. 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.

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

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.

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)

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.

active| newest | oldestfinalanswer.;-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

## 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 !

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: b

^{2}= a^{2}+x^{2}- 2*a*x*cos(θ)Then I complete the square:

(x +a*cos(θ))

^{2}= b^{2}- a^{2}+a^{2}*cos^{2}(θ)Then use a trig identity,

(x +a*cos(θ))

^{2}= b^{2}- a^{2}*sin^{2}(θ)Then take the square root of both sides,

x +a*cos(θ) = (b

^{2}- a^{2}*sin^{2}(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.

Like RT= Rmax-Rmin/2/#R or Rmax-Rmin/3/#R=RT

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.