Forces in 2D Legacy Problem #18 Guided Solution
Problem*
Charles Blondin is famous for having crossed the Niagara Falls on an 1100-foot long tight-rope at a height of 160 feet above the gorge. He accomplished the feat several times in a variety of ways – blindfolded, in a sack, carrying his manager on his shoulders, pushing a wheelbarrow and walking on stilts. Suppose the 65-kg Blondin is at the midpoint of the tight-rope and has stopped to cook and eat an omelet. If the tightrope makes an angle of 5° with the horizontal, determine the tension in the rope.
Audio Guided Solution
Charles Blondin is a famous tightrope walker who walked across the Niagara Falls on a tightrope several times doing a different stunt each time. Fascinating character, you might want to read the Wikipedia page on Charles Blondin. Now in this problem what we're told is that Charles has a mass that's estimated at 65 kilograms and he's in the exact middle of a tightrope. He's in the exact middle such that the tightrope is falling up and to the left and up and to the right, up on his body. It makes a fine degree angle with the horizontal and what we're asked to determine is the tension in the rope. The way we'd go about determining the tension in the rope is we do an equilibrium analysis knowing that the downforce on Charles Blondin's body has got to be balanced by the vertical component of the two tension forces. Now I can determine the downforce if I go 65 kilograms times 9.8. Doing that for Charles' mass would give me 637 newtons as the downforce. Now each side of this tightrope is going to pull up with one half of the downforce. So the vertical pull in either side is going to be 318.50 newtons. Now you can draw yourself a diagram in which you construct the tightrope stretched out at a fine degree angle above the vertical. Put a little arrow on the end of it to represent its pulling up and to the right. Then draw a little force triangle with a vertical side of 318.50 newtons. Now that's the side opposite the fine degree angle. So I can write the equation the sine of 5 degrees is equal to the opposite over the hypotenuse. The sine of 5 equals 318.50 newtons over F tension. Now I want to know the tension force. So if I solve this equation for F tension, I get 3654.37 newtons. That's the tension force on the left and on the right. And you don't double it, just express it as 3.7 times 10 to the 3rd newtons of tension force.
Solution
3.7 x 103 N (3654 N)
Habbits of an Effective Problem Solver
- Read the problem carefully and develop a mental picture of the physical situation. If necessary, sketch a simple diagram of the physical situation to help you visualize it.
- Identify the known and unknown quantities in an organized manner. Equate given values to the symbols used to represent the corresponding quantity - e.g., \(m = \units{1.25}{kg}\), \(µ = 0.459\), \(v_o = \units{0.0}{\unitfrac{m}{s}}\), \(θ = 41.6°\), \(v_f = \colorbox{gray}{Unknown}\).
- Use physics formulas and conceptual reasoning to plot a strategy for solving for the unknown quantity.
- Identify the appropriate formula(s) to use.
- Perform substitutions and algebraic manipulations in order to solve for the unknown quantity.
Read About It!
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