Question

Two parallel shafts 3 m apart are to be connected by a belt running over the pulleys of diamter 500 mm and 100 mm respectively. Determine the length of belt required and the angle of contacts between belt and each pulley if it is crossed. What power can be transferred if the larger pulley rotates at 220 revolutions per minute. Given: Maximum permissible tension in the belt = 1 kN and coefficient of friction between the belt and the pulley is 0.25.

Answer 1

The crossed belt drive system t is shown in Fig. From the geometry of the system,

Length of belt = Arc length DC + Arc length FE + 2BG 

= 250 θ₁ + 50 θ₂ + 2 × 3000 cos α 

= 250 × 3.342 + 50 × 3.342 + 2 × 3000 cos 5.734° 

= 6972.5 mm. 

Max. tension in the belt = 1 kN = 1000 N. 

From rope friction formula

= 5799.6 mm/sec. 

∴ Power transmitted = (T₂ – T₁) × Velocity 

= (1000 – 433.66) × 5759.6 

= 3261884 N-mm/sec. 

= 3.261884 kW.