Problem: What diameter of pulley is needed if the desired speed is 500 rpm and the driven pulley is 6.0 cm and turns at 1725 rpm?
Solution:
_ D1 x N1 _ 6.0 cm x 1725 m- 10350 ■>2 = = rev
Problem: Determine the torque available at shafts A and B in Figure 6.14.
Solution: The first step is to determine the speeds of shafts A and B. Shaft B is the first one in the series so it will be used first.
Ti x Ni 24 T x 2'200 min 52800
The speed of shaft A:
The maximum torque will be available at only at one shaft. The maximum torque will be calculated for A and B. The maximum torque delivered by the motor is:
Solving for To
9549
9549 x PkW
9549 x 37.0
2200
The motor is producing 160.6 Nm or torque. The torque at shaft A is:
5900
The maximum torque at B is:
Tox x N1
_min rev
353,320
353,320
4400
Problem: How much force will be produced by the hydraulic system in Figure 6.15 when the diameter of the first cylinder is 5 cm, the diameter of the second cylinder is 15 cm and the force on the first cylinder is 50 kN?
Solution: Force divided by area equals pressure and pressure multiplied by area equals force.
50 kN 10,000 cm2 1,000 N 1 Pa 1 kPa n x (5 cm)2 1m2 1 kN N/m2 1,000 Pa
500,000,000
The pressure on the fluid will be 25,000 kPa. The force on the second cylinder is:
4 10,000 cm2
1 kPa 1 Pa 1,000 N 417,864,669
10,000,000
441.786 or 440 kN
Problem: How far will the second piston, in the previous problem, move when the piston in the first cylinder moves 5 cm?
Solution: Area x stroke distance = area x stroke distance
176.7
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