\circ = 0$.
6. Substitute given values: $v_A = 3.25$, $v_B = \sqrt{3}$, $v_P = 1$, $\sin 30^\circ = 1/2$, $\cos 30^\circ = \sqrt{3}/2$.
7. Simplify the equation: $4\sin\theta + 8 = 13\cos\theta$.
8. Solve for $\theta$ by converting to a quadratic in $\cos\theta$: $185\cos^2\theta - 208\cos\theta + 48 = 0$.
9. Solve the quadratic equation to get $\cos\theta = \frac{4}{5}$ or $\cos\theta = \frac{12}{37}$.
10. Check both solutions in the original linear equation $4\sin\theta + 8 = 13\cos\theta$. Only $\cos\theta = \frac{4}{5}$ (which implies $\sin\theta = \frac{3}{5}$) satisfies the equation.
11. From $\cos\theta = \frac{4}{5}$ and $\sin\theta = \frac{3}{5}$, $\tan\theta = \frac{3/5}{4/5} = \frac{3}{4}$.
12. Therefore, $\theta = \tan^{-1}\left(\frac{3}{4}\right)$.