A uniform brick of length 17 m is placed over the edge of a horizontal surface.

ID: 1542319 • Letter: A

Question

a. What maximum overhang is possible for the single brick (without tipping)? Answer in units of m.

Two identical uniform bricks of length 17 m are stacked over the edge of a horizontal surface.

b. What maximum overhang is possible for the two bricks (without tipping)? Answer in units of m.

Three identical uniform bricks of length 17 m are stacked over the edge of a horizontal surface.

c. What maximum overhang is possible for the three bricks (without tipping)? Answer in units of m.

n identical uniform bricks of length 17 m are stacked over the edge of a horizontal surface.

d. What maximum overhang is possible for the n bricks (without tipping)?

17 m

part a )

The center of mass of a single brick is in its middle or L/2 length

X = 17/2 = 8.5 m

part b )

The distance between the combined centre of mass of the tower and the right hand edge of the bottom brick we'll define as X2

The distance between the centre of mass of the lower brick and right hand edge of the bottom brick we'll define as X1

X2*2M = X1*M

we know that X1 = 1/2

X2 = 1/4

overhang = 1/2 + 1/4 = 3L/4 = 12.75 m

part c )

X3 *3M = X1*M

X1 = 1/2

X3 = 1/6

over all hang point = 1/2 + 1/4 + 1/6 = 11/12

X = 11/12 *17 = 15.58 m

part d )

over hang for n

1/2 + 1/4 + 1/6 + 1/8 + 1/10 + ..+1/2i

from i = 1 to n

x = L/2 from i = i to n sum 1/i

97th option)

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