Question

A cylindrical piece of steel has a diameter of 50 mm and a length of 200 mm. Steel has a density of 7200 kg/m3 a) (3 points) Determine the mass of this piece in kg. 1. b) (3 points) If a 10 mm diameter opening is drilled through the entire length of this cylinder and that portion removed, what is the new mass of this piece? c) (4 points) The 10 mm diameter piece is pulled with a force of 500 N. What is the stress in the rod? (5 points) A wooden box is cubical in shape and has a side length equal to 20 inches. This box is to be filled with small balls, 1 inch in diameter. What is the maximum number of balls that can fit in this box? Determine the ratio of the total volume of balls in the box to the volume of the box. 2.

Answer 1

(1)
(a)
We know that
density = mass / volume
Mass = density*volume
Density of steel = 7200 kg/m³
Volume of cylinder = ($\pi$/4)D²L
where D is diameter and L is length
V = ($\pi$/4)*(0.05)²*(0.2) = 3.927*10^-4 m³
Therefore
mass (m) = 7200*3.927*10^-4 = 2.827 kg
(b)
When 10 mm diameter portion is removed then the volume would be
$V = \left ( \frac{\pi }{4} \right )\left ( D_{o}^{2} - D_{i}^{2} \right )L$
Where D_o is external dia = 50 mm
D_i is internal dia = 10 mm
$V = \left ( \frac{\pi }{4} \right )\left ( 0.05^{2} - 0.01^{2} \right )*0.2$
V = 3.77*10^-4 m³
Therefore the mass of the cylinder would be
M = density*volume = 7200*3.77*10^-4 =2.714 kg
(c)
We know that the stress is given by
Stress = Force /area
Area = ($\pi$/4)D²
where D = 10 mm = 0.01 m
A = ($\pi$/4)*0.01² = 7.854*10^-5 m²
Force = 500 N
Stress = 500 /7.854*10^-5  
Stress = 6.37*10⁶ N/m²
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