Part A Atwood Machines Please try this link:

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1 LAST NAME FIRST NAME DATE Assignment 2 Inclined Planes, Pulleys and Accelerating Fluids Problems 83, 108 & 109 (and some handouts) Part A Atwood Machines Please try this link: The Full Atwood Machine consists of two masses at the end of a single rope which is placed over a pulley. For simplicity the rope is considered to have no mass. Also for simplicity the pulley has not friction or inertia. QUESTION: For the diagram at right including the pulley and connected masses you may assume m 2 > m 1. Find an equation for the acceleration of the system (a) in terms of the masses and gravity. Your answer should be an equation not a number.

2 The Half Atwood Machine consists of two masses at the end of a single rope which is placed over a pulley. One mass (M 2 in this case is hanging from the rope) For simplicity the rope is considered to have no mass. Also for simplicity the pulley has not friction or inertia. Mass 1 sits on a horizontal surface with no friction (Yes we can still solve these with friction) QUESTION: For the diagram at right including the pulley and connected masses you may assume the surface upon which M 1 is sitting is frictionless. Find an equation for the acceleration of the system (a) in terms of the masses and gravity. You may assume the surface upon which M 1 sits if level. Your answer should be an equation not a number. WHAT IF FRICTION IS PRESENT ON THE HORIZONTAL SURFACE? Certainly if Friction is involved acceleration will not be as high. Let s see if we can quantify that. Logic: Both masses are accelerated by the same force, which is given by the equation F = W= M 2 g. In this case there is the opposing force of Friction By Newton s Second Law a = F/m, Here the mass being accelerated is (M 1 + M 2 ) F = µm 1 g M 1 + M 2 F= M 2 g Newton s Second Law gives:a = F m = M 2g+µM 1 g M 1 +M 2 = g(m 2+µM 1 ) M 1 +M 2

3 An Inclined Plane Situation For the diagram at left including the pulley and connected masses you may assume M > m. Find an equation for the acceleration of the system (a) in terms of the masses, the angle of the inclined plane and gravity. Your answer should be an equation not a number. WHAT IF FRICTION IS PRESENT ON THE RAMP SURFACE? LOGIC: The force accelerating the system is the weight of the mass hanging (m) given by mg. The force opposing this is the parallel component of the weight on the inclined plane Mg. This is given by the equation Mg sin θ. Mg sinθ M + m mg F = µf N =µmgcosθ Sum of the forces = mg - Mg sin θ - µmgcosθ Mass being accelerated = M + m By Newton s Second Law: a = mg Mgsinθ µmgcosθ M+m = g(m (M(sinθ µcosθ)) M+m

4 A skier (mass = m) is on a pair of skies with a coefficient of kinetic friction of k He is on a hill that is θ degrees. If he skis straight down the hill what will his acceleration be down the hill? (You ve been given a start by having the Free Body Diagram) Label all the forces on the FBD Given/nown Variables k, g, θ Unknown Variables m= mass of person, not given, W = weight of person = mg LOGIC: To find the acceleration down the slope the sum of the forces parallel to the slope must first be know. This will be used in Newton s Second Law to find the acceleration. First find the Weight W mg Find the Normal Force From the diagram you can see that the normal force is equal to the perpendicular component of the weight. W W cos mg cos F N This allows calculation of the Frictional Force which acts to slow the skier while going down the slope. Find the Friction Force F F W W cos mg cos N

5 Find the Component of the weight acting to accelerate the person down the slope This is called Weight parallel ; W W sin m sin g Find the sum of the forces parallel to the slope W F mg sin m cos F g *** To find the acceleration solve Newton s Second Law for acceleration F This gives a m a m m where the force is given by the equation above *** F mg sin mg cos which simplifies to a gsin cos Notice that mass has cancelled out, this means the mass of the person does not affect the acceleration. The acceleration along the slope is a function of gravity, the angle of the slope and the coefficient of friction but not mass.

6 ACCELERATING A LIQUID HORIZONTAL When the truck accelerates the liquid is pushed (not really) to the back end of the truck. What Law explains why the water is pushed to the back? Is the angle the water surface creates with a horizontal related to the rate of acceleration? (I will answer this one for you) Yes. But the bigger question is How? DERIVE THE EQUATION THAT GIVES THE ACCELERATION OF THE TRUC AS A FUNCTION OF THE ANGLE a = f(θ) Ask yourself a few questions to begin: If acceleration is larger is the angle larger or smaller? If the acceleration is negative (in the other direction) what happens to the angle? Is the water Static or Dynamic? What angle(s) might be of interest? Or allow you to get an understanding of what is happening? Draw the FREE BODY DIAGRAM TO BEGIN. This makes things very easy. When the angle is zero there is zero acceleration. If the angle were 90 degrees it would have to be infinite acceleration to eliminate the influence of gravity. Because of the relationship of zero degrees = zero acceleration and 90 degrees = infinite acceleration AND the fact that this will likely involve some trigonometry leads one to believe the solution will include the trigonometric function that produces this. Which trig function produces the following: Angle Function of the Angle Infinity Seems like the TANGENT. Draw the FREE BODY DIAGRAM TO BEGIN. This makes things very easy.. The water is accelerating in the horizontal axis but not in the vertical axis. By Newton s First Condition of Equilibrium Normal Force = W = mg By the Second Law the force accelerating the water horizontally = ma Drawing a triangle between the horizontal and vertical force (it does not matter which) the angle can be seen. To find a use the ratio of the sides of the triangle and solve for acceleration tan θ = OPP ADJ = F W = ma mg solving for a gives : a=g tan θ

7 108. As part A of the drawing shows, two blocks are connected by a rope that passes over a set of pulleys. One block has a weight of 412 N and the other has a weight of 908 N. The ropes and pulleys are massless and frictionless. A) Determine the acceleration of the lighter block? (Is it different that the acceleration of the heavier block. Why or why not?, Explain) a. The rope exerts a tension, T, acting upward on each block. Applying Newton's second law to the lighter block (block 1) gives T m 1 g = m 1 a Similarly, for the heavier block (block 2) T m 2 g = m 2 a Subtracting the second equation from the first and rearranging yields m a m m g m m/s 2 b. The tension in the rope is now 908 N since the tension is the reaction to the applied force exerted by the hand. Newton's second law applied to the block is Solving for a gives 1 T m 1 g = m 1 a 908 N 2 2 T a g 9.80 m/s 11.8 m/s m 42.0 kg c. In the first case, the inertia of BOTH blocks affects the acceleration whereas, in the second case, only the lighter block's inertia remains.

8 109. Assume pulleys and ropes are massless and frictionless. The surface with the 80 kg block has a Coefficient of inetic Friction of.1 A) Determine the acceleration of the three objects. a. The left mass (mass 1) has a tension T 1 pulling it up. Newton's second law gives T 1 m 1 g = m 1 a (1) The right mass (mass 3) has a different tension, T 3, trying to pull it up. Newton's second for it is T 3 m 3 g = m 3 a (2) The middle mass (mass 2) has both tensions acting on it along with friction. Newton's second law for its horizontal motion is T 3 T 1 µ k m 2 g = m 2 a (3) Solving Equation (1) and Equation (2) for T 1 and T 3, respectively, and substituting into Equation (3) gives a Hence, m m m g 3 1 k 2 m m m b. From part a: kg 10.0 kg kg 9.80 m/s a 10.0 kg 80.0 kg 25.0 kg 0.60 m/s 2 T 1 = m 1 (g + a) = 2 2 T 3 = m 3 (g a) = kg 9.80 m/s 0.60 m/s 104 N 25.0 kg 9.80 m/s 0.60 m/s 230 N