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A monthly mechanical engineering problem to test your knowledge or try your luck. A new problem every month. The possibilities are endless - kinematics, fluid mechanics, physics......! You never know what we'll throw at you next!

Last month's problem:

A bomber is flying at a constant horizontal velocity of 240 miles per hour at an elevation of 10,000 feet toward a point directly above it's target. At what angle of sight should the bomb be released to strike the target ?

Solution:

Initial velocity of bomb = velocity of plane. Vo = 240 miles/hr = 352 ft/sec
Angle θ  = 0

Time of fall obtained from equation: y = v0yt - 1/2gt^2 , with v0y = 0, and y = -10,000 ft

t = square root of -2y/g. t = square root of -2(-10,000) ft / 32.0 ft/sec^2 = 25.0 ft/sec

The horizontal distance traveled by the bomb in this time is: x = v0xt. x = (352 ft/sec)(25.0 sec) = 8800 ft.

The angle of sight would be f = tan^-1(x/y) = tan^-1(8800/10,000) = 41.6°.

April problem:

A 1.6 oz. golf ball is struck by a club so that a force of 18 lb is applied for 0.04 seconds. What is the velocity of the ball leaving the tee?

Solution:

Ft = mV
(18.0 lb)(0.04 sec) = (1.6 oz /16 oz /lb)(1 sec/32.2 m/sec^2)V

0.72 = (0.1)(0.0310559)V

V = (0.1) / (0.0031055) = 231.85 ft/sec
or
(231.85 ft/sec)(1 mi./5280 ft)(3600 sec/1 hr) = 158.08 mph

February problem:

A storage tank is 4 feet in diameter and 9 feet tall, filled with kerosene. What is the total weight of the kerosene? (Hint: specific gravity of kerosene is 0.81)

Solution:

V = (pD^2h)/4 = [(3.14)(4 ft)^2(9 ft)]/4 = 113.1 ft^3

Weight of water = 62.4 lb/ft^3

Weight of kerosene = (62.4 lb/ft^3)(0.81)(113.1 ft^3) = 5716.39 lb

December problem:

A 1" steel rod which is 10" long is twisted to give 1° of rotation at the free end where the torque is applied. Assume that the modulus of rigidity (G) to be 11 x 10^6 lb/in^2.

Determine the torque required to produce 1° twist.
Determine the maximum shear stress (Ss) in the rod.

Solution:

For the rod:
J = (p/32)d^4
d = 1, so J = (p/32) in^4

Torque to twist 1°:
Æs = 1° , so 1° (p/180) = 0.01745 radians
Æ = TL/JG

G = 11 x 10^6 lb/in^2
L = 10"
T = JGÆ/L = (0.01745)(p/32)(11 x 10^6)(1/10)
T = 1844.5 lb in

Maximum shear stress:
Tmax = Trmax/J = Tdo/2J
Tmax = (1884.5)(1)(32)/(2p)
Tmax = 9597.5 lb/in^2

November problem:

Here is a problem that is relevant to all of us weekly.

You go to the gas station for a fill-up and top off your tank. Now suppose that your gas tank (made of sheet Iron) holds 20-gallons of gas and you don't completely tighten the cap. Your car then sits in the sun from morning (at 40 °F) until mid-afternoon (at 85 °F).

Are you losing money from possible spillage due to expansion, and if so, how much if gas costs $1.50 per gallon?

Hint: Coefficients of volume expansion are as follows:
Gas: 900 * 10.0 ^-6 C°^-1
Iron: 36 * 10.0 ^-6 C°^-1

Solution:

Spill = DV(gas) - DV(tank)
D V = g(gas)VDT

Spill = VDT - g(tank)VDT
Since V and DT are the same for the gas and internal tank volume:

Spill = [g(gas) - g(tank)]VDT

DT = 85 °F - 40 °F = 45 °F

Converting 45 °F x 5/9 - 17.8 = 25 °C

Spill = (900 * 10.0 ^-6 C°^-1) - (36 * 10.0 ^-6 C°^-1) * (20 gallons) * (25 °C)

= (864 * 10.0 ^-6 C°^-1) * (20 gallons) * (25 °C)

=0.432 gallons spilled

Spill Cost = (0.432 gal) * ($1.50/gal) = $0.65