Presentation on theme: "Newton’s Laws and Dynamics"— Presentation transcript:
1 Newton’s Laws and Dynamics A force is a push or a pull that may affect the motion of an object.Forces can be exerted by objects in contact, or across distances (gravity, electrostatics, etc.)Forces always occur in pairs!Many pairs of forces can act at the same time on an object.When adding vector forces acting upon an object, sometimes a net force acts– an “unbalanced” force.
2 Newton’s Law of Inertia Because of inertia, a body will not change its state of motion unless a net force acts upon it!Newton’s Law of InertiaWithout a net force acting:A body at rest will stay at restA body in motion will stay in that state unchanged (no change in speed or direction)In other words, net forces cause acceleration– a change in velocityBall on the hand trickTable cloth pulled out trick
3 The acceleration of an object will depend: The amount of inertia an object contains is directly dependent upon the mass of the object!The acceleration of an object will depend:1) Directly upon the net force a α F2) Inversely upon the mass a α 1/mNewton’s Law of AccelerationWhen a net force acts upon an object, that acceleration is directly proportional to the net force and inversely proportional to the mass.The acceleration and net force are always in the same direction!
4 This is written in equation form as: or commonly:F = maThis would make the unit of force:kgm/s2This combination of fundamental units is renamed the Newton (N), a derived unit and the basic unit for force!
5 Forces due to GravityRecall that gravity is an acceleration with an accepted value of 9.80 m/s2.Force due to gravity would therefore be the product of the mass of an object and the acceleration rate due to gravity.This is what we commonly refer to as weight, or weight force.Pg 87Fw = mg
6 Difference between mass and weight amount of stuffmeasured in kgindependent of gravitational pulldoes not changefundamental quantityWeighta type of forcemeasured in Newtons (N)depends upon gravitychangesderived quantity
7 Weight-- Force due to gravity; and the Normal Force Gravity always pulls objects straight down toward the center of the earth!Surfaces always push back perpendicular to the surfaceThis force is called the Normal Force (N)NNFwFw
8 Finding Weight Forces v. Masses What is the weight of a 25.0 kg object?Fw = mg = 25.0 kg(9.80 m/s2) = 245 NWhat is the mass of an object weighing 125 N?m = Fwg= N9.80 m/s2= 12.8 kg
9 Solving problems with Newton’s Laws What force is required to accelerate a car weighing 15,200 N from rest to 25.0 m/s in 5.00 s?F = ma= 1550kg(5.00 m/s2)=7750NF = ?Fwg= 15,200 N9.80 m/s2Fw = 15,200 Nm == 1550 kgvo = 0Pg 82v - vo∆ta =v = 25.0 m/s= 25.0 m/s - 05.00 s∆t = 5.00 s= 5.00m/s2
10 1) A force of 1250 N is used to slow a car that weighs 14,200 N from 35.0 m/s to rest. How far does the car move during this time?2) A force of N is used to accelerate a box from rest to 20.0 m/s over a distance of 12.0 m. What is the weight of the box?3) A 14,500 N truck is slowed from 35.0 m/s to 12.5 m/s over a distance of 125 m. What braking force is required to do this?
11 4) An object used in an experiment on the moon is decelerated from 20 4) An object used in an experiment on the moon is decelerated from 20.0 m/s to rest in 5.00 s by a force of N. If the weight of the object on the moon is 83.5 N, what must be the acceleration due to gravity on the moon?5) A force of 2510 N is used on a 12,300 N vehicle to accelerate it from a speed of 10.0 m/s to 35.0 m/s. How far did it travel during this time and how long did it take to travel that distance?6) Make up a similar problem that involves Newton’s Laws of Motion.
13 Newton’s Law of Action-Reaction There is no such thing as a single force!For every push, there is a push back! For every pull, there is a pull back!Even with equal and opposite action-reaction forces there can be acceleration!If a trailer pulls on the hitch with the same force that the hitch pulls on the trailer, how does the vehicle speed up and slow down?Pg 83
14 How does an object accelerate? N = 400 NFa= 200 NFf = 100 NThere are actually three action-reaction pairs here!Page 86 Example diagramFw = 400 N
15 mA = 10.0 kgABmB = 20.0 kgWith how much force is A pushing on B?With how much force is B pushing on A?With how much force is B pushing on the table?With how much force is the table pushing on B?Each team pulls equally with 1200 N of force. What is the tension in the rope?
16 Composition and Resolution of Forces Forces are vectors:They add like vectors!3.0 N, East N, West = 1.0 N, West3.0 N + (- 4.0 N) = NThey have components as all vectors do:F = Fx2 + Fy2FFyøø = tan-1 (Fy / Fx)Fx
17 Resolving Gravitational Forces Forces due to gravity always act upon an object in the same direction:toward the center of the earth!On an inclined surface, gravity has components:FpøøFnFw
18 These are the standard equations for components of weight force on the incline plane. FpøøFnFwFnFwcosø =Fn = cosø(Fw)PgFpFwsinø =Fp = sinø(Fw)
19 A 25.0 kg box is pulled from rest across a frictionless, horizontal surface by a rope that makes a 35.0˚ angle with the horizontal. If the tension in the rope is 115 N, how fast will the box be moving after 5.00 s? What is the force of attraction between the box and the surface?m = 25.0 kgFw = mg= 245 NFaNø = 35.0˚øFyFa = 115 Nvo = 0Fx∆t = 5.00 sFwv = ?
20 Fx = F N + Fy = Fw N = ? N = Fw - Fy Fy = sinø(Fa) F m a = Fx = cosø(Fa)N + Fy = Fw= cos(35.0˚)(115 N)v = vo + a∆t= Fxm= (3.78m/s2)(5.00s)= 94.2 N25.0 kg= 18.9 m/s= 3.78 m/s2N = ?N = Fw - FyFy = sinø(Fa)
21 A 40.0 kg box sits on a ramp. What is the friction force that holds it on the ramp? What would a scale under the box read?A soccer ball is kicked with a force of ˚. What are the components of the applied force?A 325 N box is pulled from rest across a frictionless floor and gets up to a speed of 5.00 m/s in 3.00 s. What net force is required to achieve this?
22 A box of 50. 0 kg is held by a rope on a frictionless 12. 0˚ incline A box of 50.0 kg is held by a rope on a frictionless 12.0˚ incline. How long will it take the box to reach the bottom of the 3.00 m incline if the rope breaks?A 1050 kg car is being towed from rest by a tow truck. The tow rope has a tension of 5540 N and is inclined 65.0˚ above the horizontal. Assuming no friction forces, how far will the car have been towed after 3.50 s?