Deriving Equation #3: (b/2)θ=L(2θ)

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To derive equation #3 from equations #1 and #2, set them equal to each other, resulting in (b/2)θ = L(2θ). Dividing both sides leads to the cancellation of θ, simplifying the equation to b/2 = 2L. This can be rearranged to express the relationship between b and L, ultimately leading to the derived equation ∆x = ∆z b/4L. The key step is correctly dividing the right-hand sides of the equations while maintaining the equality of the left-hand sides. Understanding this process is crucial for successfully deriving equation #3.
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Homework Statement


Using equations #1 and #2 to derive equation #3

Homework Equations


#1- ∆x = (b/2)theta
#2- ∆z= L (2theta)
#3- ∆x= ∆z b/4L


The Attempt at a Solution


Now I set equations #1 and #2 equal to each other, but I'm not sure how the thetas end up cancelling or if I'm even supposed to set them equal to each other in the first place. HELP please!
 
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Just divide one equation by the other.
 
Does that mean that I divide (b/2)theta by L(2theta)?
 
Yes, you divide the right-hand sides of the two equations, and the result must equal the ratio of the left-hand sides of the same two equations.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

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