2What You Will LearnNewton’s Laws of MotionMotion TerminologySample Problems
3Newton’s Three Laws of Motion Newton’s First Law of Motion – An object at rest will remain at rest and an object in motion will continue in motion at a constant velocity in a straight line unless it is acted upon by a net force.This is often called the Law of Inertia because it implies that any mass offers resistance to a change in its velocity.What are other implications?No applied force to an object means no change in its velocity. Or, if velocity changes, then there has to be a net force that is making that change.If there is an acceleration, then there has to be a velocity change and a net force present.
4Newton’s Three Laws of Motion Newton’s Second Law of Motion – An acceleration is directly proportional to the applied force and inversely proportional to the mass of the object. For us; F = ma.What are the implications?You can not have a force with out an acceleration, since mass of an object does not change.Force is a vector (has magnitude and direction).
5Newton’s Three Laws of Motion Newton’s Third Law of Motion – States that all forces come in pairs that are equal in magnitude but opposite in direction. To paraphrase; For every action there is an equal but opposite reaction.What are the implications?You can not have a force with out an acceleration, since mass of an object does not change.
6Types of Forces System – An object of interest. Force – A push or a pull.Contact Force – Physically touching.Field Force – Exert without contact.Agent – It is a contact or field force and is specifically affects a given system.Free Body Diagram – Physical representation that shows the forces acting on a system.Net Force – Is the vector sum of the forces acting on a system. Symbolized by Fnet.Equilibrium – A system in which the velocity remains constant. Note: an object at rest is just a special case of velocity, namely v = 0.Weight – The gravitational force experience by an object. Fg = mg.Apparent Weight – A weight that is not true weight, but it is one you experience due to your current conditions.
7Types of Forces Weightlessness – The absents of apparent weight. Drag Force – The force exerted by a fluid on an object opposing motion through the fluid. Note: air is a fluid.Interaction Pair – Newton’s Third Law. All forces come in pairs.Tension – Force exerted by a rope, string, or cable. The rope, string, or cable is generally considered massless.
8Tension Wall or attachment point T Fg Fa Ty= T sin Tx= T cos Fy = 0 = Ty + Fg0 = T sin 120o + Fg– T sin 120o = FgT= F 𝑔 – sin 120o Fx = 0 = Tx + Fa0 = T cos 120o + Fa– T cos120o = FaT= F 𝑎 – cos 120o
9What happens to tension as gets bigger? yWall or attachment pointTFgFaTy= T sin Tx= T cos xWhat happens to tension as gets bigger?
10Tension Team Work Fg = –100 N Fax is the applied force T1 Fg T2 Fg Fa2 What is the tension in T1, T2, T3, and T4?Team Work
11Mouse or Enter to Continue TensionFg = –100 N Fax is the applied forceyxyxyxyx90oT1Fg60oT2FgFa245oT3FgFa330oT4FgFa4 F = 0 = T1 + Fg–T1 = FgT1 = – FgT1 = – (100 N)T1 = 100 N Fy = 0 = T2y + Fg0 = T2 sin 120o + Fg– T2 sin 120o = FgT 2 = (−100 𝑁) − sin 120𝑜T2 = 115 N Fy = 0 = T3y + Fg0 = T3 sin 135o + Fg– T3 sin 135o = FgT 3 = (−100 𝑁) − sin 135𝑜T3 = 141 N Fy = 0 = T4y + Fg0 = T4 sin 150o + Fg– T4 sin 150o = FgT 4 = (−100) − sin 150𝑜T4 = 200 NMouse or Enter to Continue
12Tension Fg = –100 N Fax is the applied force T1 Fg T2 Fg Fa2 T3 Fg Fa3 T1 = 100 NT3 = 141 NT2 = 115 NT4 = 200 N
13Tension m = 1.00 103 kg. Find T, no acceleration Fy = 0 = T + Fg pulley Fy = 0 = T + Fg–T = Fg𝐓=1000 𝑘𝑔 𝑚 𝑠 2 sin 270oT = – (– 9800 N)T = 9.8 103 N
14Tension m = 1.00 103 kg. Find T, anet = 3.25 𝑚 𝑠 2 Fgpulley Fy = Fnet = T + Fgmanet = T + Fg1000 𝑘𝑔 3.25 𝑚 𝑠 2 =𝐓+1000 kg 9.8 𝑚 𝑠 2 sin 270o3250 N=𝐓−9800 NT = NT = 1.3 104 N
15What You Have Hopefully Learned Newton’s Three Laws of MotionMotion TerminologySample Tension Problems
16And Now, The Crane Problem. END OF LINEAnd Now, The Crane Problem.
17Two cranes are holding a load that has a mass that will be given to you by the teacher. Find the value of T1 and T2. Show the force diagram and the vector equation needed to solve this problem.30oT1100oT2mass = 2000 kg
18Simplified Force Diagram yxSimplified Force Diagramo100oT1T2T1y = T1 sin 130oT2y = T2 sin 30oT1x = T1 cos 130oT2x = T2 cos 30oFy = 0 = T1y + T2y + Fg0 = T1 sin 130o + T2 sin 30o – N19600 N = T1 sin 130o + T2 sin 30o19600 N = T1 sin 130o + − 𝐓 1 cos 130o cos 30o sin 30o19600 N = 𝐓 1 sin 130o − cos 130o cos 30o sin 30o𝐓 1 = N sin 130o − cos 130o cos 30o sin 30o = NT1 = N = 1.7 104 NFg = mg = 2000 kg 𝑚 𝑠 2 sin 270oFg = –19600 NFx = 0 = T1x + T2x0 = T1 cos 130o + T2 cos 30o– T1 cos 130o = T2 cos 30o− 𝐓 1 cos 130o cos 30o = 𝐓 2−17000 cos 130o cos 30o = 𝐓 2N = 1.3 104 N = T2