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A uniform brick of length 21 m is placed over
the edge of a horizontal surface with a maximum overhang of 10.5 m attained without
tipping. (PICTURE ONE)
Now two identical uniform bricks of length
21 m are stacked over the edge of a horizontal
surface. (PICTURE TWO)

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

A uniform brick of length 21 m is placed over the edge of a horizontal surface with a maximum overhang of 105 m attained without tipping PICTURE ONE Now two ide class=
A uniform brick of length 21 m is placed over the edge of a horizontal surface with a maximum overhang of 105 m attained without tipping PICTURE ONE Now two ide class=

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MathPhys

Answer:

15.75 m

Explanation:

First, let's look at the top brick by itself.  In order for it not to tip over the bottom brick, its center of gravity must be right at the edge of the bottom brick.  So the edge of the top brick must be 10.5 m from the edge of the bottom brick.

Now let's look at both bricks as a combined mass.  We know the total length of this combined brick is 10.5 m + 21 m = 31.5 m.  And we know that for it to not tip over the edge of the surface, its center of gravity must be at the edge.  So the edge of the combined brick must be 31.5 m / 2 = 15.75 m from the edge of the surface.

Ver imagen MathPhys
okpalawalter8

We have that the maximum overhang length possible for

the two bricks  is mathematically given as

  • L_o=15.75m

From the question we are told

  • A uniform brick of length 21 m is placed over the edge of a horizontal surface
  • with a maximum overhang of 10.5 m attained without tipping.
  • Now two identical uniform bricks of length 21 m are stacked  over the edge of a horizontal surface.

total overhang length

Generally the equation for the value of x  is mathematically given as

X= mid point joining both bricks

Therefore

[tex]x=0.25l=\frac{10.5}{2}\\\\x=5.25m\\\[/tex]

Hence

The total overhang length

[tex]L_o=l/2+x\\\\Therefore\\\\L_o=21*5.25/2+x\[/tex]

L_o=15.75m

For more information on length visit

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