Consider a frame that is made up of two thin massless rods AB and AC as shown in the figure. A verti

Q: Why don't we use the oblique resolution formula $\mathbf{P} = \mathbf{P}_{AB} + \mathbf{P}_{AC}$?

While the wording 'resolved parallel to the arms' can be slightly ambiguous, in standard mechanics problems unless specified to resolve 'into' two specific non-orthogonal components that sum to the original vector, 'resolved component along' implies the standard orthogonal projection (dot product).

Q: What is the purpose of giving $\sin(110^\circ)$ and $\cos(110^\circ)$?

They are distractors! The angle between the downward vertical and $AB$ is $110^\circ$. The examiner provided these values to tempt you into using the complex oblique resolution formula. It's a test of your confidence in the basics.

Q: How did we get the angle $35^\circ$?

The diagram shows the angle from the *upward* vertical to $AC$ is $145^\circ$. Since force $\mathbf{P}$ points straight down (along the *downward* vertical), the angle between them is $180^\circ - 145^\circ = 35^\circ$.

Q: Does the length of the rods matter?

No, the length of the rods does not affect the resolution of the force vector at point $A$. We only care about the direction (angle) of the rods.

Q: What if the question asked for the component along $AB$?

You would find the angle between the downward vertical and $AB$. The upward vertical to $AB$ is $70^\circ$, so the downward vertical to $AB$ is $180^\circ - 70^\circ = 110^\circ$. The component would be $100 \cos(110^\circ) = 100 \times (-0.342) = -34.2\text{ N}$.

Question

Consider a frame that is made up of two thin massless rods $AB$ and $AC$ as shown in the figure. A vertical force $\mathbf{P}$ of magnitude $100	ext{ N}$ is applied at point $A$ of the frame.
Suppose the force is $\mathbf{P}$ resolved parallel to the arms $AB$ and $AC$ of the frame. The magnitude of the resolved component along the arm $AC$ is $x	ext{ N}$. The value of $x$, to the nearest integer, is ......... .
[Given, $\sin(35^\circ) = 0.573$, $\cos(35^\circ) = 0.819$, $\sin(110^\circ) = 0.939$, $\cos(110^\circ) = -0.342$]

Sigma Percentile · Accepted Answer

### The Setup: A Frame and a Force

Imagine you are looking at a rigid mechanical frame attached to a vertical wall. The frame consists of two thin, massless rods, $AB$ and $AC$, meeting at a common joint $A$. A heavy load is hung from point $A$, exerting a purely vertical downward force $\mathbf{P}$ of $100	ext{ N}$.

Our objective is to find the **resolved component** of this downward force along the specific direction of rod $AC$.

### The Core Concept: Resolving a Vector

In physics, when we are asked to find the "resolved component" of a vector along a single specific line, we are almost always talking about the **orthogonal projection** (or the dot product).

To find this projection, we only need two pieces of information:
1. The magnitude of the original vector ($P = 100	ext{ N}$).
2. The angle $	heta$ between the vector and the line of interest.

The formula is beautifully simple:
$$ P_{AC} = P \cos 	heta $$

Let's find that angle $	heta$. The diagram tells us that the angle from the **upward vertical** to rod $AC$ is $145^\circ$. However, our force $\mathbf{P}$ points straight down, along the **downward vertical**.

Since the upward and downward verticals form a straight $180^\circ$ line, the angle between the downward force $\mathbf{P}$ and rod $AC$ is simply:
$$ 	heta = 180^\circ - 145^\circ = 35^\circ $$

### The Trap: Beware of Distractors

Before we calculate the final answer, let's pause and look at the given values in the question. We are given $\sin(35^\circ)$ and $\cos(35^\circ)$, which makes sense. But why on earth did the examiner give us $\sin(110^\circ)$ and $\cos(110^\circ)$?

This is a classic JEE trap! The angle between the downward vertical and rod $AB$ happens to be $110^\circ$. The examiner provided these values to bait you into using the complex **oblique resolution** formula.

If you mistakenly assumed that $\mathbf{P}$ must be resolved *into* two non-orthogonal components $\mathbf{P}_{AB}$ and $\mathbf{P}_{AC}$ such that their vector sum equals $\mathbf{P}$, you would use the sine rule on the vector triangle, leading to a much larger, incorrect answer of $\approx 164	ext{ N}$.

Always trust the simplest interpretation of the language unless specified otherwise. "Resolved component along an arm" means orthogonal projection.

### The Final Calculation

Now, let's execute our simple, correct plan. We substitute our values into the projection formula:

$$ P_{AC} = 100 \cos(35^\circ) $$

Using the given value $\cos(35^\circ) = 0.819$:

$$ P_{AC} = 100 	imes 0.819 = 81.9	ext{ N} $$

The question asks for the value to the nearest integer. Rounding $81.9$ gives us our final, elegant answer:

$$ \mathbf{82} $$

Enter Numerical Value:

Visualized Solution

$System Setup$

$Objective$

$Finding the Angle θ$

$Component Equation$

$Substitution & Calculation$

$Final Answer$

$Beware of Distractors$

The Sigma Insight: Vector Addition, Subtraction, and Resolution

The Setup

The Core Concept

The Trap

The Final Calculation

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Enter Numerical Value:

Visualized Solution

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

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

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

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

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

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

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