**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 1**0.6 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

*= Coefficent of friction, given as 0.3*

μ

μ

*= Normal force to the ground, lb*

N

N

*f*= Frictional force of ladder, lb

*s*= Distance of McSquibb from base of ladder, when ladder begins to slip.

*= Push of wall against ladder, lb*

P

P

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,