The machinists square shown in the figure consists of a thin, rectangular blade

Question

The machinists square shown in the figure consists of a thin, rectangular blade connected to a thin rectangular handle.(Figure 1).

Part A)  Determine the x coordinate of the center of gravity.

Part B)  Determine the y coordinate of the center of gravity.

Part C)  When hanging from the point x=0, y=0, what angle would the long side of the blade make with the vertical?

The machinist's square shown in the figure consists of a thin, rectangular blade connected to a thin rectangular handle.(Figure 1). Determine the x coordinate of the center of gravity. Determine the y coordinate of the center of gravity. When hanging from the point x=0, y=0, what angle would the long side of the blade make with the vertical?

Explanation / Answer

A.) I shall assume the centre of gravity coincides with the centre of mass. Let's find the individual centre of masses first. The centre of mass of the upper block is in it's centre. It's coordinates are r = (11/2 i + (12 + 4/2) j) = (5.5i + 14j) cm
The centre of mass of the other block is r = ((8 + (3/2)) i + (12/2) j) = (9.5i + 6j) cm
And the total centre of mass = (m1r1 + m2r2)/(m1 + m2) = ((40)(5.5i + 14j) + (80)(9.5i + 6j))/(40 + 80) = (8.17i + 10.7j) cm

B.) wut...it can't even pivot about the origin

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