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Transcript
Ch 4: Newton’s Laws Demo time: Do you remember your Newton’s Laws? Coin on a paper over a paper cup Dropping a hammer Roll of paper towel Glass bottles and dollar bill Table cloth pull Inertia weight Dropping ball Two people on a rope Air puck Van de Graff Generator Break a bottle by hitting the top 4-1 Newton’s 1st Law (N1): The Law of Inertia Newton’s Laws are only valid within an inertial reference frame. See the definition below. 4-2 Force, Mass and Newton’s 2nd Law (N2) Most people think of N2 as F = ma, but really, what the law says is that the amount to which something will change it’s velocity (or accelerate) is in direct proportion to how much (external) force is applied. On the flip side, if you apply an external force to a mass, you will cause that mass to change its current velocity, or accelerate. If you subsequently triple that force, all other things remaining equal, you will accelerate it three times as much. Thus, force is morr accurately proportional to acceleration. Q: Is F always in the same direction as a? A much better interpretation is: F α ma Algebraically manipulating this equation to express it in terms of F a m a, , we see that the acceleration of an object is directly proportional to its net force, and inversely proportional to its mass. When the net force on an object is zero, there is no net force acting on it, so it will not accelerate. Q: Can it still be moving? A: Yes; with constant velocity. Q: If there is a net force, how does the object move? A: In the direction of the of the net force, which is the same as the direction of the net acceleration Q: What do you call an object’s resistance to the change in its motion? A: Inertia. Actually, it may also be thought of as an objects resistance to acceleration. Q: How do you measure an object’s resistance to the change in its motion? A: By measuring its mass. Before going on, 1. What is the SI unit of Force? Mass? Acceleration? 2. What is a Newton, conceptually and with respect to units? 3. How do you convert: Lb → kg? Kg → N? Lb → N? 1 kg = 2.20 lbs 1 N = 1 kg ∙ m/s2 9.81 N = 2.20 kg Imagine you accelerate? on a 12 kg with a force of 2N. How much will it If you apply the same , but you double the mass of the to 24 kg, how much will it accelerate now? How does this new value compare to the original acceleration? Explain this in terms of N2. This all boils down to the following law which allows you to explore the comparative masses and / or accelerations of two objects undergoing an equivalent force: See Ex 4-1, p 88 A given force produces an acceleration of 5.0 m/s2 on mass m1 = 1.0 kg (the standard). When an equal force is applied to a carton of ice cream of mass m 2, it produces an acceleration of 11 m/s2. What is the mass of the carton of ice cream? What is the magnitude of the force? F1 = m1a1 F2 = m2a2 If F1 = F2, set the two equations equal to each other m1a1 = m2a2 and use the above law m2 = a1 = 5 m/s2 m1 a2 11 m/s2 m2 = 5 m1 = 5 (l kg) = 0.45 kg 11 11 So, if F = m1a1 = (1.0 kg)(5.0 m/s2) = 5.0 N, for either force See Ex 4-2, p 88-89 You’re stranded in space away from your spaceship. (Eeeeek!!) Fortunately, you have a propulsion unit that provides a constant force of F for 3.0 seconds. After 3.0 seconds, you have moved 2.25 m. If your mass is 68 kg, find F. Assume the propulsion unit pushes you to the right. If the force acting on you is constant, your acceleration is constant. Remember, there is force acting on you. Don’t confuse this with constant velocity! Use the equations from ch 2 to find acceleration. Then, use acceleration to find force. Can you make an assumption about v0? Is the acceleration in the x-direction zero this time? Why / why not? xf = x0 + v0t + ½ axt2 = ½ axt2 ax = 2∆x = 2(2.25 m) = 0.50 m/s2 t2 (3.0 s)2 Fx = max = (68 kg)(0.50 m/s2) = 34 N See Ex 4-3, p 89. A particle of mass 0.40 kg is subjected simultaneously to two forces: F1 = - 2 Ni – 4 Nj and F2 = - 2.6 N i + 5 N j. If the particle is at the origin and starts from rest with t = 0, find: a) Its position vector r b) Its velocity v at t = 1.6 s a) → → → → r = r0 + v 0 t + ½ a t 2 → → a =F m → → → F = F1 + F2 = -4.6 N i + 1.0 N j → → a = F = -11.5 m/s2 i + 2.5 m/s2 j m → → → → rf = r0 + v0 t + ½ a t2 → → r = ½ a t2 = ½ axt2 i + ½ ayt2 j = -14.7 m i + 3.20 m j evaluate at t = 1.6 s b) → → v = a t = (-11.5 m/s2 i + 2.5 m/s2 j ) t evaluate at t = 1.6 s = -18.4 m/s i + 4.00 m/s j 4-3 The Force Due to Gravity: Weight Q: Is weight a force? How do you know? A: Yep; it causes an object to accelerate. The reason for this is gravity. If the only force acting on an object is its weight, the object is said to be in true free-fall. Q: How does N2 look in terms of weight and free-fall? W = mg For all intensive purposes, g is assumed to be the same for all objects within an inertial reference frame. Thus, weight within that reference frame is the same as mass. We assume g = 9.81 m/s2 = 9.81 N/kg (check the units!!) Q: Is g always 9.81 m/s2 everywhere on earth? A: Nope. Objects weigh an eenie-meanie bit less at high elevations than at sea level. G actually varies inversely with the square of the distance to the center of the earth (Newton’s Law of Gravitation). Earth is not a perfect sphere, but instead, is slightly flattened at the poles. Q: Is weight an intrinsic property of an object? Why / why not? A: Nope; it varies with location. Q: What determines the sense of weight that you observe? A: The forces that balance it out. DEF: Apparent weight = the feeling weight as compared to the other force(s) that counteract that weight. DEF: Weightlessness = the condition in which there is no force to balance your weight; or apparent weight = 0. Astronauts orbiting in satellites experience weightlessness. Q: Are astronauts in an orbiting satellite in free-fall? A: Yep. The only force acting on them is their own weight, which produces an acceleration of g. Since there is no force balancing the force of gravity, their apparent weight is zero (unless you consider the weight of the diaper acting against them……err, no .) Units of Force and Mass Kg, m, s, N, amu = u (remember this from freshman chemistry?) = 1/12 the mass of a carbon-12 atom. Old-timers still use ft, lbs, so g might be expressed as 32.2 ft/s2 See Ex 4-4, p 91 If the net force acting on a 130-lb student is 25 lbs. What is her acceleration? a = F Since w = F due to gravity = mg, F__ , 25 lb_____ = 6.19 ft/s2 m w/g (130 lb)/(32 ft/s2) Do Ch 4 HW Day 1