Question

Assume that the connecting rod and the crank of an engine forms as two sides of the triangle. If the area included in the triangle is maximum when the crank is at $60^\circ$, then the ratio of length of connecting rod to the radius of the crank is

• A. $1$
• B. $1.414$
• C. $1.732$
• D. $0.577$

Hint:

When the area is maximized, the crank and connecting rod form a right angle ($90^\circ$). This creates a standard $30^\circ-60^\circ-90^\circ$ right triangle, where the ratio of the opposite side to the adjacent side is always $\sqrt{3} \approx 1.732$.

Answer 1

The Correct Option is C

Concept: Let the slider-crank kinematic mechanism be represented by a triangle $\triangle OBC$, where:
• $OB = r$ represents the crank radius length.
• $BC = l$ represents the long connecting rod length.
• Vertex $C$ is constrained to slide along the horizontal line of stroke line $OC$. The trigonometric area of any triangle formed by two adjacent sides and their included angle $\theta$ is given by: \[ \text{Area} = \frac{1}{2} \cdot a \cdot b \cdot \sin\theta \] For a slider-crank mechanism with a fixed horizontal path, the area enclosed by the links reaches its maximum value when the angle between the crank and the connecting rod is exactly a right angle ($90^\circ$).

Step 1: Setting up the geometric conditions for maximum area.
The area of $\triangle OBC$ can be written using sides $OB$ and $BC$ along with their included interior angle $\angle OBC$: \[ \text{Area} = \frac{1}{2} \cdot r \cdot l \cdot \sin(\angle OBC) \] Since the lengths $r$ and $l$ are constant geometric parameters, the area is maximized when the sine function reaches its maximum value of 1. This occurs when the angle between the crank and the connecting rod is exactly $90^\circ$: \[ \sin(\angle OBC) = 1 \quad \Rightarrow \quad \angle OBC = 90^\circ \] This means $\triangle OBC$ is a right-angled triangle with the hypotenuse along the slider path line $OC$.

Step 2: Using trigonometry to find the ratio of lengths.
The problem states that this maximum area configuration occurs when the crank angle relative to the line of stroke is $60^\circ$: \[ \angle BOC = 60^\circ \] In our right-angled triangle $\triangle OBC$ (where $\angle OBC = 90^\circ$), we look at the tangent of the crank angle ($\angle BOC = 60^\circ$): \[ \tan(\angle BOC) = \frac{\text{Opposite Side}}{\text{Adjacent Side}} = \frac{BC}{OB} \] Substitute the lengths $BC = l$ and $OB = r$ into the tangent equation: \[ \tan(60^\circ) = \frac{l}{r} \]

Step 3: Computing the numerical value.
Recall the standard trigonometric value for $\tan(60^\circ) = \sqrt{3}$: \[ \frac{l}{r} = \sqrt{3} \approx 1.732 \] Thus, the ratio of the length of the connecting rod to the crank radius is exactly $1.732$, which matches Option (C).