Say each lift cames up to 600 ly (that's the car and a few people) The it is operated by a motor and a courterweight as

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Say Each Lift Cames Up To 600 Ly That S The Car And A Few People The It Is Operated By A Motor And A Courterweight As 1 (28.19 KiB) Viewed 43 times

Say Each Lift Cames Up To 600 Ly That S The Car And A Few People The It Is Operated By A Motor And A Courterweight As 2 (45.41 KiB) Viewed 43 times

Say Each Lift Cames Up To 600 Ly That S The Car And A Few People The It Is Operated By A Motor And A Courterweight As 3 (20.12 KiB) Viewed 43 times

Say Each Lift Cames Up To 600 Ly That S The Car And A Few People The It Is Operated By A Motor And A Courterweight As 4 (12.56 KiB) Viewed 43 times

Say each lift cames up to 600 ly (that's the car and a few people) The it is operated by a motor and a courterweight as shown in Figure 1 We want to calculate the force F the DC motor must provide to power the taccording to our cent's specifications. We wort to specity the power rating of the motor т FEN m = 200 kg FRN 200g m = 600 K 600g Figure 1. Force diagram for the litt system, adapted from 0) T is the tension in the cables, gis the acceleration dire to gravity which we'll take as 10 m F is the force exerted by the motor Here's the specification • Total lift mass is 600 kg, counterweight is 200 kg • Accoleration of 2 ms ---2 over 0.5 to a speed of 1 ms .-1 Steady spoed between the doors of 1 ms -1 for 48 Decelerate at 2 ms over 0.5 s to a stop

Speed (ms) 1 1 2 3 4 5 Time (5) Fome 2 Speed-time characteristic of the poor to lose the Mbylor We need to work out the recurements for the motor over the 5 seconds to move up on or Q1. Foroo and power required for acceleration towards We can write broneous quons one for the lit going up, one for the counter weight going When the Congo, acceleration is in the positive direction (upwards). The counterweight is cong dow words at the same rate its acceleration is nogave because of the direction downwards For the counterweight T - 300 200 For the litt T + P 600g = 600 d 2) Song, we find 71600 N and hence R-5600 How much power do we noodEnter your worning [1 mark]

Q3. Force and power required for decceleration: You need to rewrite equations (1) and (2) with change of sign for the acceleration The tonsion in the cable is Nimber N [1 mark] The force required from the motor is Number N [1 mark] The power required from the motor is Number w [1 mark] Conclusions: The greatest power is required from the motor when accelerating upwards, the smallest power is regured when slowing down Event steady speed the motor has to do work, to counter the acceleration due to granty We're ignoring friction in all of these calculations in reality we would also need to add a safety factor, so the motor is not at a capacity al the time

Q4. Imagine we now go down a floor with the same acceleration and speed charateristics What is the net power supplied through the motor? The net power supplied through the motor for the whole process (up and down) Is Number W [1 mark]