Question
Download Solution PDFThe power transmission takes place in shaft rotating at 400 rpm and this rotating shaft drives another shaft at 600 rpm. The smaller pulley has the diameter of 0.5 m. The centre distance between pulleys is 4m. If the angle of contact on the smaller pulley for the open belt drive is 1.8° then calculate the diameter of larger pulley (in m) and angle of contact (in rad)?[Take sin-1 (0.03125) = 1.79°]
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFN1 = 400 rpm, N2 = 600 rpm, D2 = 0.5m, x = 4m, θ2 = 1.8°
\(\frac{{{N_2}}}{{{N_1}}} = \frac{{{D_1}}}{{{D_2}}} \Rightarrow {D_1} = {D_2}\frac{{{N_2}}}{{{N_1}}} = 0.5 \times \frac{{600}}{{400}} = 0.75\;m \)
\(\sin \alpha = \frac{{{r_1} - {r_2}}}{x} = \frac{{{D_1} - {D_2}}}{{2x}} = \frac{{0.75 - 0.5}}{{2 \times 4}} = 0.03125\)
α = 1.79°
Angle of contact or lap,
\({\rm{\theta }} = \left( {180^\circ + 2{\rm{\alpha }}} \right)\frac{{\rm{\pi }}}{{180}}{\rm{rad}} = 3.2{\rm{\;rad\;}}\)
Last updated on May 28, 2025
-> SSC JE ME Notification 2025 will be released on June 30.
-> The SSC JE Mechanical engineering application form will be available from June 30 to July 21.
-> SSC JE 2025 CBT 1 exam for Mechanical Engineering will be conducted from October 2 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.