A Part A By what distance will the center of mass of this part move horizontally

Question

A Part A By what distance will the center of mass of this part move horizontally and vertically if the vertical bar is pivoted counterclockwise through 90 ° o make the entire part horizontal A machine part consists of a thin, uniform 4.00-kg bar that is 1.50 m long, hinged perpendicular to a similar Find the magnitude of horizontal displacement. vertical bar of mass 3.00 kg and length 1.80 m. The longer bar has a small but dense 2.00-kg ball at one end Figure 1) Isal. Submit My Answers Give U incorrect; Try Again; 5 attempts remaining Check your signs. Part B Find the direction of horizontal displacement. O to le left O to the right Submit My Answers Give Up Part C Find the magnitude of vertical displacement. Figure 1 of 1 Hinge k 50 m Ay m. 4.00 kg Submit My Answers Give U Part D 3.00 kg 1.80 m Find the direction of vertical displacement. O upward 2.00 8-Y kg O downward

Explanation / Answer

m1 = 4         x1 = -1.5/2 = -0.75 m     y1 = 0

m2 = 3kg         x2 = 0                y2 = -1.8/2 = -0.9m

m3 = 2kg         x3 = 0               y3 = -1.8 m

xcm = ((m1*x1)+(m2*x2)+(m3*x3))/(m1+m2+m3)

xcm = (-(4*0.75)+(3*0)+(2*0))/(4+3+2) = -0.333 m

ycm = ((m1*y1)+(m2*y2)+(m3*y3))/(m1+m2+m3)

ycm = ((4*0)-(3*0.9)-(2*1.8))/(4+3+2) = -0.7m

after the rod becoming horizantal

x1' = -0.75 m    y1' = 0

x2' = +0.9m      y2' = 0

x3' = 1.8m       y3' = 0

XCm' = ((m1*x1')+(m2*x2')+(m3*x3'))/(m1+m2+m3)

xcm = (-(4*0.75)+(3*0.9)+(2*1.8))/(4+3+2) = 0.367 m

ycm = ((m1*y1')+(m2*y2')+(m3*y3'))/(m1+m2+m3)

ycm = ((4*0)-(3*0)-(2*0))/(4+3+2) = 0m

part(A)

dx = xcm - xcm' = 0.333 - (-0.367) = + 0.7 m

part(B)

to the left

xcm' is positive xcm is negative

part(C)

dy = ycm - ycm' = 0.7

part(D)

y has moved from -0.7 to 0

direction is upward

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