Question

The crank radius and connecting rod length of an IC engine are 250 mm and 1000 mm, respectively. If the crank turns 100° from the head dead centre and the net force acting on the piston along its direction of motion is 35 kN, the turning moment of the crank shaft at that instant in kN m is _____. \textit{[Round off to two decimal places]}

Hint:

The turning moment (torque) is the product of the force, the crank radius, and the sine of the angle between the crank and the connecting rod.

Answer 1

Correct Answer: 8

The turning moment (torque) can be calculated using the formula:
\[ T = F \times r \times \sin(\theta) \] where:
- \( F = 35 \, \text{kN} \) is the net force acting on the piston,
- \( r = 250 \, \text{mm} = 0.25 \, \text{m} \) is the crank radius,
- \( \theta = 100^\circ \) is the angle between the connecting rod and the crank.
Substitute the values into the formula:
\[ T = 35 \times 0.25 \times \sin(100^\circ) \] \[ T = 35 \times 0.25 \times 0.9848 = 8.00 \, \text{kN m}. \] Thus, the turning moment of the crank shaft is approximately \( \boxed{8.00} \, \text{kN m} \) (rounded to two decimal places).