Center of Mass Center of Gravity

Body equilibrium & CentroidsENGR 103Lecture 20

http://studypoints.blogspot.com/2011/05/equilibrium-of-industry-under-perfect_545.html

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Review from last time:

25158180 lb100 lbNeglect the weight of the beam.

What happens if we cant neglect the weight of beam?

25158180 lb100 lbWeight of the beam is 400 lb.

DefinitionsCentroidThe average position of shape distribution, where a shape can consist of a single shape or distribution of shapes. (Line, area, or volume)

Center of GravityThe average position of weight distribution

Center of MassThe average position of mass distribution

In Class Example #1We will use a 1-D center of gravity example to develop a definition of average position that will be applied to a variety of different distributions. Consider a tray with wine and pasta: x represents the average position of the weight distribution.

Q: What are F and x ?

A: Use equivalent force system concepts to determine these.

Let:

Then:

Generalization for an arbitrary number of objects

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For average x position of weight Wi:

For average x position of mass mi :

For average y position of weight Wi :

Etc .....Generalization for other distributions

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For average x position of areas Ai :

For average x position of volumes Vi :

For average y position of areas Ai :

Etc .....Generalization for other distributions

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Centroids Center of Area of ShapeArea found by summing smaller regular areas:

First moment of area found by summing product of center distance and smaller areas:

Centroid (center) of the composite shape found by

In Class Example #2Calculate the total weight and center of gravity of the pipe assembly shown. Each piece of pipe is 6.00 inches in diameter and 60.0 inches long and weighs 2 lbf per foot.

In Class Example #2Calculate the total weight and center of gravity of the pipe assembly shown. Each piece of pipe is 6.00 inches in diameter and 60.0 inches long and weighs 2 lbf per foot.

A (reference point)60606612

In Class Example #2

In Class Example #2

A (reference point)60606612

What happens if we cant neglect the weight of beam?

25158180 lbf100 lbfWeight of the beam is 400 lbfFor a symmetrical beam with constant weight distribution, the weight of the beam can simply be represented as a single force vector located at the center of the beam. Solve for the reaction forces at the roller and the pin.

Center of GravityWhy is the CG so useful? How is it used?An objects weight can be represented as a single force vector at one position the center of gravity

W

The 2nd is a lot easier to deal with mathematically16

What happens if we cant neglect the weight of beam?2578180 lbf100 lbf8400 lbfAyByBxAB

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Now solve for the reactions:

Practice Problem #1

Find the center of gravity and the total weight of this plywood piece in the x and y directions. The density of the plywood is 20.0 lb/ft3 and is 1 thick.

Step 1: Divide object into shapes you can easily find area and center.

Area 1Area 2Assumption: The plywood has a uniform thickness and density.

Step 2: Calculate area of each shape.

Area 1Area 2Area 1 = 1/2bh = (2 ft)(2 ft -7.5 in) *watch units! =1.375 sq ft

Area 2 = bh = (2 ft)(7.5 in) = 1.25 sq ft

Total Area = A1 + A2 = 1.375 sq ft + 1.25 sq ft = 2.625 sq ft

Step 3: Find Xc of each shape measured back to origin.

Area 1Area 2XC,1 = 1/3(b) = 1/3(2 ft) = 0.667 ft

XC,2 = (b) = (2 ft) = 1 ftShapeArea (sq ft)Xc (ft)Yc (ft)#11.375#21.25

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Step 4: Find Yc of each shape measured back to origin.

Area 1Area 2YC,1 = 1/3(h) + height of area 2 = 1/3(2 ft -7.5 in) + 7.5 in = 1.083 ft

YC,2 = (h) = (7.5 in) = 0.313 ftShapeArea (sq ft)Xc (ft)Yc (ft)#11.3750.667#21.251

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Step 5: Calculate the Xc,t and the Yc,t

Area 1Area 2Xc,t = (Ai*Xc,i)/(Ai) = [(1.37 ft2*0.667 ft) + (1.25 ft2*1 ft)]/(1.375 ft2+1.25 ft2) = 0.826 ft = 9.9ShapeArea (sq ft)Xc (ft)Yc (ft)#11.3750.6671.083#21.2510.313

Yc,t = (Ai*Yc,i)/(Ai) = [(1.37 ft2*1.083 ft)+(1.25 ft2*0.313 ft)]/(1.375 ft2+1.25 ft2) = 0.714 ft = 8.6

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Step 6: Calculate the Total Weight

Area 1Area 2ShapeArea (sq ft)Xc (ft)Yc (ft)#11.3750.6671.083#21.2510.313

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Homework Due Fri 2/27/15

Basic homework submission format (20 pts). Neatness (10 pts). Legitimate, concerted effort to solve all the problems (20 pts)

Drawing Plates Iso 5 & Plate Number 16

Pipe BC, BD, and BA have weights of 20 lbf, 10 lbf, and 25 lbf, respectively, and the diameter of each pipe is 30 cm. Assume the small gray supporting brace has negligible weight. Find the x and y location of the center of gravity of the pipe structure. For the pipe structure to balance on point A (the left side of the base), what must Q be equal to?

RubricPointsCorrect x CG of each pipe (3 pipes)6Correct y CG of each pipe (3 pipes)6Correct x CG of entire pipe structure6Correct y CG of entire pipe structure6Correct FBD10Correctly solve for force Q16Total50

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