The load A is being pulled with the help of two inextensible strings that pass over two fixed pulley

Q: Why can't we just add the velocities $u_1$ and $u_2$ directly?

Velocities are vectors, not scalars. The strings are pulling in different directions, so their effects don't just add up linearly. The load's actual velocity is a single vector whose *projections* onto the strings match $u_1$ and $u_2$.

Q: Why is the component of velocity along the string equal to the string's speed?

Because the string is inextensible (it doesn't stretch). If the load moved faster along the string's direction, the string would go slack. If it moved slower, the string would break. So, they must match exactly.

Q: What if the angle $\theta$ is 90 degrees?

If $\theta = 90^\circ$, then $\cos(90^\circ) = 0$ and $\sin(90^\circ) = 1$. The formula simplifies to $\sqrt{u_1^2 + u_2^2}$, which makes perfect sense as the two velocity components would be completely independent and perpendicular.

Q: Why did we align the X-axis with String 1?

It's a mathematical trick to save time! By aligning the axis with one string, we immediately know the X-component of the velocity ($v_x = u_1$). If we chose an arbitrary axis, we'd have to solve a messy system of two equations.

Q: Can the load move upwards if both strings are pulled downwards?

Yes! The diagram shows the strings going over pulleys. Pulling the ends downwards means the segments attached to the load are getting shorter, which forces the load to move generally upwards towards the pulleys.

Question

The load A is being pulled with the help of two inextensible strings that pass over two fixed pulleys as shown in the figure. At an instant velocities of the ends of the string being pulled are $u_1$ and $u_2$ and the angle between the strings connected to the load is $	heta$, what is speed of the load?

Sigma Percentile · Accepted Answer

### The Art of Constrained Motion: Unraveling the Two-Pulley Problem

Welcome to a beautiful exploration of constrained motion! Imagine you are the load $A$ in this setup. You are being tugged by two strings simultaneously, each passing over a fixed pulley. You can't just move anywhere you please! Your actual velocity must perfectly agree with how fast each string is being pulled.

This is the essence of a **constraint**. Because the strings are inextensible (they cannot stretch or go slack), the component of your velocity along the direction of each string must exactly match the speed at which that string is being pulled.

### Setting the Stage: A Smart Coordinate System

To solve this elegantly, let's set up a smart coordinate system. We'll align our $X$-axis right along String 1.

Now, if the load moves with a velocity vector $\vec{v}$, its shadow, or projection, along String 1 must exactly equal $u_1$. Mathematically, this means the $x$-component of our velocity is simply:

$$v_x = u_1$$

Now look at String 2. It's sitting at an angle $	heta$ relative to String 1. The exact same logic applies here. The projection of our velocity vector $\vec{v}$ onto String 2 must be exactly $u_2$. If we drop a perpendicular from the tip of $\vec{v}$ onto the line of String 2, that length is $u_2$.

### The Master Equation: Resolving the Vectors

Let's bring in the math. We can write our velocity vector $\vec{v}$ in terms of its $x$ and $y$ components:

$$\vec{v} = v_x \hat{i} + v_y \hat{j}$$

We already found that $v_x = u_1$. Now, what about String 2? We can define its direction using a unit vector $\hat{e}_2$ that incorporates the angle $	heta$:

$$\hat{e}_2 = \cos	heta \hat{i} + \sin	heta \hat{j}$$

To find the projection mathematically, we take the dot product of the velocity vector and the unit vector of String 2. This dot product must equal $u_2$:

$$\vec{v} \cdot \hat{e}_2 = u_2$$

Expanding this out gives us a simple linear equation relating our components to $u_2$:

$$v_x \cos	heta + v_y \sin	heta = u_2$$

### Isolating the Unknowns

We are almost there! We know $v_x$ is $u_1$. Let's substitute that into our equation:

$$u_1 \cos	heta + v_y \sin	heta = u_2$$

Now, we have a straightforward equation where the only unknown is $v_y$. With a quick rearrangement, we isolate $v_y$:

$$v_y = \frac{u_2 - u_1 \cos	heta}{\sin	heta}$$

Look at that! We now have both the $x$ and $y$ components of the load's velocity.

### Final Calculation: The Speed of the Load

The question asks for the **speed**, which is the magnitude of the velocity vector. We just use the good old Pythagorean theorem. Square the $x$-component, square the $y$-component, add them up, and take the square root:

$$|\vec{v}|^2 = v_x^2 + v_y^2$$

Let's plug in our expressions:

$$|\vec{v}|^2 = u_1^2 + \left(\frac{u_2 - u_1 \cos	heta}{\sin	heta}ight)^2$$

Now for some algebra gymnastics. We expand the numerator of our squared fraction:

$$|\vec{v}|^2 = u_1^2 + \frac{u_2^2 - 2u_1 u_2 \cos	heta + u_1^2 \cos^2	heta}{\sin^2	heta}$$

To combine the terms, we take $\sin^2	heta$ as the common denominator:

$$|\vec{v}|^2 = \frac{u_1^2 \sin^2	heta + u_2^2 - 2u_1 u_2 \cos	heta + u_1^2 \cos^2	heta}{\sin^2	heta}$$

Notice how we have a $u_1^2 \sin^2	heta$ and a $u_1^2 \cos^2	heta$? That's the magic of trigonometry! Using the identity $\sin^2	heta + \cos^2	heta = 1$, those two terms combine beautifully into just $u_1^2$:

$$|\vec{v}|^2 = \frac{u_1^2 + u_2^2 - 2u_1 u_2 \cos	heta}{\sin^2	heta}$$

Finally, we take the square root of the whole thing, and we arrive at our elegant expression for the speed of the load:

$$|\vec{v}| = \frac{\sqrt{u_1^2 + u_2^2 - 2u_1 u_2 \cos	heta}}{\sin	heta}$$

A perfect blend of physics constraints and vector math!

Visualized Solution

$C o n s t r ain t R e l a t i o n s hi p$

$P r o j ec t i o n o n S t r in g 1$

$P r o j ec t i o n o n S t r in g 2$

$R eso l v in g t h e V e l oc i t y V ec t or$

$A ppl y in g t h e C o n s t r ain t o n S t r in g 2$

$F in d in g v_{y}$

$M a g ni t u d e o f V e l oc i t y$

$E x p an d in g t h e S q u a r es$

$T h e F ina l S p ee d E x p r ess i o n$

The Sigma Insight: Newton's Laws of Motion

The Art of Constrained Motion

Setting the Stage

The Master Equation

Isolating the Unknowns

Final Calculation

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Constraint Relationship

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Projection on String 1

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Projection on String 2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Resolving the Velocity Vector

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Applying the Constraint on String 2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Finding vy​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Magnitude of Velocity

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Expanding the Squares

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Final Speed Expression