# Center of pressure

1. Jan 29, 2016

### goldfish9776

1. The problem statement, all variables and given/known data
pls refer to photo 2 , (circled part) How to transform equation in 3-20 to become equation in 3-22a ?

2. Relevant equations

3. The attempt at a solution
i have tried in photo 3 , yet can get the same form as in 3-22a . In step 2 , i divided the y_p with Fr

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2. Jan 29, 2016

### haruspex

I can see an error in your working. In the left hand term, you substituted FR=P0A (wrong), but in the right hand term you correctly used PcA.
I'll continue to study the text, but it is hard to read in places.

Edit:
I eventually figured out that the denominator in the last term of 3-22a reads $(y_c+P_0/(\rho g \sin(\theta))A$, with which I agree.
Do you get that with the above correction?

Last edited: Jan 29, 2016
3. Jan 29, 2016

### goldfish9776

no , i still gt stucked here

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4. Jan 29, 2016

### haruspex

Put the denominator in the second term back into the form PcA for now.
Take the second term in the numerator (the one with yc2; you missed out the c subscript) and shift it into the first term after the equals sign. In the numerator there, you can extract a factor yc. The other factor is PcA, so cancels with the denominator.

5. Jan 29, 2016

### goldfish9776

how to facrorise yc ? there are 3 terms there , but there are only 2 terms contain yc .... and i cant factorise PoA ...

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6. Jan 29, 2016

### haruspex

You have successfully factored out the yc. In the other factor, you have two terms each with a denominator PcA. So you can add together the numerators. I think you'll find those numerators add up to PcA.

7. Jan 29, 2016

### goldfish9776

ok , i get my yc now , how to transform the ( (rho)g sin theta ( I xxc) / PcA )into the same form as 3-22a?

8. Jan 29, 2016

### haruspex

In that term, expand the Pc in the denominator into P0+ etc. form, then divide through numerator and denominator by (rho)g sin theta.