Center of Gravity and Centroid

Question: Determine the approximate amount of paint needed to cover the surface of the water storage tank. Assume that a liter of paint covers 2.5 m2. Also, what is the total inside volume of the tank.
[A].

27.6 liters of paint, V = 52.6 m3

[B].

20.1 liters of paint, V = 50.3 m3

[C].

26.4 liters of paint, V = 56.5 m3

[D].

25.1 liters of paint, V = 55.0 m3

Answer: Option C

Explanation:

No answer description available for this question.

Question: Determine the volume of concrete needed to construct the circular curb.
[A].

V = 1.083 m3

[B].

V = 1.309 m3

[C].

V = 1.756 m3

[D].

V = 8.67 m3

Answer: Option A

Explanation:

No answer description available for this question.

Question: Determine the distance to the centroidal axis of the beam’s cross-sectional area.
[A].

= 112.3 mm

[B].

= 125.0 mm

[C].

= 100.0 mm

[D].

= 91.7 mm

Answer: Option D

Explanation:

No answer description available for this question.

Question: Determine the distance to the centroid axis of the beam’s cross-sectional area. Neglect the size of the corner welds at A and B for the calculation.
[A].

= 75.2 mm

[B].

= 97.5 mm

[C].

= 85.9 mm

[D].

= 102.5 mm

Answer: Option C

Explanation:

No answer description available for this question.

Question: Locate the center of gravity of the homogeneous “bell-shaped” volume formed by revolving the shaded area about the y axis.
[A].

= 3.33 ft

[B].

= 2.80 ft

[C].

= 3.20 ft

[D].

= 3.00 ft

Answer: Option A

Explanation:

No answer description available for this question.

Question: Locate the center of gravity of the volume generated by revolving the shaded area about the z axis. The material is homogeneous.
[A].

= 2.80 ft

[B].

= 2.50 ft

[C].

= 2.67 ft

[D].

= 3.00 ft

Answer: Option C

Explanation:

No answer description available for this question.

Question: Locate the centroid of the shaded area.
[A].

= 0.667 m, = 2.40 m

[B].

= 0.500 m, = 2.80 m

[C].

= 0.800 m, = 2.00 m

[D].

= 0.600 m, = 2.60 m

Answer: Option C

Explanation:

No answer description available for this question.

Question: Locate the centroid of the exparabolic segment of area.
[A].

= -4a/5, = -b/4

[B].

= -3a/4, = -3b/10

[C].

= -2a/3, = -b/3

[D].

= -5a/7, = -3b/8

Answer: Option B

Explanation:

No answer description available for this question.