One could use any point, but I'll set the origin at the left end of the horizontal bar. "Weight" each mass by its horizontal or vertical distance from the origin and divide by the total mass.

total mass M = (4.00 + 3.00 + 2.00) kg = 9.00 kg

Bent:

x_cm = (4.00*1.50/2 + 3.00*1.50 + 2.00*1.50)kg·m / 9.00kg = 1.17 m

y_cm = (0 - 3.00*1.80/2 - 2.00*1.80)kg·m / 9.00kg = -0.700 m

Straight:

x'_cm = (4.00*1.50/2 + 3.00*(1.50 + 1.80/2) + 2.00*(1.50 + 1.80))kg·m / 9.00kg

x'_cm = 1.87 m

y'_cm = 0 (by observation)

Δx = (1.87 - 1.17) m = 0.700 m → to the right

Δy = 0.700 m → up

Hope this helps!

# Physics bars?

A machine part consists of a thin, uniform 4.00-kg bar that is 1.50 m long, hinged perpendicular to a similar 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)

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 ∘ to make the entire part horizontal?

Find the magnitude of horizontal displacement.

b)Find the direction of horizontal displacement.

c)Find the magnitude of vertical displacement.

d)Find the direction of vertical displacement.

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 ∘ to make the entire part horizontal?

Find the magnitude of horizontal displacement.

b)Find the direction of horizontal displacement.

c)Find the magnitude of vertical displacement.

d)Find the direction of vertical displacement.

## 1 Answers

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