A cube of wood: mass calculation puzzle?

Question

-A wooden cube of uniform density has sides of length 12 cm and a mass of 1.296 kg.

-A circular hole, 4 cm in diameter, is drilled into the middle of each face of the cube. 

-Each hole is drilled perpendicular to the face and tunnels all the way through to the opposite side of the block.

-The end result is a block of wood with 3 perpendicular intersecting tunnels through it's centre.....oh, and a pile of sawdust on the floor.

The puzzle is this:

What is the mass of the drilled block of wood?

String · Accepted Answer

EDIT 2: I finally found my mistake - see calculation of T at the bottom. Thanks for the Steinmetz-link! Here is another (and more trustworthy) link on it:

http://mathworld.wolfram.com/SteinmetzSolid.html

So here is my final solution:

Let the centre of the cube be at (0,0,0) and the direction of the drilled cylinders be parallel to the axes. Then if (x,y,z) lies within one of the cylinders, symmetry implies that (x',y',z') with x'=±x, y'=±y and z'=±z will be within the same cylinder. Hence one point corresponds to 8 points as a total. Due to this we need only consider x,y,z≥0 to calculate any volume related to the cylinders which then provides 1/8 of the volume.

Let the volume of the points contained in a single cylinder be denoted S, the volume of the intersection of two cylinders D, and points within all three T. Then D-T denotes the volume of points contained in exactly two, not all three. With this the volume of the three intersecting cylinders will add up to:

3S-3(D-T)-2T = 3S-3D+T

The challenge is now to determine D and T:

To determine D consider the two cylinders parallel to the z- and y-axis and restrict to x,y,z≥0 to determine D/8. The cylinders are defined as:

x²+y² ≤ 4 (z-axis cylinder)
x²+z² ≤ 4 (y-axis cylinder)

which is satisfied simultaneously exactly when 0≤x≤2 and 0≤y,z≤√(4-x²). With these limits we have the triple integral:

D/8 = ∫∫∫ 1 dz dy dx = ∫ (4-x²) dx = 16/3

Now T is a bit more tricky... It will be convenient with an additional symmetry. Let us assume that x is at least as big as y and z. Then we are only considering 1/3 of the volume already referred to since symmetrically either y or z could be the biggest. Thus we find T/24. For x to be the bigger we must have √2≤x≤2 and if we still say that 0≤y,z≤√(4-x²) the third cylinder equation

y²+z² ≤ 4 (x-axis cylinder)

will automatically be fulfilled since y,z≤√(4-x²)≤√2. Now since only the lower limit for x has been altered, this leads to a calculation very similar to that of D/8:

T/24 = ∫∫∫ 1 dz dy dx = ∫ (4-x²) dx = (16-10√2)/3

But here I just forgot that for 0≤x≤√2 it is clearly still possible to have x as the biggest of the three coordinates just by requiring 0≤y,z≤x. Assuming this we still have all three cylinder equations fulfilled so this adds a volume to T/24 of

∫∫∫ 1 dz dy dx = ∫ x² dx = (2√2)/3

which then leads to T/24 = (16-8√2)/3 like you rightfully stated. With this the mass becomes 1.02459 kg.

? · Answer

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sexton · Answer

1

Shashank Kapare · Answer

Total Volume of Cube = 12^3 = 1728 cc
Mass of Cube = 1.296 kg

So, Density of Cube is 0.75 gm/cc

Now,
Volume of 1st hole drilled = (pi/4)*(4*4)*(12) = 150.8 cc

Considering all 3 holes will share same central cavity.
Volume of common central cavity = 4*4*4 = 64 cc (approx)

So, effective drilled volume of 2nd and 3rd hole will be = 150.8 - 64 = 86.8 cc

So, total volume of wood removed from cube = 150.8 + 86.8 + 86.8 = 324.4 cc

Mass of wood removed = vol. of wood removed * density = 324.4 * 0.75 = 243.3 gm

Hence, Remaining mass of drilled block = 1.296 - .2433 = 1.053 kg.

JOE90 · Answer

You need to find the volume of the holes which are drilled into the wood.   Make sure you do not do any double-counting nor triple-counting.

Tom · Answer

OK. I'll post the exact formula tomorrow. The integration is a bit messy. Your comments about the other answers is correct.

Pranil · Answer

mass = volume × density
volume of full cube = 12³ = 1728 cm³
density of wood used = 1.296/1728 = 0.00075 kg/cm³

volume of remaining cube 
= volume of cube – 3 × volume of cylinders removed + 2 × volume of cylinders in the middle of cube

]addition of 2 times volume of cylinders required as it is already included in 3 × volume of cylinders removed. actually we have to subtract once only.]
volume of 1 cylinder removed = πR² h
= 3.14 × 2² × 12
= 150.72

volume of 3 cylinder removed = 3 πR² h
= 3 × 150.72
= 452.16 cm³

volume of cylinder crossing in the middle of cylinder = πr²h ------- (here r = h = 2)
= 3.14 × 2² × 2
= 25.12
volume of such 2 cylinders = 2 × 25.12 = 50.24 cm³

volume of piece after cutting = 1728 – 452.16 + 50.24 
= 1326.08 cm³

mass = 1326.08 × 0.00075 
= 0.995 kg
----Edit
what I consider, there are 3 small cylinders having radius 2 cm and height 2 cm seems to be wrong.
height is 4 cm
volume of cylinder crossing in the middle of cylinder = πr²h ------- (here r = 2 and h = 4)
= 3.14 × 2² × 4
= 50.24
volume of such 2 cylinders = 2 × 50.24 = 100.48 cm³

volume of piece after cutting = 1728 – 452.16 + 100.48 
= 1376.32 cm³

mass = 1376.32 × 0.00075 
= 1.032 kg
----

A cube of wood: mass calculation puzzle?

7 Answers