Practical application of moment calculation If the power of the reducer is 0.75KW, the output speed is 5R / min, t = 1432.5n. M. that is to say, the output torque of the shaft is 1432.5n. M (if the shaft meets the strength requirements). The reducer is directly connected with the shaft, how to calculate how much weight can be put on the chain conveyor belt with this 1432.5n. M torque?

Practical application of moment calculation If the power of the reducer is 0.75KW, the output speed is 5R / min, t = 1432.5n. M. that is to say, the output torque of the shaft is 1432.5n. M (if the shaft meets the strength requirements). The reducer is directly connected with the shaft, how to calculate how much weight can be put on the chain conveyor belt with this 1432.5n. M torque?

When p = F.V, P = 0.75KW, v = 5R / min, the speed should be converted into m / s, which lacks one condition: the distance r from the conveyor belt to the axis of the shaft; then v = 5 × 2 × 3.14r/60 = 3.14r/6; the distance r between the conveyor belt and the axis of the shaft; the distance r between the conveyor belt and the axis of the shaft is 5 × 2 × 3.14r/60 = 3.14r/6;
T (max) = f (max). L (arm of force), t = 1432.5n. M
To solve this problem, we need to assume that the working mode of conveyor belt is horizontal conveying;
F (1) = P / v = 6p / 3.14r = (calculated according to power)
F (2) = t (max) / L = 1432.5/r (calculated according to torque)
Comparing f (1) with F (2), the minimum value is the weight that can be placed on the conveyor net;
I calculated that the two values are exactly the same;
The above is my personal opinion. I haven't done physics for many years. Please correct it