Forces Newton’s Second Law

Newton’s Second Law States that the greater the force applied, the greater the acceleration and the greater the mass the less the acceleration F=ma

Weight Mass and weight are not the same! Weight is equal to the force of gravity F=ma Acceleration due to gravity is 9.8 m/s2 Force due to gravity (weight)=mass x 9.8 m/s2

Example of Weight A boy has a mass of 50 kilograms, what is his weight? An object exerts a force of 40 N on the ground, what is the mass? 490 N 5.88 kg

Example of Weight Tina exerts 513.4 Newtons of force, what is her weight in pounds? 1 pound=.45 kilograms F=ma F=mg M=F/g M= 513.4/9.8 M=52.387 kg 116.4 lbs

Net Force When you have two forces Add them- when they are working together Subtract them- when they are opposing each other

Newton’s Second Law Arnold needs to lift a 35.0 kg rock. If he exerts an upward force of 502 Newtons on the rock, what is the rock’s acceleration? F net= F applied- Fg 502- 35* 9.8= 502-343=159 N F=ma a=F/m A=159/35= 4.54 m/s2

An Example Using a Constant Velocity A student raises their 15 kg backpack from the floor at a constant velocity of 5.0 m/s. How much force must the student apply?

Example Annie is holding a stuffed dog, with a mass of 0.30 kg, when Sarah decides she wants it and tries to pull it away. If Sarah pulls horizontally on the dog with a force of 10.0 N and Annie pulls with a horizontal force of 11.0 N, what is the horizontal acceleration of the dog? Fnet= F annie- F Sarah Fnet= 1 N towards Annie A=F/m A= 1/0.30= 2.2 m/s2

Apparent Weight How does a scale work? Apparent weight is the force exerted by the scale Weightlessness is when there are no contact forces pushing up on the object Draw a free body diagram of a scale. There are two forces, Fg and Fspring going up No Fg will be different

Elevator Problem Your mass is 75.0 kg and you are standing on a bathroom scale in an elevator. Starting from rest the accelerates 2.0 m/s2 for 2.00 seconds and then continues at a constant speed. Is the scale reading during acceleration greater than, equal to, or less than the scale reading when the elevator is at rest? F net= F scale- Fg so Fscale= Fnet+ Fg At rest F scale=Fg F scale= Mg F scale= 75(9.8) Fscale 735 N at rest When accelerating up the Fn is greater because you are moving up A net = 2+ 9.8= 11.8 F= 75 (11.8) F = 885 N

Elevator Problems Tips Apparent weight changes when you are accelerating or decelerating, NOT when you are at rest OR going at a constant speed Accelerating upwards= Weight is greater Accelerating downwards= weight is less Decelerating upwards=weight is less Decelerating Downwards= Weight is greater

Drag Force and Terminal Velocity Drag Force- force exerted by a fluid (liquid OR a gas) on an object moving through the fluid Depends on Motion of the object Properties of object and fluid As speed increases for the object the drag force also increases When drag force is equal to the force of gravity a terminal velocity is reached Terminal Velocity is the highest velocity can reach when falling

An example with UAM A ball is kicked and slides to a stop. The ball has an initial velocity of -4.5 m/s and the force of friction is 1.4 N. The ball has a mass of 47 grams. What is the displacement of the ball. Draw a free body diagram- Force of friction- right Fg- down Fn-up No applied force because it is sliding to a stop Use Ff to find acceleration Ff=ma a= 1.4/.047 a = 29.7 m/s Vf2= Vi2 + 2 a d D=.34 m

Newton’s 3rd Law

Newton’s Thrid Law Recap For every force there is an opposing force in the opposite direction F(a on b)=-F(b on a)

Action Reaction Pairs A string pulls a book A tennis racket hits a tennis ball Chalk writes on a chalk board

Example When a softball with a mass of 0.18 kg is dropped, its acceleration toward Earth is equal to g, the acceleration due to gravity. What is the force on Earth due to the ball and what is Earth’s resulting acceleration? Earth’s mass is 6.0 x 10 24 kg. Known- m (ball)=0.18 kg m(earth)=6.0 x 10 24 kg g = 9.80 m/s 2 Unknown F (earth) a (earth) F ball= F earth F (ball)= .18(9.8) = 1.8 N Technically it is negative F ball= - F earth 1.8=6 x 1024 (a) a = 2.9 x 10 -25 m/s2

Example of 3rd Law and Tension A 50.0 kg bucket is being lifted by a rope. The rope will not break if the tension is 525 N or less. The bucket started at rest, and after being lifted 3.0 m, it is moving at 3.0 m/s. If the acceleration is constant, if the rope in danger of breaking? Known- m=50.0 kg Vi=0 Vf=3.0 m/s d=3.0 m Ft=? Ft is up and Fg is down so subtract to get the net force F net= Ftension – Fg Ftension=Fnet + Fg A =1.5 m/s 2 Fg= 50 (9.8)= 490 N Ft= 50 (1.5)= 75 N Fnet= 490+ 75= 565 N

Friction

Static Friction Versus Kinetic Friction Friction- force exerted by two surfaces rubbing against each other Static Friction- No motion Acts in response to other forces, but there is a limit to how large static friction can be (maximum) Once the maximum point has been met and movement starts static friction kinetic friction Kinetic Friction- Motion Is less than the static friction

Equations Kinetic Friction Static friction fk=μk Fn Kinetic Friction= kinetic coefficient of friction times the normal force Static friction Fs=μs Fn Static friction= static coefficient of friction time the normal force

Example A force of 98 N is exerted on a 25.0 kg box and the box is moving. If the coefficient of friction (kinetic) is 0.20, what is the acceleration? The normal force is equal to the weight. Fn=25 (9.8)=245 Ff= .2(245)=49 Fnet=98-49=49 F=ma 49=25a a=1.96

Tips for friction problems Use static coefficient when not moving Use kinetic coefficient when moving The normal force is equal to the weight of the object