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Consider the triangle formed by the connecting rod and the crank of an IC engine as the two sides of the triangle. If the maximum area of this triangle occurs when the crank angle is 75°, the ratio of connecting rod length to crank radius is
- a)5
- b)4
- c)3.73
- d)3
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
Consider the triangle formed by the connecting rod and the crank of an...
Area will be max imum when A = 90°
i.e. PQR is a righ tan gled traingle.
∴ Ratio of connecting rot length to crank radius,
Most Upvoted Answer
Consider the triangle formed by the connecting rod and the crank of an...
Solution:
Given, maximum area of the triangle occurs when the crank angle is 75°.
Let the length of connecting rod be l and radius of crank be r.
Let us assume that the connecting rod makes an angle θ with the horizontal.
Then, the position of the piston can be given by
x = r cos θ + l cos (θ + φ)
y = r sin θ + l sin (θ + φ)
where φ is the angle made by the connecting rod with the line perpendicular to the crank.
Area of the triangle is given by
A = 0.5 × base × height
where base is the length of the crank and height is the distance between the crank and the connecting rod.
∴ A = 0.5 × r × y
= 0.5 × r × (r sin θ + l sin (θ + φ))
= 0.5r² sin θ + 0.5rl sin (θ + φ)
Differentiating A w.r.t θ, we get
dA/dθ = 0.5r² cos θ + 0.5rl cos (θ + φ)
For maximum area, dA/dθ = 0
⇒ 0.5r² cos 75 + 0.5rl cos (75 + φ) = 0
⇒ r cos 75 + l cos (75 + φ) = 0.7266r
Dividing throughout by r cos 75, we get
1 + l/r cos φ = 0.7266/ cos 75
⇒ 1 + l/r cos φ = 3.73
⇒ l/r cos φ = 3.73 – 1
⇒ l/r cos φ = 2.73
⇒ l/r = 2.73/cos φ
As cos φ is always less than or equal to 1,
⇒ l/r ≥ 2.73
Therefore, the ratio of connecting rod length to crank radius is 3.73.
Hence, option (c) is the correct answer.
Given, maximum area of the triangle occurs when the crank angle is 75°.
Let the length of connecting rod be l and radius of crank be r.
Let us assume that the connecting rod makes an angle θ with the horizontal.
Then, the position of the piston can be given by
x = r cos θ + l cos (θ + φ)
y = r sin θ + l sin (θ + φ)
where φ is the angle made by the connecting rod with the line perpendicular to the crank.
Area of the triangle is given by
A = 0.5 × base × height
where base is the length of the crank and height is the distance between the crank and the connecting rod.
∴ A = 0.5 × r × y
= 0.5 × r × (r sin θ + l sin (θ + φ))
= 0.5r² sin θ + 0.5rl sin (θ + φ)
Differentiating A w.r.t θ, we get
dA/dθ = 0.5r² cos θ + 0.5rl cos (θ + φ)
For maximum area, dA/dθ = 0
⇒ 0.5r² cos 75 + 0.5rl cos (75 + φ) = 0
⇒ r cos 75 + l cos (75 + φ) = 0.7266r
Dividing throughout by r cos 75, we get
1 + l/r cos φ = 0.7266/ cos 75
⇒ 1 + l/r cos φ = 3.73
⇒ l/r cos φ = 3.73 – 1
⇒ l/r cos φ = 2.73
⇒ l/r = 2.73/cos φ
As cos φ is always less than or equal to 1,
⇒ l/r ≥ 2.73
Therefore, the ratio of connecting rod length to crank radius is 3.73.
Hence, option (c) is the correct answer.
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Consider the triangle formed by the connecting rod and the crank of an IC engine as the two sides of the triangle. If the maximum area of this triangle occurs when the crank angle is 75°, the ratio of connecting rod length to crank radius isa)5b)4c)3.73d)3Correct answer is option 'C'. Can you explain this answer?
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