Engineering Mechanics 11th Ed. Chapter 4 Problem 146 The sketch is located in th

Question

Engineering Mechanics 11th Ed. Chapter 4 Problem 146 The sketch is located in that ploblem under textbookhelp. on step 5, when summing the moment about point A. (-11.25kN) X = -1.5kN/m x 3m x 1.5m -1kN/m x 3m x (3 +1.5)m - 2.5kN/m x 1.5m x (6 + 0.75)m i was wondering where the 1.5 m , (3 + 1.5)m , and the (6 +0.75)m came from. Engineering Mechanics 11th Ed. Chapter 4 Problem 146 The sketch is located in that ploblem under textbookhelp. on step 5, when summing the moment about point A. (-11.25kN) X = -1.5kN/m x 3m x 1.5m -1kN/m x 3m x (3 +1.5)m - 2.5kN/m x 1.5m x (6 + 0.75)m i was wondering where the 1.5 m , (3 + 1.5)m , and the (6 +0.75)m came from.

Explanation / Answer

it seems that they just used the centroid of the distributedload section i.e
MA = w1a*a/2 + w2*b(a+b/2)+... remember that w is kN/m so if you do w1*a you get kN/m *m =kN... and a moment is kN*m i.e. w2 b(a+b/2) is just w2*bthe force and the a+b/2 term is where the force acts... this is just a simplier way of thinking about distributedloads (of any shape) that the point-mass of a object acts at thecentriod of the object (like your center of gravity if someonmoves that they move you as a whole...) I hope this clears things up for you...

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