1111ProblemSet06_solns.pdf

From our diagram we can write out the total force in

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From our diagram, we can write out the total force in each direc- tion: X F x = f s X F y = F N - F g = F N - mg. Next we apply Newton’s Second Law. In the vertical direction, there’s no acceleration, so X F y = ma y = 0 = F N - mg = 0 = F N = mg. In the horizontal direction, X F x = ma x = ma x = f s . We want to find the maximum acceleration, so we need the static friction to be its maximum value, f s = μ s F N . When we combine this with the other equations above, we obtain ma x = f s ma x = μ s F N ma x = μ s mg a x = μ s g a x = 0 . 9(9 . 81) = 8 . 83 m/s 2 . (4) Pulling a Sled: f k N F α F T F g y x The free-body diagram is shown in the figure at right. From the diagram we can write down expressions for the net hori- zontal and vertical forces: X F x = F T cos α - f k X F y = F T sin α + F N - F g . Now, we know that kinetic friction can be written as f k = μ k F N . Furthermore, the sled isn’t accelerating in the vertical direction, so F y = ma y = 0. This lets us answer the two multiple-choice questions. The normal force is less than the sled’s weight because the tension in the rope also helps to Copyright c 2017 University of Georgia. Unauthorized duplication or distribution prohibited.
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Physics 1111 Fall 2017 PS #6 Solutions Page 4 of 10 counteract the weight, in the vertical direction. The tension won’t depend on the velocity of the sled, because none of the other forces do, and the tension will be related to those forces. Now, for the quantitative part of the problem. In the vertical direction, since a y = 0, we obtain F T sin α + F N - mg = 0 = F N = mg - F T sin α. In the horizontal direction, we get X F x = ma x F T cos α - f k = ma x F T cos α - μ k F N = ma x . We can now substitute the normal force that we solved for using the equation for the vertical direction, and then find the tension: F T cos α - μ k ( mg - F T sin α ) = ma x F T cos α + μ k F T sin α = ma x + μ k mg F T (cos α + μ k sin α ) = m ( a x + μ k g ) F T = m ( a x + μ k g ) cos α + μ k sin α . All that’s left is the numbers: F T = 420(7 . 95 + 1 . 3(9 . 81)) cos 40 + 1 . 3 sin 40 = 8695 N 1 . 602 = 5430 N . Copyright c 2017 University of Georgia. Unauthorized duplication or distribution prohibited.
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Physics 1111 Fall 2017 PS #6 Solutions Page 5 of 10 (5) Ramp Kinematics: x y m θ g f k N F The free-body diagram for the sliding box is shown in the figure. We choose a coordinate system that’s aligned with the ramp, because it matches the direction of the box’s acceleration. (The acceleration is a value we can calculate from the given information.) From this diagram, we can write out the force components: X F x = mg sin θ - f k X F y = F N - mg cos θ. When we apply Newton’s Second Law, we realize that the y component of the acceleration is zero (the box doesn’t move off the ramp), so X F x = ma x = mg sin θ - f k = ma x X F y = m (0) = F N - mg cos θ = 0 . We can solve the y equation for the normal force, and then substitute it into the x equation, remembering that f k = μ k F N . This gives us F N = mg cos θ.
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