March 17th, 2018, 01:14 AM  #1 
Newbie Joined: Dec 2017 From: australia Posts: 2 Thanks: 0  Hydrostatics Question
Hey, I have been having trouble working out why they solved the vertical force the way they did. can someone explain why they broke it down the way they did and how the sum calculates the hydrostatic force? Thanks, Swift 
March 17th, 2018, 07:47 AM  #2 
Senior Member Joined: Jun 2015 From: England Posts: 795 Thanks: 233 
Who are "they"? Is this your working or the working in the book? It is incorrect. However one mistake (luckily !?!) cancels out the other. Last edited by skipjack; March 18th, 2018 at 08:28 AM. 
March 18th, 2018, 04:02 AM  #3 
Senior Member Joined: Jun 2015 From: England Posts: 795 Thanks: 233 
As there have been so many views since yesterday, I thought I would indicate the error in the posted working. This is an error in Mathematics, not Physics. If the OP ever returns and is still interested, I will explain the Physics properly. Essentially, the working incorrectly shows that the vertical distance to the contact point  AB  is (R + Rsina) where a is the slope angle. (I have used alpha in the diagram.) Of course it should be (R + Rcosa). Now with the stated problem values this is not fatal, as a is given as 45 and sin 45 = cos 45. But this is the only value for which this is true, so for any other slope angle the working would produce the wrong values. This affects the first part of the problem, the calculation of the Horizontal force. However, the reverse may well be true in the vertical calculation, but it is not stated explicitly which trigonometric function is used and triangle 3 is isosceles. Last edited by skipjack; March 18th, 2018 at 08:28 AM. 

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