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(a)
ToShow: Successive bricks must extend no more than (starting at the top)
12,14,16 and 18 Of their length beyond the one below.
(a)
Explanation of Solution
Given data:
System is in equilibrium.
Formula used:
To remain in equilibrium, centre of mass of above system must lie just above the right edge.
For one brick over 2
To remain in equilibrium, centre of mass of brick 1 must lie above the right edge of brick 2 or before. So, for maximum extension overhanging part must be equal to L2 .
Brick is of uniform thickness, so centre of mass is at mid-point.
For two bricks (1,2) over 3
For equilibrium of brick 1 and 2, and for maximum extension,centre of mass of brick 1 and 2 must lie above the edge of 3 bricks.
xcm(1,2,3)=3L4×M+L4×M+0.M3M=L3
Distance of centre of mass of 1, 2 and 3 from the right edge of brick 3:
=L2−L3=L6
Centre of mass of brick 1,2 and 3must lie above the right edge of brick 4, so distance between right edges of brick three and 4=16
Above system remains in equilibrium when C.M of mass of bricks 1,2,3,4 lies above the right edge of table.
First find the C.M of brick 1,2,3,4
W.r.t C.M of brick 4 .
Let C.M of brick is origin.
x− Coordinate of C.M of brick 4:xcm(4)=0
x− Coordinate of C.M of brick 3:xcm(3)=L6
x− Coordinate of C.M of brick 2:xcm(2)=L6+L4
=5L12
x− Coordinate of C.M of brick 1:xcm(1)=L6+L4+L2
=1112L
With the help of previous solution, you can say that distance between C.M of brick 1 and 2 must be equal to R2 . For the center of mass of brick 1 and 2 to lie just abovethe right edge of brick 3 , centre of mass of brick 1 and 2 must be equidistance to vertical line. Hence the overhanging part of brick 2 should be equal to L4 .
For equilibrium of 1,2,3 over 4 . These three brick remain in equilibrium with maximum overhanging part only when C.M of brick 1 , brick 2 and brick 3 must be above the right edge of brick four.
Let C.M of brick 3 as origin
xcm(1,2,3)=Mxcm(1)+Mxcm(2)+Mxcm(3)3M
xcm(1): ( x -coordinate of centre of mass of brick 1) =3L4
xcm(2): ( x -coordinate of centre of mass of brick 2) =L4
xcm(3): ( x -coordinate of centre of mass of brick 3) =0
M : Mass of each brick.
xcm(1,2,3,4)=M.xcm(1)+Mxcm(2)+Mxcm(3)+Mxcm(4)4M=(M)×(11L2)+M(5L12)+M(L6)+M(0)4M=3L8
Distance between right edges of brick 4 and xcm(1,2,3,4)
=L2−3L8=L8
Hence distance between right edge of table and right edge of 4th brick =L8
Hence hanging part of brick 1,2,3,4th are L2,L4,L6,L8 to remain in equilibrium with maximum extend.
Conclusion:
Over hanging part of brick 1,2,3,4th are L2,L4,L6,L8 , etc. to remain equilibrium and with maximum extend.
(b)
Whether the top brick is completely beyond the base.
(b)
Answer to Problem 38P
Solution:
Yes
Explanation of Solution
(c)
Ageneral formula for the maximum total distance spanned by n bricks if they are to remain stable.
(c)
Answer to Problem 38P
Solution:
The general formula of maximum total spanned by n bricks =n∑i=1(L2i) .
Explanation of Solution
Given data:
You can use previous result.
Formula used:
Find out the general term of the series with the help of previous result.
Overhanging part of 1st brick =L2=L2×1
Overhanging part of 2nd brick =L4=L2×2
Overhanging part of 3rd brick L6=L2×3
Overhanging part of 4th brick =L8=L2×4
From above it becomes clear that overhanging part of nth brick = L2n
Hence, maximum span length obtained with the help of n brick.
=L2+L2×2+L2×3+L2×4+...L2n=n∑i=1L2i
Hence, maximum span length obtained with the help of n brick
=n∑i=1(L2i)
Conclusion:
Maximum span length obtained with the help of n brick
=n∑i=1(L2i)
(d)
The minimum number of bricks each 0.30 m long and uniform is needed if the arch is to span 1.0 m .
(d)
Answer to Problem 38P
Solution:
35 bricks
Explanation of Solution
Given data: Span length =1.0 m
Length of brick =0.30 m
Formula used:
Maximum span length (overhanging part)
Obtain from n brick
=n∑i=1L2i
Calculation:
The span length of arch =1.0 m
Length of one side span length =1.0 m2=0.5 m
Let number of required bricksbe n to develop 0.5 m overhanging part.
n∑i=1L2i≥0.5L=0.30 m(given)n∑i=10.302i≥0.50n∑i=112i≥53
12+14+16+18+110+112... to n≥53
Do the sum of left side until you obtain the sum is either greater than or equal to 5/3
12+14+16+18+110+112+114+116+118+120+122+124126+128(n=14)=1.625
If n=15 15∑i=112i=1.659
If n=16 16∑i=112i=1.690=53=1.666
Form the above solution it becomes clear that
For n=15, overhanging part is less 0.5
For n=16, overhanging is just greater than 0.5
Hence, required number of bricks =2×16+(1)+(2)
Multiply 2 both sides
1 is for brick on the top and 2 is for the base of each side.
2×16+(1)+(2)
=35
Conclusion:
Minimum no of brick to developed span length 1.0 m and arch as given in figure =35 .
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