Physics for Scientists and Engineers, Technology Update (No access codes included)
Physics for Scientists and Engineers, Technology Update (No access codes included)
9th Edition
ISBN: 9781305116399
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 12, Problem 12.7P
To determine

The centre of gravity for the three object system.

Expert Solution & Answer
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Answer to Problem 12.7P

The centre of gravity for the three object system is (2.54m,4.75m) .

Explanation of Solution

Given info: The mass of the rod is m1=6.00kg , the mass of the right triangle is m2=3.00kg , the mass of the square is m3=5.00kg .

The centre of gravity of the rod is,

(x1,y1)=(xr1+xr22,yr1+yr22)

Here,

xr1 is the first x -coordinate of the rod.

xr2 is the second x -coordinate of the rod

yr1 is the first y -coordinate of the rod.

yr2 is the second y -coordinate of the rod.

Substitute 2m for xr1 , 9m for xr2 , 7m for yr1 and 7m for yr2 in the above equation.

(x1,y1)=(2m+9m2,7m+7m2)=(5.5m,7m)

The centre of the gravity of the right angled triangle is at distance two third of the length of the base and one third of the height of the triangle from the base.

The centre of the gravity of the right angled triangle is,

(x2,y2)=(xt1+23(xt2xt1),yt1+13(yt2yt1))

Here,

xt1 is the first x -coordinate of the base of the right angled triangle.

xt2 is the second x -coordinate of base of the right angled triangle.

yt1 is the first y -coordinate of the right angled triangle.

yt2 is the second y -coordinate of the right angled triangle.

Substitute 8m for xt2 , 4m for xt1 , 5m for yt2 and 1m for yt1 in the above equation.

(x2,y2)=(4m+23(8m4m)+,1m+13(5m1m))=(6.67m,2.33m)

The centre of gravity of the square is,

(x3,y3)=(xs1+xs22,ys1+ys22)

Here,

xs1 is the first x -coordinate of the square.

xs2 is the second x -coordinate of the square.

ys1 is the first y -coordinate of the square.

ys2 is the second y -coordinate of the square.

Substitute 5m for xs1 , 2m for xs2 , 5m for ys1 , 2m for ys2 in the above equation.

(x3,y3)=(5m2m2,5m+2m2)=(3.5m,3.5m)

Formula to calculate center of gravity of three masses is,

(xcm,ycm)=(m1x1+m2x2+m3x3m1+m2+m3,m1y1+m2y2+m3y3m1+m2+m3)

Here,

xcm is the x coordinate of the centre of the gravity of complete system.

ycm is the y coordinate of the centre of the gravity of complete system.

x1 is the positions of centre of gravity of the mass m1 .

x2 is the positions of centre of gravity of the  mass m2 .

x3 is the positions of centre of gravity of the mass m3 .

y1 is the position of the centre of gravity of the  mass m1 .

y2 is the position of the centre of gravity of the mass m2 .

y3 is the position of the centre of gravity of the mass m3 .

Substitute 6.00kg for m1 , 3.00kg for m2 , 5.00kg for m3 , 5.5m for x1 , 6.67m for x2 , 3.5m for x3 , 7m for y1 , 2.33m for y2 , 3.5m for y3 in the above equation.

(xcm,ycm)=((6.00kg)(5.5m)+(3.00kg)(6.67m)+(5.00kg)(3.5m)6.00kg+3.00kg+5.00kg,(6.00kg)(7m)+(3.00kg)(2.33m)+(5.00kg)(3.5m)6.00kg+3.00kg+5.00kg)

Simplify the above equation.

(xcm,ycm)=(35.51kgm6.00kg+3.00kg+5.00kg,66.49kgm6.00kg+3.00kg+5.00kg)=(35.51kgm14.00kg,66.49kgm14.00kg)=(2.536m,4.749m)(2.54m,4.75m)

Conclusion:

Therefore, the centre of gravity for the three object system is (2.54m,4.75m) .

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Chapter 12 Solutions

Physics for Scientists and Engineers, Technology Update (No access codes included)

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