# Forces and Motion.

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Forces and Motion

What You Will Learn Newton’s Laws of Motion Motion Terminology Sample Problems

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.

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).

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.

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.

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.

Tension Wall or attachment point T Fg Fa Ty= T sin  Tx= T cos 
 Fy = 0 = Ty + Fg 0 = T sin 120o + Fg – T sin 120o = Fg T= F 𝑔 – sin 120o  Fx = 0 = Tx + Fa 0 = T cos 120o + Fa – T cos120o = Fa T= F 𝑎 – cos 120o

What happens to tension as  gets bigger?
y Wall or attachment point T Fg Fa Ty= T sin  Tx= T cos  x What happens to tension as  gets bigger?

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

Mouse or Enter to Continue
Tension Fg = –100 N Fax is the applied force y x y x y x y x 90o T1 Fg 60o T2 Fg Fa2 45o T3 Fg Fa3 30o T4 Fg Fa4  F = 0 = T1 + Fg –T1 = Fg T1 = – Fg T1 = – (100 N) T1 = 100 N  Fy = 0 = T2y + Fg 0 = T2 sin 120o + Fg – T2 sin 120o = Fg T 2 = (−100 𝑁) − sin 120𝑜 T2 = 115 N  Fy = 0 = T3y + Fg 0 = T3 sin 135o + Fg – T3 sin 135o = Fg T 3 = (−100 𝑁) − sin 135𝑜 T3 = 141 N  Fy = 0 = T4y + Fg 0 = T4 sin 150o + Fg – T4 sin 150o = Fg T 4 = (−100) − sin 150𝑜 T4 = 200 N Mouse or Enter to Continue

Tension Fg = –100 N Fax is the applied force T1 Fg T2 Fg Fa2 T3 Fg Fa3
T1 = 100 N T3 = 141 N T2 = 115 N T4 = 200 N

Tension m = 1.00  103 kg. Find T, no acceleration  Fy = 0 = T + Fg
pulley  Fy = 0 = T + Fg –T = Fg 𝐓=1000 𝑘𝑔 𝑚 𝑠 2 sin 270o T = – (– 9800 N) T = 9.8 103 N

Tension m = 1.00  103 kg. Find T, anet = 3.25 𝑚 𝑠 2
Fg pulley  Fy = Fnet = T + Fg manet = T + Fg 1000 𝑘𝑔 3.25 𝑚 𝑠 2 =𝐓+1000 kg 9.8 𝑚 𝑠 2 sin 270o 3250 N=𝐓−9800 N T = N T = 1.3 104 N

What You Have Hopefully Learned
Newton’s Three Laws of Motion Motion Terminology Sample Tension Problems

And Now, The Crane Problem.
END OF LINE And Now, The Crane Problem.

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. 30o T1 100o T2 mass = 2000 kg

Simplified Force Diagram
y x Simplified Force Diagram o 100o T1 T2 T1y = T1 sin 130o T2y = T2 sin 30o T1x = T1 cos 130o T2x = T2 cos 30o Fy = 0 = T1y + T2y + Fg 0 = T1 sin 130o + T2 sin 30o – N 19600 N = T1 sin 130o + T2 sin 30o 19600 N = T1 sin 130o + − 𝐓 1 cos 130o cos 30o sin 30o 19600 N = 𝐓 1 sin 130o − cos 130o cos 30o sin 30o 𝐓 1 = N sin 130o − cos 130o cos 30o sin 30o = N T1 = N = 1.7  104 N Fg = mg = 2000 kg 𝑚 𝑠 2 sin 270o Fg = –19600 N Fx = 0 = T1x + T2x 0 = 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 2 N = 1.3  104 N = T2