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Wiley Plus Reminder! Assignment 1 6 problems from chapters 2 and 3 Kinematics Due Monday October 5 Before 11 pm ! Chapter 4: Forces and Newton’s Laws •Force, mass and Newton’s three laws of motion •Newton’s law of gravity •Normal, friction and tension forces •Apparent weight, free fall •Equilibrium Problem 4.17: A 325 kg boat is sailing 15.0 degrees north of east at 2.00 m/s. 30 s later, it is sailing 35.0 degrees north of east at 4.00 m/s. During this time, three forces act on the boat: Auxiliary engine : F1 31.0 N to 15.0 degrees north of east Water resistance: F2 23.0 N to 15.0 degrees south of west Wind resistance : Fw Find the magnitude and direction of the force Fw . Give the direction with respect to due east. Apply Newton’s Second Law: F1 F2 Fw ma First, find the average acceleration : north vi 2 m/s 15 vf 4 m/s 35 east t 30 sec vf vi v vf vi Then use Newton’s second law: north Fw ? m a 24.1 N F1 31 N ? east 15 F1 F2 8N F2 23 N Newton’s Third Law of Motion When you exert a force on an object, it exerts a force back on you (otherwise you fall over). The ball hits the block and exerts a force F1 on it. F1 The block exerts an equal and opposite force F2 on the ball. F2 F1 F2 Action and reaction forces are equal in magnitude and opposite in direction. Action and reaction forces act on DIFFERENT OBJECTS! F1 All of the forces acting on the block All of the forces acting on the ball wblock F2 wball As action and reaction forces act on different objects, the forces do not cancel or add … you can not apply both in the same sum for net forces … calculate net forces only for each individual object. Newton’s Third Law of Motion An astronaut of mass ma = 92 kg exerts a force P = 36 N on a spacecraft of mass ms = 11,000 kg. What is the acceleration of each? Force on the spacecraft, P = 36 N to the right. Second Law: Acceleration of craft: P as ms 36 N 0.0033 m/s 2 11000 kg Reaction force of spacecraft on the astronaut is –P (Newton’s 3rd law). Second Law: Acceleration of astronaut: P 36 N aa 0.39 m/s 2 ma 92 kg The Fundamental Forces of Nature • Strong Nuclear Force: the strongest of all. Responsible for binding neutrons and protons into the nuclei of atoms. Acts over only very short distances of about 10-15 m. • Electroweak Force -a combination of: – electromagnetic force: binds electrons to nuclei to form atoms and molecules. – weak nuclear force: responsible for nuclear beta (plus)-decay (primary mechanism for energy generation in a star) • Gravity: the weakest force of all. Significant because all matter (we believe) generates an attractive gravitational force. • Perhaps, a repulsive gravitational force acting at long distances (distant galaxies appear to be moving away faster than they should if only normal gravity acts). Action and Reaction Forces: also for action at a distance Why is g=9.80 m/s2 ? The earth exerts a gravitational force on the moon. The moon exerts an equal and opposite force on the earth. This means that the force must depend on the product of both masses: F ME MM Newton’s Law of Gravitation (deduced from observations of the motion of the planets) Every particle in the universe exerts an attractive force on every other particle. The gravitational force between two masses, m1 and m2, is proportional to the product of the masses and inversely proportional to the square of the distance between their centers. Gm1m2 Fg r2 Distance between centers of gravity G is the universal gravitational constant: G 6.673 10 11 Nm2 /kg 2 m1, m2 are “gravitational masses”. In all cases seen, they are equal to the inertial masses. Gm1m3 Gm2m3 Fg 3 0 2 2 r13 r23 Three objects of mass m1, m2, m3 are located along a straight line. m2 is greater then m1. The net gravitational force acting on mass m3 is zero. Which drawing correctly represents the locations of the objects? (Remember gravity attracts m3 toward both m1 and m2, and F ~ 1/r2 ) Newton’s Law of Gravitation Newton’s law of universal gravitation can be applied to extended (finite size) objects such as planets. The prove requires the use of calculus (not done here). The general equation: Gm1m2 Fg r2 works as long as the mass of the object is distributed around the center with spherical symmetry. In this case r is the distance between the centers of the spheres (not the surfaces). Why is g=9.80 m/s2 ? Earth m w ME 5.98 10 24 kg R R 6.38 10 6 m ME What is the gravitational force (weight, w) of a mass m on the earth’s surface? 24 GmME m ( 5 . 98 10 kg ) 11 2 2 w (6.637 10 Nm / kg ) 2 R (6.38 10 6 m )2 w 9.80 m mg GME g 2 R At the earth’s surface Above earth’s surface, weight decreases with distance r from the centre of the earth as 1/r2. Weight and Gravitational Acceleration m Due to gravity – acceleration = a w mg is downwards gravitational mass Newton’s second law: w ma (w accelerates the mass) inertial mass So, the weight and g a w mg ma independent of mass All objects fall equally fast in the absence of air resistance If gravitational and inertial masses were not equal, this would not be the case! Problem 4.19: A bowling ball (mass m1 = 7.2 kg, radius r1 = 0.11 m) and a billiard ball (mass m2 = 0.38 kg, radius r2 = 0.028 m) may be treated as uniform spheres. What is the magnitude of the maximum gravitational force between them? 0.38 kg r Problem 4.26 (6th edition): The weight of an object is the same on planets A and B. The mass of planet A is 60% that of planet B. Find the ratio of the radii of the planets. m w M r Problem 4.28: Three uniform spheres are located at the corners of an equilateral triangle with sides of 1.2 m. Two of the spheres have a mass of 2.8 kg. The third sphere is released from rest. Considering only the gravitational forces the spheres exert on each other, what is the magnitude of its initial acceleration? m2 ? F1 F2 1.2 m 30 30 1.2 m a m1 2.8 kg m1 2.8 kg