2 What You Will LearnNewton’s Laws of MotionMotion TerminologySample Problems
3 Newton’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.
4 Newton’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).
5 Newton’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.
6 Types 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.
7 Types 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.
8 Tension 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
9 What happens to tension as gets bigger? yWall or attachment pointTFgFaTy= T sin Tx= T cos xWhat happens to tension as gets bigger?
10 Tension 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
11 Mouse 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
12 Tension 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
13 Tension 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
14 Tension 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
15 What You Have Hopefully Learned Newton’s Three Laws of MotionMotion TerminologySample Tension Problems
16 And Now, The Crane Problem. END OF LINEAnd Now, The Crane Problem.
17 Two 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
18 Simplified 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