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### Re: Simple Machines B/Compound Machines C

Posted: January 25th, 2015, 2:59 pm
I've done simple machines in middle school, but does anyone knew how to calculate the ratio between 2 masses with the compound lever? My problem is that the mass of the arm of the class 2 lever complicates the equation(l1*w1 = l2*w2) without knowing at least one mass.

### Re: Simple Machines B/Compound Machines C

Posted: January 25th, 2015, 5:09 pm
You do not need to know any of the weights. You are finding the ratio of one to another. d1wt1 = d2 wt2 when they are in balance, so the ratios are really the relationship of d1 and d2

### Re: Simple Machines B/Compound Machines C

Posted: January 25th, 2015, 6:12 pm
Need help. How to find ratio of mass when you have 2 unknown masses?   ### Re: Simple Machines B/Compound Machines C

Posted: January 25th, 2015, 6:47 pm
I understand that, but the problem is that the ratio between d1 and d2 is that the mass of the bar prevents finding an equal ratio. I've tried tests with 2 known masses, and I found measurements to create equilibrium with 2 200g masses, but not 2 20g masses. Does anybody else know how to figure out the ratios? Thanks.

### Re: Simple Machines B/Compound Machines C

Posted: January 25th, 2015, 8:53 pm
I understand that, but the problem is that the ratio between d1 and d2 is that the mass of the bar prevents finding an equal ratio. I've tried tests with 2 known masses, and I found measurements to create equilibrium with 2 200g masses, but not 2 20g masses. Does anybody else know how to figure out the ratios? Thanks.
The mass of the bar shouldn't prevent you from your results. Is it at equilibrium before putting masses on? If not, then get it there! Use a counterbalancing mass.

### Re: Simple Machines B/Compound Machines C

Posted: January 26th, 2015, 2:41 pm

You do not require a mass to determine the ratio. Instead, you can derive the ratio by setting the torques of each side of the lever equal to each other such that τ(1)=τ(2). Then, rmgsinΘ=rmgsinΘ, and simplification yields r(1)m(1)=r(2)m(2), which can be rearranged to r(1)/r(2)=m(2)/m(1), where r(1) is the distance mass A is away from the fulcrum and r(2) is the distance mass B is away from the fulcrum. Simplifying r(1)/r(2) so the denominator equals 1 will yield the ratio m(2):m(1), and no further calculation is necessary.

### Re: Simple Machines B/Compound Machines C

Posted: January 26th, 2015, 4:09 pm

You do not require a mass to determine the ratio. Instead, you can derive the ratio by setting the torques of each side of the lever equal to each other such that τ(1)=τ(2). Then, rmgsinΘ=rmgsinΘ, and simplification yields r(1)m(1)=r(2)m(2), which can be rearranged to r(1)/r(2)=m(2)/m(1), where r(1) is the distance mass A is away from the fulcrum and r(2) is the distance mass B is away from the fulcrum. Simplifying r(1)/r(2) so the denominator equals 1 will yield the ratio m(2):m(1), and no further calculation is necessary.
Expanding on what you've said for compound machines.(And apologies for ms paint)
Let's say that we have a mass Y, and we're trying to figure out the mass R. This pictured lever are at equilibrium. As stated before, the main equation of levers is m1d1=m2d2.

The left lever is pushing upwards on the right lever at the point of contact.If we look at the left lever alone, what mass would need to be on the other side of the lever to have it at equilibrium? From the equation, we get m=Y*d1/d2.
The right lever is pushing downwards on the left lever at the point of contact. If we look at the right lever alone, what "mass" would need to be pushing upwards on the right lever? From the equation, we get n=R*d4/d3
Now, these two "masses" are equal (this comes from newtons third law, if we simply turn them into forces).so, we get that Y*d1/d2=R*d4/d3. Solving, the ratio Y/R=d4*d2/(d3*d1).
Now the best part of this equation is that d2 and d3 can be measured in advance, and you simply have to plug in your distances.
I'm pretty tired right now (school and stuff) so please correct me if I made a mistake.

### Re: Simple Machines B/Compound Machines C

Posted: January 27th, 2015, 5:48 pm
For all of you in Simple Machines, what are some of the harder problems you've encountered? It seems that the rules make the event so limited in terms of scope, and I'm wondering what I should study.

### Re: Simple Machines B/Compound Machines C

Posted: January 27th, 2015, 6:23 pm
For all of you in Simple Machines, what are some of the harder problems you've encountered? It seems that the rules make the event so limited in terms of scope, and I'm wondering what I should study.
From my experience last year in simple machines:
Being able to take an object and figure out ALL simple machines that are in it.
For example, take a bicycle: you have the wheels (wheel/axle), the gear chains (pulley), lever (handlebars), etc.
Make sure that you are able to solve basic questions about Work, Energy, forces, etc.
Also history of simple machines. Make sure to read up on that.

### Re: Simple Machines B/Compound Machines C

Posted: January 28th, 2015, 4:19 am
For all of you in Simple Machines, what are some of the harder problems you've encountered? It seems that the rules make the event so limited in terms of scope, and I'm wondering what I should study.
From my experience last year in simple machines:
Being able to take an object and figure out ALL simple machines that are in it.
For example, take a bicycle: you have the wheels (wheel/axle), the gear chains (pulley), lever (handlebars), etc.
Make sure that you are able to solve basic questions about Work, Energy, forces, etc.
Also history of simple machines. Make sure to read up on that.
At Dodgen Invitational we had some very difficult problems involving levers with multiple balance points and masses of parts of the machine. The hardest I've seen in Div B are slightly under the level of Fizix's Div C test on the test exchange.