# Lifting and lowering and hold

hi ya'all

i am not in technical and not a student.

just wanna do a small project on my own for my small farm.
'
what i want to do :

1. a lifting device to lift about 100 kg up - about 350 mm

2. then it will lower it down and the hold it in position ( need to press at this point,means exert the force of the motor here - about 190 kgf)

speed of travel is about 3 to 5 seconds.

after checking the net for standard parts and so on,i will be using timing belt with timing pulley.

so,may i know what size of motor and gear ratio to use ?

thanks all !!

BvU
Homework Helper
Hello TM, welcome to PF !

Well I do hope you are practical, then. I can give you a few physics equations but your solution depends a great deal more on the practical implementation .

To raise 100 kg against gravity 10 m/s2 requires a force of at least 100 kgf or 1000 Newton
To lift it by 1 m requires 1000 Newton x 1 m = 1000 Joule, so to lift it by 0.35 m requires 350 J.
If that has to be done in 3.5 seconds, that is a power of 350 J / 3.5 sec = 100 Watt.

To accelerate 100 kg from 0 to 0.1 m/s in, say 0.5 sec, requires 100 x 0.1 / 0.5 = 20 Newton.
Lifting and accelerating is less than what's required for pressing, so that 195 kgf is the determining quantity.

You mention timing belt and pulley, but that yields torque, rotation force. To convert to up and down motion you need a ball screw or a rack and pinion. They then determine a gear ratio. Friction from all those items must be overcome as well.

To me the physics look simple, I hope that for you the practical side is simple !

CWatters
Homework Helper
Gold Member
Perhaps google for Motorised car jack?

Might not be fast enough?

100 watt . isn't it too small ?

my neighbor built one ti lift 50 kg and he uses a 0.37 kw motor and it struggles. my system is 100 kg...

i will try to provide a sketch and see what is the suggestions.

thanks !!

ok...here's what i can 'sketch''...lol..

#### Attachments

• Sk 1.pdf
4.4 KB · Views: 118
BvU
Homework Helper
The 0.37 kW motor struggling with 50 kg goes to show a good neighbor is worth more than a far-away physicist !

Apparently the power required is a lot more than the net physics calculation, which I stil think is correct.

Can't say i really understand your picture. Does the timing belt really go around a corner ? What makes it lift the load ? Where and how does the pressing down take place ? Frame plate is fixed to the ground and mounting plate moves up and down ?

maybe this will provide a clearer picture...sorry mate.

Frame fixed to the ground. mounting plate moves up and down and carry the load and press at it's lower position.

#### Attachments

• Sk 1.pdf
4.5 KB · Views: 121
BvU
Homework Helper
Somewhat clearer, indeed. So the timing belt is attached to the mounting plate.
Is the load also attached to the mounting plate ? How does it get lifted ?
What is being pressed ? in what direction ? What's holding it in place on the other side ?

ok...here goes. again..lol

timing belt and load - yes attached to the mounting plate which will move up and down

then the load will rest on my 'product' and then press down ,say a few mm. it's a little only .

i got a friend somehow to try to calculate for me, I need 0.6 kw motor ... far off from 100 watt, right ?

BvU
Homework Helper
If 0.37 kW struggles with 50 kg, then 0.6 kW will struggle even harder with 100 kg ?
My guess is that the physical power needed for lifting isn't really the bottleneck for this device.

Is it clear to you that the angle between the lower end of the belt and the ground should remain 90 degrees ?
The pulling force from the motor on the belt has one vertical component and one horizontal component.
When that angle is,e.g. 30 degrees, only half the pulling force is doing the lifting.

Same for pressing down; now the high end of the belt should remain reasonably vertical.

hahaha...sorry.

the sketch is drawn in a way that the belt can be seen !! it's vertical / parallel to the up -down motion.