Image

Fun With Fundamentals: Problem 205

May 1, 2000
A penny saved can be two pennies earned as this month’s problem by Andy Ulrich of Stillwater, Minn., demonstrates

Cast of characters
Problem 205 — A penny saved can be two pennies earned as this month’s problem by Andy Ulrich of Stillwater, Minn., demonstrates.

“Just check out my new lure,” said Lucius McSkeat. “It cost me $5 and it’s guaranteed.”

“Guaranteed to what?” replied Marius McSneed.

The two men were practicing their casting off the back porch in McSkeat’s backyard. The yard was 100 ft deep from the end of the porch and bordered by a fence. On the other side of the fence lived a big dog.

As McSkeat cast his rod the heavy lure, which was at the rod’s tip, flew off the line and was nowhere to be seen on the ground. The two men recreated the accident to see where the lure was.

The lure left the rod when McSkeat had it at the 10:00 position. Also, it takes 1/6 of a second for McSkeat’s cast to make a quarter circle, going from the horizontal to the vertical. McSkeat’s fishing rod measures 5 ft.

Assume constant angular velocity during the quarter circle cast. Since the lure is heavy, and no wind was stated, neglect air resistance. The center of rotation of the lure is 6.66 ft and directly above the end of the porch and 3.33 ft below the top of the fence when the lure flies off the line. Does the lure clear the fence?

Send your answer to:

Fun With Fundamentals
POWER TRANSMISSION DESIGN
1100 Superior Ave.
Cleveland, OH 44114-2543

Deadline is June10. Good luck!
Technical consultant, Jack Couillard, Menasha, Wis.

Solution to last month’s problem 204 — You know your limits if you answered 10.6 ft. Here’s how McSquibb received a cleaning detail that also was not in his job description.

Let:

W = Weight of McSquibb, given as 150 lb
w
= Weight of ladder, given as 80 lb
μ
= Coefficent of friction, given as 0.3
N
= Normal force to the ground, lb
f = Frictional force of ladder, lb
s = Distance of McSquibb from base of ladder, when ladder begins to slip.
P
= Push of wall against ladder, lb

Since the system is in equalibrium, the sum of the forces and moments must equal zero. Thus, the upward forces equal the downward forces.

N = W + w = 230 lb

f = μN = 69 lb

The forces towards the right equal the forces toward the left:

P = f

Taking the moments about the bottom of the ladder,

Sponsored Recommendations

The entire spectrum of drive technology

June 5, 2024
Read exciting stories about all aspects of maxon drive technology in our magazine.

MONITORING RELAYS — TYPES AND APPLICATIONS

May 15, 2024
Production equipment is expensive and needs to be protected against input abnormalities such as voltage, current, frequency, and phase to stay online and in operation for the ...

Solenoid Valve Mechanics: Understanding Force Balance Equations

May 13, 2024
When evaluating a solenoid valve for a particular application, it is important to ensure that the valve can both remain in state and transition between its de-energized and fully...

Solenoid Valve Basics: What They Are, What They Do, and How They Work

May 13, 2024
A solenoid valve is an electromechanical device used to control the flow of a liquid or gas. It is comprised of two features: a solenoid and a valve. The solenoid is an electric...

Voice your opinion!

To join the conversation, and become an exclusive member of Machine Design, create an account today!