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. Plot the magnitude of the couple M required to hold the system in equilibrium as a function of θ for 0 ≤ θ ≤ 2π
The magnitude of the force P applied to the piston of an engine system during one revolution of crank AB is shown in the figure below. Plot the magnitude of the couple M required to hold the system in equilibrium as a function of θ for 0 ≤ θ ≤ 2π.
Comments
7 years ago
No really. This is my final answer.
;-P
Note: I think Instructables is no longer allowing me to upload files with certain file extentions. So the attached octave script has been renamed from .m to .m.txt
7 years ago
I was immediately stymied by the word
equilibrium
how can that possibly occur except at rust-out....The whole concept is is to extract excess torque [M] from shaft [A]
Is this a gender language difficultie ?
BTW kudos to those great math mecinations being presented here !
7 years ago
I'm going to try this one more time, then I promise I'll leave it alone.
This time I'm going to try to solve this using work (or energy) balance. I start by renaming the side lengths of that triangle to some shorter names:
AB = a = constant = 90 mm = 0.09 m
BC = b = constant = 200 mm =0.20 m
AC = x is a variable that depends on θ
For the energy balance trick, I assume
M*dθ = P*dx
and I divide both sides by dθ and get
M = P*(dx/dθ)
So that derivative (dx/dθ) is what I want to find. I need to start with some equation that connects x and theta. For that I use Law of Cosines.
According to the Law of Cosines: b2 = a2 +x2 - 2*a*x*cos(θ)
Then I complete the square:
(x +a*cos(θ))2 = b2 - a2 +a2*cos2(θ)
Then use a trig identity,
(x +a*cos(θ))2 = b2 - a2*sin2(θ)
Then take the square root of both sides,
x +a*cos(θ) = (b2 - a2*sin2(θ))(1/2)
Then take the derivative of both sides, with respect to θ, and if I did that right, I get :
(dx/dθ) = a*sin(θ) ( 1 - cos(θ)*(b2/a2 - sin2(θ))(-1/2) )
Then that's pretty much it. Now I can substitute (dx/dθ) into
M = P*(dx/dθ) = P*a*sin(θ) ( 1 - cos(θ)*(b2/a2 - sin2(θ))(-1/2) )
and I think that's the answer. P is P(θ), a function of θ, and the plot M(θ) is just the product of multiplying those two functions of θ together.
Answer 7 years ago
I think they are looking for a quick tech route.
Like RT= Rmax-Rmin/2/#R or Rmax-Rmin/3/#R=RT
Answer 7 years ago
I don't see why you are using an energy balance ? There's a GRAPH showing the function, all you have to do is fit it to something, and for some reason, its piece-wise linear.
7 years ago
This attempt had a serious error in it: the wrong formula for torque. So I am going to delete it. My other answer is better, and I think reading both would be confusing.
7 years ago
Lemme try this again...
Point B traces out a circle with A the center.
I choose A as the origin, and the point B can be written in terms of its x y components as:
Bvector = (AB)*cos(theta)*i + (AB)*sin(theta)*j
And the moment Mvector is pointing into the page, so
Mvector = -M*k
where {i,j,k} are unit vectors in the {x,y,z} directions. Mvector is the negative z direction, pointing into the page. M is the magnitude of Mvector. (AB) = 90 mm = the length of AB
The moment Mvector is the cross product of Fvector, the force at pointB, crossed with Bvector
Mvector = Fvector x Bvector = -M*k = (Fx*i + Fy*j) x ( (AB)*cos(theta)*i + (AB)*sin(theta)*j)
-M*k = Fx*(AB)*sin(theta)*k - Fy*(AB)*cos(theta)*k
Divide both sides by -M
k = -(1/M)*Fx*(AB)*sin(theta)*k + (1/M)*Fy*(AB)*cos(theta)*k
And I think that has the solution: Fx= -(M/(AB))*sin(theta), Fy= (M/(AB))*cos(theta)
Next, I am guessing that rod BC just sort of magically connects the x component of Fvector and x-component of Pvector, which is all in the x direction. So that the magnitude of Pvector and the magnitude of the x-component of Fvector are equal to each other, like so:
(M/(AB))*sin(theta) = (1/(AB))*M*sin(theta) = P
I divide both sides by (1/(AB))*sin(theta) and get:
M = P*(AB)/(sin(theta))
Now I have an expression for M as a function of theta. Since P is a function of theta, and sin(theta) is a function of theta, and (AB)=90 mm is a constant.
Notice M(theta) equals infinity when theta=0 or theta=pi. I hope that's not a problem.
Also curiously, I never used the length of that other rod, BC, which is kind of weird, but like I was saying before: This is just a guess. I am not totally confident this is the right answer.
7 years ago
M=PK (K=coefenciency)
7 years ago
Good Luck!