Why is Theta 2 Independent in Solving for Theta 3 in a 4-Bar Mechanism?

[EastWindBreaks]

Homework Statement

Homework Equations

The Attempt at a Solution

it seems like because theta 2 is independent, therefor, you can solve theta 3 by just using one equation from the system of equation? on a previous problem where its a 4 bar mechanism( which it didn't specify that theta 2 is independent), you have to combine the system of equation into one single equation :

and then solve it from there using some substitutions. but for this problem, it didn't do that, is it because theta 2 is independent?

 

[kuruman]

To answer the title question, $\theta_2$ is independent because you can change it at will. Once you choose a value for the independent variable $\theta_2$, the dependent variables $\theta_3$ and $d$ acquire unique values. Did I answer your question? I am not familiar with what you refer to as the "previous problem."

 

[EastWindBreaks]

To answer the title question, $\theta_2$ is independent because you can change it at will. Once you choose a value for the independent variable $\theta_2$, the dependent variables $\theta_3$ and $d$ acquire unique values. Did I answer your question? I am not familiar with what you refer to as the "previous problem."

thank you, I guess my question boils down to this:

if given a system of equations like this one below,

and if $\theta_2$ is not independent variable, then you can not simply solve $\theta_2$ by just using one of the equations, correct?

another question is that, for the example below where $\theta_2$ is independent variable:

can we solve both $\theta_3$ and d from just

?

 

[kuruman]

... and if $\theta_2$ is not independent variable, then you can not simply solve $\theta_2$ by just using one of the equations, correct?

Correct. You only have two equations which means that you cannot find $\theta_2$ in addition to $\theta_3$ and $d$ because that makes 3 unknowns.

 

[EastWindBreaks]

You only have two equations which means that you cannot find $\theta_2$ in addition to $\theta_3$ and $d$ because that makes 3 unknowns.

I am guessing you were referring to this example:

$\theta_3$ and d are unknown, one independent variable ( $\theta_2$), yet d is solved using only 1 equation from the system of equations, containing 2 unknowns on the right side of the equation. so my question is if you can do that with d, then why can't $\theta_3$ = arccos( (a*cos$\theta_2$ - d)/b) ?

 

[kuruman]

... why can't $\theta_3 = \arccos( (a*\cos\theta_2 - d)/b)$ ?

Suppose you replaced $d$ on the right side with the expression you found, $d=a*\cos \theta_2-b *\cos \theta_3$. If you do that, you get $$\theta_3 = \arccos( (a*\cos\theta_2 - a*\cos \theta_2+b *\cos \theta_3)/b)=\arccos(\cos \theta_3))=\theta_3.$$ So the answer to your question is, it can but you won't get anything new.

 

[EastWindBreaks]

Suppose you replaced $d$ on the right side with the expression you found, $d=a*\cos \theta_2-b *\cos \theta_3$. If you do that, you get $$\theta_3 = \arccos( (a*\cos\theta_2 - a*\cos \theta_2+b *\cos \theta_3)/b)=\arccos(\cos \theta_3))=\theta_3.$$ So the answer to your question is, it can but you won't get anything new.

okay, I got it, thank you!