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Transcript
Using Newton’s 2nd Law We have learned that Newton’s 2nd law is F = ma We have learned some of the forces that can be acting on an object: weight, tension, normal, friction. We have learned how to draw free body diagrams to help us solve problems. Now it is time to put it all together. Newton’s 2nd Law in 3D F = ma must be true in all directions. Fx = max, Fy = may, Fz = maz. If the object isn’t moving in the ydirection, for example, then ay = 0 and Fy = 0. If there is more than 1 force acting in the y-direction, for example, then ∑Fy=may Mass on a plane with friction Forces acting on the pink box are: Weight – pointing straight down Normal – pointing up perpendicularly from the 2 surfaces. friction – pointing in the opposite direction of motion (opposing motion). Mass on a plane with friction The Weight has a component parallel to the plane and pointing down the plane, that we will call Wparallel: Wparallel = mgsinθ For an object on a non-horizontal surface, we know that N = mgcosθ. And we know that f=μN= μmgcosθ. Mass on a plane with friction- not moving If the mass is not moving, we know that the force pulling it down the plane (Wparallel) equals the force opposing the motion (friction). mgsinθ = μmgcosθ But which μ is it? Static or kinetic? Since the mass isn’t moving, it must be static. Mass on a plane with friction- not moving Therefore: mgsinθ = μsmgcosθ Notice that mass and g will cancel out, and this equation becomes: sinθ = μscosθ Rearranging this equation becomes: μs = cosθ/sinθ = tanθ This can be used to find μs for a non-moving object on an inclined plane. Mass on a plane with friction- not moving Example What is the friction acting on a 15 kg mass resting on a plane inclined at 25° to the horizontal? Remember if it isn’t moving, then Friction = Wparallel (and Wparallel = mgsinθ) Therefore: Friction = mgsinθ Friction = (15 kg)(9.8 m/s2) sin 25 ° Friction = 62 N Mass on a plane with friction- not moving Example 2 What is the coefficient of static friction acting on the same mass? Remember Friction = μsmgcosθ 62 N = μs (15 kg)(9.8 m/s2) cos 25 ° μs = .47 Remember coefficients of friction are unitless! Mass on a plane with friction- moving But what if the mass is accelerating? If we sum up the forces acting parallel to the line of motion (and choosing our signs to indicate up or down direction): Σ F = -mgsinθ + μmgcosθ But which μ is it? Static or kinetic? Because the object is moving, it will be kinetic friction. Mass on a plane with friction- moving Σ F = -mgsinθ + μkmgcosθ By Newton’s 2nd Law, F = ma, so -mgsinθ + μkmgcosθ = ma Notice mass can cancel out (but not g): -gsinθ + μkgcosθ = a This equation can be used to find the acceleration parallel to the plane of a mass on an inclined plane. Note that acceleration should turn out to be negative, indicating motion down the plane. Mass on a plane with friction- moving Example What is the acceleration parallel to the plane of an object sliding down a plane that is inclined at 48°? The coefficient of kinetic friction between the object and the plane is .26. a = -gsinθ + μkgcosθ a = - (9.8 m/s2)sin48° + (.26)(9.8 m/s2)cos48° a = -5.6 m/s2 Mass on a plane with friction- moving Example 2 How long will it take the same object to travel down .50 meters of the plane? Assume the acceleration is constant. a = -5.6 m/s2 vi = 0 m/s d = -.50 m (negative cuz “down”) t=? d = vit + ½ at2 -.50 m = (0) + ½ (-5.6 m/s2) t2 t = .42 seconds 2 attached masses on an inclined plane Mass 1 is the mass on the plane. Mass 2 is the hanging mass. Now we have an added force of Tension, pointing away from each mass along the rope. 2 attached masses on an inclined plane The forces acting on mass 1 (the mass on the plane) are the same as for a mass on a plane with the addition of Tension: -m1gsinθ + μkm1gcosθ + T = m1a Where a is acceleration parallel to the plane. The forces acting on mass 2 (the hanging mass) are only in the y direction. If we ignore air friction, they are: Σ Fy = -m2g + T = m2a If we assume a massless, non-stretching rope, we can assume that this a is the same as the acceleration of mass 1. 2 attached masses on an inclined plane Solving one equation for T and plugging into the other equation yields the following “Big Ugly” Equation: For the magnitude of the acceleration down a plane: a = (m1gsinθ - μkm1gcosθ - m2g) / (m1 + m2) For the magnitude of the acceleration up a plane: a = (-m1gsinθ - μkm1gcosθ + m2g) / (m1 + m2) 2 attached masses on an inclined plane But how do you determine if acceleration will be up or down the plane? Compare the values of m1gsinθ and m2g to determine which is bigger. If m1gsinθ is bigger it will pull the mass down the plane. If m2g is bigger it will pull the mass up the plane. In general, a big m1 and a big θ lead to acceleration down the plane. A small m1 and a small θ lead to acceleration up the plane. 2 attached masses on an inclined plane Example If m1 is 15 kg, m2 is 35 kg, μk is .13, and θ is 12°, what will be the acceleration of the mass? First determine if mass 1 will accelerate up or down the plane by comparing m1gsinθ and m2g. m1gsinθ = (15kg)(9.8 m/s2)sin 12° = 31 N m2g = (35 kg )(9.8 m/s2) = 343 N So it will accelerate up the plane. a = (-m1gsinθ - μkm1gcosθ + m2g) / (m1 + m2) a = 5.9 m/s2 2 attached masses on an inclined plane Example 2 For the previous example, what will be the tension in the rope? For problems such as these, the easiest way to solve is to sum up the forces acting on the hanging mass. This way we can ignore friction and angles. The forces acting on the hanging mass are tension pointing up, and weight pointing down. Note that acceleration will be negative since mass 2 is falling! Σ F = T – m2g = m2a T – (35 kg) (9.8 m/s2) = (35 kg) (-5.9 m/s2) T = 140 N 2 attached masses in an Atwood machine An Atwood machine is simply 2 masses attached to a string that passes thru a pulley. Notice the FBD’s for each mass have been drawn for you. Atwood Machine If mass 1 is smaller than mass 2, mass 1 will accelerate upwards and mass 2 will accelerate downwards. Again assuming a massless non-stretching rope, we can say that these accelerations will be equal. Sum up the forces acting on mass 1: Σ F1 = T – m1g = m1a Note that this acceleration is positive because mass 1 is moving up. Sum up the forces acting on mass 2: Σ F2 = T – m2g = -m2a Note that this acceleration is negative because mass 2 is moving down. Atwood Machine Solving each equation for Tension and setting them equal to each other: m1g + m1a = m2g - m2a This equation can be manipulated to solve for m1, m2, or a. Remember that in our derivation of this equation, mass 2 was the heavier mass. Atwood Machine - Example If mass 1 is 12 kg and mass 2 is 15 kg, what will be the acceleration of each mass? m1g + m1a = m2g - m2a Rearranging this equation so that a is on one side: m2a + m1a = m2g – m1g Factoring out an a: a(m2 + m1) = m2g – m1g Solving for a: a = (m2g – m1g) / (m2 + m1) Atwood Machine – Example 2 a = (m2g – m1g) / (m2 + m1) Plug in your values: a =[(15 kg)(9.8.m/s2) – (12 kg)(9.8 m/s2)] / (15 + 12 kg) a = 1.1 m/s2 Note that for Atwood machines, if you get a = 9.8 m/s2 you have done something wrong! Atwood Machine - Example What is the tension in the rope? Use one of the Σ equations derived earlier: Σ F1 = T – m1g = m1a T = 12 kg(9.8 m/s2) + (12 kg)(1.1 m/s2) T = 130 N Atwood Machine – Example 3 How fast will mass 2 be falling at the end of 3.0 seconds if released from rest? a = - 1.1 m/s2 (negative cuz mass 2 is falling) vi = 0 m/s t = 3.0 s vf = ? vf = vi + at vf = 0 m/s + (- 1.1 m/s2 )(3.0 s) vf = - 3.3 m/s Elevator problems Elevators are a classic problem in physics. Weight will still be a force, since gravity will still be attracting the body. But if the elevator is accelerating (not moving at a constant velocity) then the acceleration will also be a force. The question is: will it be a positive or a negative force? Elevator accelerating upwards A man, mass 50 kg, is inside an elevator which begins to accelerate upward from rest at 2 m/s2. There are 2 forces acting on the man. His weight pulls down. But the scale also pushes up on the man. Since the net acceleration is 2 m/s2 upward, Newton's Second Law says Fnet = ma Fnet = (50 kg)(2 m/s2) Fnet = 100 N (upward because accel is upward) Elevator accelerating upwards Since the net force is 100 N, the upward force (of the scale) and downward force (of the weight) must equal a 100 N upward force. Fnet = F scale – W 100 N = F scale - (50 kg)(9.8 m/s2) F scale = 600 N Therefore, the scale will read 600 N Note that this is a heavier reading than if the elevator were at rest. This is often called the “apparent weight”. Elevator accelerating downwards A man, mass 50 kg, is inside an elevator which begins to accelerate downward from rest at 2 m/s2. The same 2 forces are acting on the man (weight pulls down and scale pushes up) Since the net acceleration is 2 m/s2 downward, Newton's Second Law says ΣF = ma ΣF = (50 kg)(-2 m/s2) ΣF = -100 N (downward) Elevator accelerating downwards Since the sum force is -100 N, the upward force (of the scale) and downward force (of the weight) must equal a 100 N downward force. ΣF = F scale – W -100 N = F scale - (50 kg)(9.8 m/s2) F scale = 400 N Therefore, the scale will read 400 N Note that this is a lighter reading than if the elevator were at rest. Elevator The force that the scale exerts on the man is the normal force. To sum up: N = mg if the elevator is at rest or moving at constant velocity N = mg + ma if the elevator has an upward acceleration N = mg - ma if the elevator has a downward acceleration Attached masses – unequal tension Consider 3 masses, each attached to a massless, nonstretching string, then pulled upward so that they accelerate. The accelerations will be equal (a1=a2=a3). But the tensions will NOT be equal. Attached masses – unequal tension 2 Sum up the forces on each mass: ΣF3: T3 – m3g = m3a Mass 2 will have 2 tensions acting on it, T3 pulling down and T2 pulling up, so: ΣF2: -T3 + T2 – m2g = m2a Similarly for mass 1: ΣF1: -T2 + T1 – m1g = m1a Attached masses – unequal tension 3 Solve each summation for T: ΣF3: T3 – m3g = m3a T3 = m3g + m3a T3 = m3(g + a) ΣF2: -T3 + T2 – m2g = m2a T2 = T3 + m2g + m2a T2 = T3 + m2(g + a) But T3 = m3(g + a) So T2 = m3(g + a) + m2(g + a) Attached masses – unequal tension 4 Solve each summation for T (cont’d) ΣF1: -T2 + T1 – m1g = m1a T1 = T2 + m1g + m1a T1 = T2 + m1(g + a) But T2 = m3(g + a) + m2(g + a) So T1 = m3(g + a) + m2(g + a) + m1(g + a) See the pattern? Attached masses – unequal tension 5 Each mass adds a tension m(g+a) to the string pulling the box above it. References for images http://content.answers.com/main/content/wp/en/thumb/d/d8/1 80px-Free_body_diagram_mod.png http://panda.unm.edu/Courses/Price/Phys160/F11-2.jpeg http://www.batesville.k12.in.us/Physics/PhyNet/Mechanics/Newt on2/ElevatorProblem.html http://dept.physics.upenn.edu/courses/gladney/mathphys/java/ sect3/phys_lecture_3.html