Here are the answers, forgive the lack of boxes and system boundary as these are not possible in a comment box! All parts of the physical system should be in boxes and in the system boundary, the real life inputs and outputs are not in boxes and outside the system boundary.
electrical energy > motor > Mechanical system > bucket > linear motion
2. The comment also won't properly space out the working! But here are the equations etc you should use.
VR = 10/40 x 10/15 = 1/6 (or 0.167)
SPEEDdrum = SPEEDmotor x VR = 500 x 1/6 = 83.3 rev/min = 1.39 rev/sec
C = Pi x D = 3.14 x 250 = 785mm = 0.785m
SPEEDrope = SPEEDdrum x C = 1.39 x 0.785 = 1.09 m/s
VR = 1/2 with 2 ropes
SPEEDload = SPEEDrope x VR = 1.09 x 1/2 = 0.55 m/s
Here are the answers, forgive the lack of boxes and system boundary as these are not possible in a comment box! All parts of the physical system should be in boxes and in the system boundary, the real life inputs and outputs are not in boxes and outside the system boundary.
ReplyDeleteelectrical energy > motor > Mechanical system > bucket > linear motion
2. The comment also won't properly space out the working! But here are the equations etc you should use.
VR = 10/40 x 10/15
= 1/6 (or 0.167)
SPEEDdrum = SPEEDmotor x VR
= 500 x 1/6
= 83.3 rev/min
= 1.39 rev/sec
C = Pi x D
= 3.14 x 250
= 785mm
= 0.785m
SPEEDrope = SPEEDdrum x C
= 1.39 x 0.785
= 1.09 m/s
VR = 1/2 with 2 ropes
SPEEDload = SPEEDrope x VR
= 1.09 x 1/2
= 0.55 m/s