Discussion on: Problem 512 | Friction
Following is a discussion on the Reviewer item titled: Problem 512 | Friction.
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how we got ph=40(w sin x)+20(w cos x)
please tell me ,how we got this ph=40(w sin x)+20(w cos x)
Re: how we got ph=40(w sin x)+20(w cos x)
In reply to how we got ph=40(w sin x)+20(w cos x) by Mohammed Ahmed
The equation you mentioned is the moment about point A.
W sin (theta) = x-component of W or Wx and
W cos (theta) = y-component of W or Wy.
Re: how we got ph=40(w sin x)+20(w cos x)
In reply to how we got ph=40(w sin x)+20(w cos x) by Mohammed Ahmed
IT'S VARIGNON'S THEOREM AS IT STATES MOMENT SUM ABOUT ANY PT. =SUM OF ALL MOMENTS DUE TO ALL FORCES =>FORCE*DIST = MOMENT
Re: Discussion on: Problem 512 | Friction
then why dont u take w cos [theta] multiply with perpendicular distance to get moment about A
if y- component is wcos[thetA] then whAt About normAl reAction
Re: Discussion on: Problem 512 | Friction
In reply to Re: Discussion on: Problem 512 | Friction by Sharvani Reddy
Analyze the forces involved when the block starts to tip over point A and you will find it why.