# Thread: calculating class 2 leverage

1. ## calculating class 2 leverage

Hi there - I'm new to this forum and hoping for some guidance from experts

leverage.jpg

I'm working on a little leverage project. A motor pulling up the end of a lever (class 2). The weight is in the middle - between a fulcrum and the end (pulled up by the motor).

The motor will be geared, with an output torque of 2kgcm. I've had a stab at calculating it begin able to lift a weight (in blue) of around 4kg based on the lever size/distances. (both distances are in cm's)

Is it right?

(I am unsure if 2kgcm motor torque translates directly into 4kg of weight)...

many thanks / simon

2. Check forum policies on homework.

I will give you a hint though: Your motor power is expressed as torque (a rotary thing). The force required to lift the load is a linear thing. How does that motor torque get converted to a linear force?

3. Adding to JB's comments on Homework, I would like to add a comment to your note on the drawing. In 40+ years of Engineering, I do not ever recall using a "guess" for anything. You may want to drop that word from your vocabulary or things break, fall down and people even die.

I am sure your project has been covered in a broader sense in the course notes and lectures. You did take notes during the lectures, didn't you?

Hard work and dedication is the only way to success. Getting someone on the Internet to answer your questions is not going to do you any good as the course gets harder and harder. You need to understand what is going on, not just provide correct answers.

4. Yes I see, you've been hammered with home work requests.

Not to worry, I'm more of a designer than an engineer and working on a simple mechanism I needed to understand this.
Made a quick prototype using wood and some tins of tomatoes to simulate + test with a 1kgcm geared motor which worked fine using 1.55kg
I needed to know the numbers to understand what geared motor to next buy and test the following stage.

The lever will be pulled up with steel wire a via small flywheel on the motor (less than 1cm dia). I understand the output of the flywheel will be 2kgcm, as opposed to the motor being that spec. The flywheel output (which lifts the lever) will be 2kgcm.

So:

2kgcm x 8cm > 4kg x 4cm = balance

I will either increase the right side, or reduce the left side of the equation.

many thanks / simon

5. Homework ---- probably though see:

http://www.engineersedge.com/calcula...e_levers_3.htm

6. Originally Posted by oem_odm
The lever will be pulled up with steel wire a via small flywheel on the motor (less than 1cm dia).
The answer to your question lies within your statement above. Wrap a wire around a 1cm dia drum. Drive that drum a 2kgcm torque. How much force will that wire produce? Look up winch calculations.

7. thanks for the link Kelly - have been using that over the past few days.

the output of the flywheel (which wire is connected to), that flywheel will have output of 2kgcm (for arguments sake). The motor (+ it's gearing) will probably be in the region of 2.8kgcm. Thanks for the note jboggs

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