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Class notes on the topic of the coefficient of kinetic friction in Physics 204A. It explains the concept of frictional forces, the difference between static and kinetic friction, and how to calculate the acceleration, frictional force, and coefficient of kinetic friction using examples. The notes also include a discussion on the relationship between the coefficients of static and kinetic friction.
Typology: Exercises
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Example 12.1: A 750kg car traveling at 90.0km/h (25.0m/s) locks its brakes and skids to rest after traveling 35.0m. Find (a)the acceleration (b)the frictional force and (c)the coefficient of kinetic friction. (a)Use the kinematic equation without the time, v^2 = vo^2 + 2 a ( x − xo ) ⇒ 0 = vo^2 + 2 ax ⇒ a = − vo^2 2 x
a = −8.93 m^ s 2. (b)Given: m = 750kg and a = - 8.93m/s^2 Find: Ffr =? and μk =? Thanks to Newton, we know what causes this acceleration. It is the forces that act on the car which are drawn in the sketch above. To make applying Newton’s Second Law easier, the forces are redrawn on the coordinate system with their tails at the origin. This is called the “free body diagram.” Now apply the Second Law to the x-direction only, Σ Fx = max ⇒ − Ffr = ma ⇒ Ffr = − ma = −(750)(−8.93) ⇒ Ffr = 6700 N. Notice that the frictional force is in the minus x-directon. (c)Applying the Second Law to the y-direction only, Σ Fy = may ⇒ Fn − Fg = 0 ⇒ Fn = Fg = mg = (750)(9.80) ⇒ Fn = 7350 N (c)Using the definition of the coefficient of kinetic friction, μ k ≡ Fkf Fn
⇒ μ k =^ 0. COMMENT ON PROBLEM SOLVING: We need to add a step when the force vectors involve more than one coordinate axis. When you want to use the Second Law to solve a problem:
Example 12.3: The block of example 2 is on a ramp where angle of incline is slowly increased until the block begins to slide. Given this angle is 52.0˚, find the coefficient of static friction. Given: m = 2.00kg, θ = 52.0˚, and a ≅ 0 Find: μs =? Notice that the gravitational force points downward, but that the normal force points perpendicular to the surface. The gravity and normal forces are no longer co-parallel and therefore can’t be equal. In this problem, two of the three forces will be along one of the axes if we choose the x-axis to be along the ramp instead of horizontal. Then we will only have to find vector components for gravitational force. A careful free body diagram is essential here. The free body diagram is always draw by rotating the coordinates and all the vectors until the y-axis is vertical. Then it is easier to wrap our brains around finding the vector components. Note that when this rotating is done, the angle of the ramp is equal to the angle between the y- axis and the force of gravity. Using the free body diagram, apply the Second Law to each direction, ΣFx = ma (^) x ⇒ Fg sinθ − Fsf ,max = 0 ⇒ Fsf , max = Fg sinθ. ΣFy = may ⇒ Fg cosθ − Fn = 0 ⇒ Fn = Fg cosθ. We have assumed that the acceleration is very near zero. Using the definition of the coefficient of static friction, μs ≡ Fsf ,max Fn
Fg sinθ Fg cosθ
sin θ cosθ ⇒ μs = tan θ. Plugging in the numbers, μs = tan 52.0˚⇒ μs = 1.28. COMMENT ON PROBLEM SOLVING: Choosing a convenient coordinate system is an interesting issue. The behavior of an object canʼt possibly be determined by choosing a coordinate system, because it is essentially and imaginary device. However, the mathematics of solving a problem can be greatly simplified. In the previous problem, using the traditional horizontal and vertical coordinates would have resulted in having to find components for both the frictional and normal forces. The result would have been the same, but the math far more ugly. Here are some tips for choosing a convenient coordinate system. Generally, it is best to choose coordinates where most of the vectors are along an axis. This includes, not just the force vectors, but the acceleration vector as well. Free Body Diagram y x F g F n θ F sf,max F sf,max θ x Fn^ y m F g
Example 12.4: Assume that the coefficient of kinetic friction is 90.0% of the coefficient of static friction. Find the acceleration of the block down the incline once it is started. Given: m = 2.00kg, θ = 52.0˚, and μk = 0.9μs = 0.9(1.28) = 1. Find: a ≅? The free body diagram is the same, but now there is acceleration, so the equations generated by the Second Law will be a bit different. Applying the Second Law to each direction, ΣFx = ma (^) x ⇒ Fg sinθ − Fsf ,max = ma ⇒ a = Fg sin θ − Fsf , max m
ΣFy = may ⇒ Fg cosθ − Fn = 0 ⇒ Fn = Fg cosθ. Using the definition of the coefficient of kinetic friction, μk ≡ Fsf ,max Fn ⇒ Fsf ,max = μkFn = μkFg cosθ. Substituting into the expression for the acceleration and using the mass/weight rule, a = Fg sinθ − μkFg cosθ m
m/gsin θ − μk m/gcosθ m /
a = g(sin θ − μk cosθ) Plugging in the numbers,
The math is a bit messier than example 12.3, but the process is the same. Example 12.5: A student drags a 20.0kg laundry bag up a 30.0˚ ramp at a constant speed by exerting a force of 150N along the ramp. Find the forces that act on the laundry bag. Given: m = 20.0kg, θ = 30.0˚, Fp = 150N, and a = 0 Find: Fg = ?, Fn = ?, Ffr = ?, and Fp =? The velocity is constant, so the acceleration is zero. The four forces on the laundry bag are the pull of the student upward along the ramp, the weight of the bag (gravity) vertically downward, normal force perpendicular to the ramp, and friction opposite the motion along the ramp. The pull of the student is given, Fp = 150N (^). Gravity can be found using the mass/weight rule, Fg = mg = (20.0)(9.80) ⇒ Fg =^196 N^. To find the other forces, we must apply the Second Law. We’ll choose coordinates where the x-axis is along the ramp and draw the free body diagram. Using the free body diagram to carefully apply the Second Law to each direction separately, ΣFx = ma (^) x ⇒ Fg sinθ + Ffr − Fp = 0 ⇒ Ffr = Fp − Fg sinθ. Putting in the numbers, Ffr = 150 − 196 sin30.0˚⇒ Ffr = 52.0 N (^). Free Body Diagram y x F fr F g F n ! Fp Fp ! x y F fr F g F n m Free Body Diagram y x F g F n θ F kf θ x Fn y m F g Fkf a