a machine must be driven at 100 r/s from a belt pulley of 600 mm diameter revolving at 72 r/s . the maximum tensile force on the belt must not exceed 3000n .assume that the ratio of the tensile force is 2,5:1
To determine the size of the pulley required to drive the machine at 100 revolutions per second (r/s), we can use the concept of belt pulley ratios.
The pulley ratio can be calculated as the ratio between the rotational speeds of two pulleys. In this case, we have a speed ratio of 100 r/s for the driven machine and 72 r/s for the driving pulley.
Let's call the diameter of the driven machine pulley D1 and the diameter of the driving pulley D2. The equation that relates the pulley sizes to the speed ratio is:
Speed Ratio = (n1 / n2) = (D2 / D1)
Given D2 = 600 mm and n1 = 72 r/s, we can solve for D1:
D1 = (D2 * n1) / n2
= (600 mm * 72 r/s) / 100 r/s
= 43200 mm^2/s / 100 r/s
= 432 mm
So, the diameter of the driven machine pulley (D1) should be approximately 432 mm to achieve the desired speed of 100 r/s.
Now, let's consider the maximum tensile force on the belt. We are given that the maximum force should not exceed 3000 N, and the ratio of the tensile force is given as 2.5:1.
We can assume that the larger pulley (driving pulley) experiences a force of 2.5x, while the smaller pulley (driven pulley) experiences a force of x. Therefore:
Total Force = 2.5x + x = 3000 N
Simplifying the equation, we have:
3.5x = 3000 N
Solving for x, we find:
x = 3000 N / 3.5
≈ 857.14 N
The tensile force on the smaller pulley (driven pulley) is approximately 857.14 N, and the tensile force on the larger pulley (driving pulley) is 2.5 times that value:
2.5 * 857.14 N ≈ 2142.86 N
Therefore, the maximum tensile force on the belt would be within the desired limit of 3000 N when the driving pulley has a force of approximately 2142.86 N and the driven pulley has a force of approximately 857.14 N.