Forces in 2D Legacy Problem #27 Guided Solution
Problem*
Ethan recently found out that the sledding hill on the west side of Lake Bluebird is more of a shredding hill than a sledding hill. He boarded his 60-inch sled and descended the steepest section near the tall trees. Before his ride ended at the location of a tree stump buried a few inches below the snow, he was accelerating at 2.5 m/s/s down the 19° incline. Determine the coefficient of friction between the sled and the snow. (Suggestion: Begin with a free body diagram.)
Audio Guided Solution
This is a very difficult problem that becomes all about visualizing and reasoning. There's not a lot of numbers here. In fact, there's only two. And we need to use these two numbers in order to calculate the coefficient of friction for a sled that is screaming down a hill with a 19 degree incline and accelerating at 2.5 meters per second per second. It's advisable that you use the free body diagram that you see here. Now what we know about this sled is we know that there are three forces which act upon it. There's the force of gravity which goes straight down. There's the normal force which goes perpendicular to the hill and upwards upon the sled. And there's the friction force which resists the sled down this icy hill. Now one of these forces is not real useful to us, and that's the gravity force. It's not useful as drawn. We really need to find the perpendicular and the parallel components of this gravity force. So since we don't have a value for the mass, it's impossible to actually get a numerical value for these parallel and perpendicular components of the weight vector. So what I'm going to do is I'm going to work the whole problem out using variables, symbols, like m and g and mu and etc. So for the F perpendicular force, it's simply going to be mg cosine theta. Now if I knew my m value, I could plug it into that equation, and I could actually calculate a numerical value for F perpendicular. But I don't have an m value, so I'm just going to call it mg cosine theta. For the parallel component of the gravity force, it's just mg sine of theta. Now the purpose or the usefulness of knowing mg cosine theta is that it's equal to F normal. F normal is going to balance F perpendicular since there's no acceleration perpendicular at L. And so F normal is also equal to mg cosine theta. And the value of knowing that is friction depends upon it. The friction force is mu times F normal, or mu times mg cosine theta. Now of course we still don't have a mass, so we can't calculate a numerical value for the friction force, nor for the parallel component of gravity. So we'll have to continue to work the problem out in variable form. So what I've determined so far is that parallel to the inclined plane, there are two forces which are not balancing, and it's this imbalance that causes the acceleration. One of the forces, F parallel, is going down the L, and its value is mg sine theta. And the other force is F friction. It resists this acceleration, and its value would be given by mu mg cosine theta. So now I'm going to use F net equal m a in order to solve for the coefficient of friction. And this is where things get real interesting. For F net, I know that if I had the values of F parallel and F friction, I could plug them in and I could calculate F net as F parallel minus F friction. But since I don't know values, I'm going to now say simply F parallel minus F friction equal m a. Now for F parallel, I'm going to write mg sine theta. And for F friction, I'm going to write mu mg cosine theta. So now my equation becomes mg sine theta minus mu mg cosine theta is equal to m a. Now you'll notice there's an m in every term of this three-term equation. So I can divide through by m, and I can cancel my m's. The equation now becomes g sine theta minus mu g cosine theta is equal to a. And I wish to solve for mu. I know everything in this equation except for mu. So I'm going to get the mu term by itself on one side of the equation. What I'm going to do is add mu mg cosine theta to each side of the equation, and I'm going to subtract m a from each side of the equation. The equation would then become g sine theta minus a is equal to mu g cosine theta. Now you can divide both sides of the equation through by g cosine theta, and you can solve for mu. The equation would become the quantity g sine theta minus a all divided by g cosine theta. Plug in your numbers for g, 9.8, your numbers for theta, 19 degrees, and your number for a, 2.5. And you can solve for mu. I get 0.0745, etc. And I can round that to two digits. It's 0.075 as the coefficient of friction value.
Solution
0.075
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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