Determine the Acceleration of Collar B as a result of Collar A

[Northbysouth]

Homework Statement

Collars A and B slide along the fixed right-angle rods and are connected by a cord of length L = 7 m. Determine the acceleration of collar B when y = 2.2 m if collar A is given a constant upward velocity v_A = 2.66 m/s. The acceleration of B is positive if to the right, negative if to the left.

I have attached an image

Homework Equations

x²+y² = l²


3. The Attempt at a Solution [/b

I found the distance x, from the base of the support to B to be 6.64529 m

I took the derivative of the Pythagorean to get:

2xx' + 2yy' = 0

Then I took the derivative again:

2x'x' + 2xx" + 2y'² + 2yy"

Solving for x" = -x'² - y'² -yy"/x

x" = -(-yy'/x)² - y'²/x

x" = -0.9057

But it says it's wrong. Where is my mistake?

 

[SammyS]

Homework Statement

Collars A and B slide along the fixed right-angle rods and are connected by a cord of length L = 7 m. Determine the acceleration of collar B when y = 2.2 m if collar A is given a constant upward velocity v_A = 2.66 m/s. The acceleration of B is positive if to the right, negative if to the left.

I have attached an image

Homework Equations

x²+y² = l²

3. The Attempt at a Solution [/b

I found the distance x, from the base of the support to B to be 6.64529 m

I took the derivative of the Pythagorean to get:

2xx' + 2yy' = 0

Then I took the derivative again:

2x'x' + 2xx" + 2y'² + 2yy"

(I'm assuming you are using the prime to mean differentiation with respect to time.)

I'm with you to here, although, for this problem, y'' = 0 .

Solving for x" = -x'² - y'² -yy"/x

This should be

x'' = (-x'² - y'² - yy'')/x

  = (-x'² - y'²)/x​

x" = -(-yy'/x)² - y'²/x

x" = -0.9057

But it says it's wrong. Where is my mistake?