0
19kviews
Two pulleys, one 450 mm diameter and the other 200 mm diameter are on parallel shafts 1.95 m apart are connected by a cross belt.

Find the length of the belt required and angle of contact between the belt and each pulley.

What power can be transmitted by the belt when the larger pulley rotates at 200 rev/min, if the maximum permissible tension in the belt is 1 KN, and the coefficient of friction between the belt and pulley is 0.25?

Subject: Kinematics of Machinery

Topic: Belts, Chains and Brakes

Difficulty: High

1 Answer
0
2.7kviews

Given :- $d_{1}$ = 450 mm= 0.45 m or $r_{1}$ =0.225 m; $d_{2}$ = 200 mm= 0.2 m; or $r_{2}$ =0.1 m; x =1.95 m, $N_{1}$ =200 rpm; $T_{1}$ = 1kN= 1000N ; $\mu$ =0.25

We know that speed of the belt,

$v=\frac{\pi d_{1}N_{1}}{60}=\frac{\pi\times 0.45\times200}{60}$ = 4.714 m/s.

Length of …

Create a free account to keep reading this post.

and 4 others joined a min ago.

Please log in to add an answer.