Wiley Plus Reminder! Assignment 1 6 problems from chapters 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 35 kg boat is sailing 15.0 degrees north of east at.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 : F 1 31.0 N to 15.0 degrees north of east Water resistance: F 3.0 N to 15.0 degrees south of west Wind resistance : F w Find the magnitude and direction of the force. Give the direction with respect to due east. F w Apply Newton s Second Law: F 1 F F w ma
First, find the average acceleration : north v i m/s 15 v f 4m/s 35 east t 30sec v i v f v v f v i
Then use Newton s second law: north m a F w? 4.1 N 15 F 3N F 1 F? 8N F 1 31N east
Newton s Third Law 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 F 1 on it. The block exerts an equal and opposite force F on the ball. F 1 F F 1 F Action and reaction forces are equal in magnitude and opposite in direction.
Action and reaction forces act on DIFFERENT OBJECTS! All of the forces acting on the block F 1 All of the forces acting on the ball F w block w ball 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 An astronaut of mass m a = 9 kg exerts a force P = 36 N on a spacecraft of mass m s = 11,000 kg. What is the acceleration of each? Force on the spacecraft, P = 36 N to the right. Second Law: Acceleration of craft: a s P m s 36 11000 N kg 0. 0033 m/s Reaction force of spacecraft on the astronaut is P (Newton s 3rd law). Second Law: Acceleration of astronaut: a a P m a 36N 0. 39m/s 9kg
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-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).
The earth exerts a gravitational force on the moon. The moon exerts an equal and opposite force on Earth. This means that the force must depend on the product of both masses: Action and Reaction Forces: also for action at a distance Why is g=9.80 m/s? F M E M M
Newton s Law of Gravitation (deduced from observations of the motion of the planets) Every particle in the universe exerts an attractive gravitational force on every other particle. The gravitational force between two masses, m 1 and m, is proportional to the product of the masses and inversely proportional to the square of the distance between their centers. Gm1m F g r Distance between centers of gravity G is the universal gravitational constant: G 6. 67310 11 Nm /kg m 1, m are gravitational masses. In all cases seen, they are equal to the inertial masses.
Gm m Gm m 1 3 3 F g 3 r13 r3 0 Three objects of mass m 1, m, m 3 are located along a straight line. m is greater then m 1. The net gravitational force acting on mass m 3 is zero. Which drawing correctly represents the locations of the objects?
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: Gm1m F g r 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/s? m w R Earth M E 5.9810 R 6.38 10 4 6 kg m M E What is the gravitational force (weight, w) of a mass m on earth s surface? GmM w R E ( 6. 637 10 11 Nm / kg 4 m ( 5. 9810 kg ) ) 6 ( 6. 3810 m) GM w 9. 80 m mg g R E At earth s surface Above earth s surface, weight decreases with distance r from the centre of the earth as 1/r.
Weight and Gravitational Acceleration m w mg Due to gravity acceleration = a is downwards gravitational mass F ma Newton s second law: ( accelerates the mass) So, the weight inertial mass w mg ma F F and g a 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 m 1 = 7. kg, radius r 1 = 0.11 m) and a billiard ball (mass m = 0.38 kg, radius r = 0.08 m) may be treated as uniform spheres. What is the magnitude of the maximum gravitational force between them? r 0.38kg