Question
Download Solution PDFThe given figure shows a quick return mechanism. The crank OA rotates clockwise uniformly. OA = 2 cm, OO' = 4 cm. The ratio of time taken for forwarding motion to that for return motion is:
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
A cutting stroke occurs when the crank rotates through an angle β and the return stroke occurs when the crank rotates through an angle α or (360° - β) in the clockwise direction.
Calculation:
Given:
Length of the crank, AO = 2 cm
Distance between the center of the crank and slotted lever, OO' = 4 cm
Consider ΔOO'A,
Therefore, α = 120°
The cutting stroke angle β,
β = 360 - α
β = 360 - 120 = 240°
Last updated on Jul 15, 2025
-> SSC JE ME Notification 2025 has been released on June 30.
-> The SSC JE Mechanical engineering application form are activated from June 30 to July 21.
-> SSC JE 2025 CBT 1 exam for Mechanical Engineering will be conducted from October 27 to 31.
-> SSC JE exam to recruit Junior Engineers in different disciplines under various departments of the Central Government.
-> The selection process of the candidates for the SSC Junior Engineer post consists of Paper I, Paper II, Document Verification, and Medical Examination.
-> Candidates who will get selected will get a salary range between Rs. 35,400/- to Rs. 1,12,400/-.
-> Candidates must refer to the SSC JE Previous Year Papers and SSC JE Civil Mock Test, SSC JE Electrical Mock Test, and SSC JE Mechanical Mock Test to understand the type of questions coming in the examination.