
#1
Nov2307, 08:52 AM

P: 3

I was having trouble calculating the equilibrium of this object. I thought I had to solve it using the scalar cross product...but i'm not certain I am on the right track. Can some one help me? This is the question.
A carpenter's square has the shape of an L, where d1 = 19.0 cm, d2 = 6.00 cm, d3 = 6.00 cm, d4 = 11.0 cm. Locate its center of gravity. (Hint: Take (x, y) = (0, 0) at the intersection of d1 and d4) 



#2
Nov2307, 09:36 AM

P: 197

Can you please draw this carpenter.I'll try to help you.
Like this? 



#3
Nov2407, 06:24 PM

P: 3

Sort of, actually d3 is connected to d1, and d1 is the longer side of the carpenter square. The picture looks more like this...




#4
Nov2407, 06:57 PM

Sci Advisor
HW Helper
PF Gold
P: 5,966

Equilibrium question.
Rather than using cross products, break up the shape into 2 rectangles, and sum moment areas about each axis separately , where the moment of an area is the area times the perpendicular distance from its centroid to the axis, and where
(A1)x1 + A2(x2) = (A1 + A2)(X_), etc., (which you are familiar with?). 



#5
Nov2407, 08:23 PM

P: 3

Well I was using the equation ((m1)(x1)+(m2)(x2))/(m1+m2)= but i keep getting the wrong answer...using this method how do i find y1, y2, and x1, x2. I know it seems like a real fundamental question but I am still having trouble.




#6
Nov2407, 09:40 PM

Sci Advisor
HW Helper
PF Gold
P: 5,966





#7
Nov2407, 11:23 PM

P: 197

Centre of mass is between two centres of mass,so in the red line.Where the red lines coincide,there is the centre of mass of the carpenter.



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