Machines B/C

[CPScienceDude]

I built a prototype with a meter stick as the beam. Would this be a good material for the event as well, or will I need to use metal? From the rules, I understand the maximum weight the beam should bear is 800 grams. Will a wooden beam be fine for carrying this weight?

I’m using a wooden yard stick from Home Depot and it’s working great.

[JoeyC]

It depends on what type of setup you have and how tough your meter stick is. Check if it can hold 2 pounds (approximately 800 grams). If it can't, use a different material. I use wood (I think 2 by 1s but I'm not sure) - which is notably thicker than a meter stick.

[VansBuilders]

Thank you for your help!

[sciolymachine]

Can someone please help me get started on my lever designs? I’ve made two prototypes, both of which do not work as intended. I am in division C, and my question is do I need counterweights to balance the class 2 or class 3 lever? And also how do I ensure my levers are in equilibrium and will be precise in measuring?
Thanks

[JoeyC]

Yes, you will need something to counterweight the class 2/3 level to ensure that it's center of gravity is at the fulcrum - otherwise you'll be dealing with a level which is already unbalanced at the start, and therefore unable to be used effectively as a scale (though you could if you needed to by just raw accumulation of data).

Basically, you've probably overlooked (as I did) that your level is not massless, and so just put the fulcrum at the end of a stick.
An easy solution to this is to just move it so that you put the fulcrum in the center of your level (it'll look like a class 1 level now) but then use it as a class 2 or 3 level by placing all the effort and loads on the same side.

[sciolymachine]

Thanks for that tip! I have another question though, in what way would you connect the two levers? Would you put them side by side and how would you calculate the ratios with the two levers?
Thanks

[JoeyC]

I just use string to connect them, along with most people (I think).

You could theoretically put them side to side, but then you'd receive an extra horizontal component of force you don't want to have to deal with; just attach one above the other.

Calculating ratios is best done empirically; don't use the formulas, just gather test data and make an estimating formula off of that with linear regression analysis - there's no good way to calculate all the factors necessary to get the theoretical equations to work in our situation.

[waterboy]

I have found many examples of questions asking for the IMA of an inclined plane with only the angle measurement. How would you solve this?

Ex: What is the IMA of an inclined plane with an angle of 35 degrees?

[Umaroth]

I have found many examples of questions asking for the IMA of an inclined plane with only the angle measurement. How would you solve this?

Ex: What is the IMA of an inclined plane with an angle of 35 degrees?

The IMA of an inclined plane is csc(θ) or 1/sin(θ), so plug that into a calculator and you get 1.743. This is because sine is the ratio of the leg opposite to the angle, which is the height, to the hypotenuse, which is the length of the plane. This would give you h/L. Since what you have probably learned is that it is L/h, you would take the reciprocal of the sine.

[waterboy]

I have found many examples of questions asking for the IMA of an inclined plane with only the angle measurement. How would you solve this?

Ex: What is the IMA of an inclined plane with an angle of 35 degrees?

The IMA of an inclined plane is csc(θ) or 1/sin(θ), so plug that into a calculator and you get 1.743. This is because sine is the ratio of the leg opposite to the angle, which is the height, to the hypotenuse, which is the length of the plane. This would give you h/L. Since what you have probably learned is that it is L/h, you would take the reciprocal of the sine.

Thanks. That helps a lot!

Also, what if the angle is 45°? If you put that in the calculator you would get 1.414, but since the legs of the inclined plane are opposite two 45° angles shouldn't the length of the legs be the same causing the IMA to be 1?