Problem 164 — The old bore can be right on the money, as this month’s problem by Stephen Cayton of Alpena, Mich., demonstrates.
The Megalopolis Amusement Park had just closed for the season; and as evening drew on, lights burned bright in the Park’s main office. The planning session for next year’s new rides was underway, and was threatening to become the usual shouting match.
“... and as I’ve been saying all along,” said J. Montgomery Drone, “just use the old drum brake from the discontinued roller coaster for ‘The Bungee Jump.’ We’ll save the expense of a new brake. The 4-in.-diam air cylinder on the old brake will be just fine. Why this reminds me of the time ...”
“Ridiculous!” shouted project leader, Nick Blowhard.
‘The Bungee Jumper’ consists of a harness attached to a steel cable that is controlled by a drum-type winch with a speed governor. The rider travels to the top of a 200-ft tower, dons the harness and jumps. On the tower, a photoelectric beam located 25 ft from the ground detects the rider and triggers a brake on the winch to slow the descent from a maximum 22 fps to 0 fps over a vertical distance of 20 ft. Maximum weight for the rider is assumed to be 250 lb. The brake is a drum type with a friction block applied in a radial direction by pneumatic cylinder.
The suspended portion of the cable that has unwrapped at the time of the brake application weighs 400 lb. The brake drum and remaining wrapped cable weigh 100 lb. The radius of gyration of the drum and wrapped cable is 1.5 ft, as is the drum’s radius, and the coefficient of friction between the brake pad and drum is 0.30. The brake drum has a 1-ft radius, and compressed air for the brake cylinder is regulated at 112 psi.
Neglect friction losses in the air cylinder, winch bearings, and any pulleys that the cable travels over. Also neglect the braking effect of the governor during the final 20 ft of travel. Assume the air is applied to the full area of the cylinder piston. Will a cylinder with a 4-in. bore be able to slow a 250-lb rider down to 0 fps over the distance indicated?
Technical consultant, Jack Couillard, Menasha, Wis.
Solution to last month’s problem 164 — You live life in the fast lane if you answered no. Here’s how Puff saved the day:
There are two pieces of information that we must find to determine whether the box falls into the incinerator.
A. How far does the box slide after the conveyor comes to a stop?
B. If the box slides 65 ft or more, how long does it take to reach the incinerator? Let:
d = The distance the box slides when the conveyor stops, ft
w = Weight of the box, given as 50 lb
v = Initial velocity of the box, given as 25 mph (36.67 fps).
m = coefficient of friction, given as 0.30
g = acceleration due to gravity, 32.174 ft/sec2
First solve A by using the formulas for work and kinetic energy.
Fd = (½)mv
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Here F is the force of friction, equal to μw. Since m =w/g, we can set up the following:
Plugging in the values for v, m and g, we get d = 69.66 ft. Therefore the box will certainly reach the incinerator unless Puff is able to get there first.
To find the time it takes the box to reach the incinerator, set the kinetic energy of the box equal to the impulse of the frictional force, Ft.
Puff will reach the box in time to save it from destruction.
Contest winner — Congratulations to Bob Forsythe of Wichita Falls, Texas, who won our July contest by having his name drawn from the 251 contestants who answered correctly out of a total of 288 for that month. A TI-68 calculator is in the mail to him. Further congratulations to those who figured out the open cells were the prime numbers!
The TI-68 Advanced Scientific Calculator by Texas Instruments can solve five simultaneous equations with real and complex coefficients and has 40 number functions that can be used in both the rectangular and polar coordinate systems. Other functions include formula programming, integration, and polynomial root finding. The calculator also features a last-equation replay function that lets you double-check your work.