Mass training of trainers General Physics 1

Mass training of trainers General Physics 1
May 20-22, 2017 Application of Newton’s Law to a Single and Multiple Bodies Prof. MARLON FLORES SACEDON Physics Instructor Visayas State University Baybay City, Leyte

The Power of Imaginations
Activity #3: The Power of Imaginations 𝑵 @ rest position 𝒘

Activity #1: The Power of Imaginations 𝟐𝑵 @ rest position 𝟐𝑵

Activity #1: The Power of Imaginations 𝑭 𝟏 𝟐𝑵 Moving Frictionless table

Activity #1: The Power of Imaginations 𝟐𝑵 Moving 𝑭 𝟏 Frictionless table 𝟐𝑵

Activity #1: The Power of Imaginations 𝟐𝑵 Moving 𝟓𝑵 𝟔𝑵 Frictionless table 𝟐𝑵

Activity #1: The Power of Imaginations 𝟓𝑵 𝟐𝑵 𝟔𝑵 Moving Frictionless table

Activity #1: The Power of Imaginations 𝟓𝑵 𝟐𝑵 𝟔𝑵 Moving Frictionless table 3𝑵

Activity #1: The Power of Imaginations 𝟓𝑵 𝟐𝑵 𝟔𝑵 3𝑵 Moving Frictionless table

Objectives To study Free-body diagram.
To apply Newton’s Laws of motion on One-object and multiple body system.

Law of Action - Reaction (Inter-Action)
Recall: THE THREE LAWS OF NEWTON Law of Inertia “There is no change in the motion of a body unless a resultant force is acting upon it. If the body is at rest, it will continue at rest. If it is in motion it will continue in motion with constant speed in straight line unless there is net external force acting.” Σ 𝐹 =0 Law of Acceleration “If a net external force acts on a body, the body accelerates. The direction of acceleration is the same as the direction of the net force. Acceleration is inversely proportional to the mass of moving particle.” Σ 𝐹 =𝑚 𝑎 Law of Action - Reaction (Inter-Action) “If body A exerts a force on body B (an “action”), then body B exerts a force on body A (a “reaction”). These two forces have the same magnitude but are opposite in direction. These two forces act on different bodies.” Σ 𝐹 𝑎𝑐𝑡𝑖𝑜𝑛 =Σ 𝐹 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛

Recall: Friction Friction 𝑓 - refers to actual forces that are exerted to oppose motion - a resistance that opposes every effort to slide or roll a body over another Kinds of Friction: 1. Static friction - force that will just start the body. 2. Kinetic friction - force that will pull the body uniformly. Coefficient of friction 𝜇 - the ratio of the force necessary to move one surface over the other with uniform velocity to the normal force pressing the two surfaces to other. 𝜇= 𝑓 𝑁 𝑓=𝜇𝑁 Where: N is the normal force exerted by the contact surface to the object. 𝑦 𝑥 𝑦 𝑥 𝑦 𝑥 𝑁 𝑁 𝑁 𝑃 𝑓 𝑃 m m 𝑓 m 𝑓 𝑤 𝑤 𝜃 𝜃 𝑤

Free-Body Diagram (FBD)
FBD is a vectors diagram showing all forces acting on the body. 𝑦 𝑥 m 𝑁 𝑃 𝑤 𝑓 𝜃

FBD is a vectors diagram showing all forces acting on the body. Σ 𝐹 𝑦 =𝑁−𝑤=0 (for static equilibrium) 𝑦 𝑥 𝜮 𝑭 Σ 𝐹 𝑥 =𝑃−𝑓 (Net External force) 𝒂 N FBD of Furniture 𝜮 𝑭 = 𝑃 𝑥 −𝑓 Net external force 𝒇=𝝁𝑵 Note: If pulling force 𝑃 is less than friction 𝑓 then furniture is at rest otherwise it will move to the right.. 𝑷 𝒘=𝒎𝒈

FBD is a vectors diagram showing all forces acting on the body. FBD of box 𝑦 𝑥 𝜮 𝑭 Σ 𝐹 𝑦 =𝑁+ 𝑃 𝑦 −𝑤=0 (for static equilibrium) N 𝒂 Σ 𝐹 𝑥 = 𝑃 𝑥 −𝑓 (Net External force) P 𝑷 𝒚 𝜮 𝑭 = 𝑃 𝑥 −𝑓 𝒇=𝝁𝑵 Net external force 𝒘=𝒎𝒈 𝑷 𝒙 Note: If Σ 𝐹 𝑦 is not equal to zero then box didn’t touch the ground surface. Therefore Net external force is not equal to Σ 𝐹 𝑥 .

Exercise Problem: Construct the free-body diagram (FBD) FBD of box 𝜮 𝑭 𝒂 𝑦 𝑥 N 𝒇=𝝁𝑵 𝒘=𝒎𝒈 𝒘 𝒙 𝒘 𝒚 20 𝑜

Formulas Average velocity: 𝑣 𝑎𝑣𝑒 = 𝑠 𝑡 Average velocity: 𝑣 𝑎𝑣𝑒 = 𝑣 𝑖 + 𝑣 𝑓 2 Acceleration: 𝑎= 𝑣 𝑓 − 𝑣 𝑖 𝑡 Combined formulas a. 𝑣 𝑓 = 𝑣 𝑖 +𝑎𝑡 b. 𝑠= 𝑣 𝑖 𝑡 𝑎 𝑡 2 c. 𝑣 𝑓 2 = 𝑣 𝑖 2 +2𝑎𝑠 A force of 60 dynes acts upon a mass of 15g a) What acceleration is imparted to the body, b) What velocity will the body acquire in 8s? c) What distance will the body cover in these 8s? 𝑎 =? 𝑦 𝑥 Σ𝐹 =60 𝑑𝑦𝑛𝑒𝑠 𝑣 𝑓 =? 15g 15g 60 dynes 𝑆=? b) Solving for velocity acquire in 8s 𝑣 𝑓 a) Solving for acceleration 𝑎 c) Solving for 𝑠 𝑣 𝑖 =0 𝑣 𝑖 =0 Σ 𝐹 =𝑚𝑎 Second Law of Newton 𝑡=8𝑠 𝑡=8𝑠 𝑎=4𝑐𝑚/ 𝑠 2 𝑎=4𝑐𝑚/ 𝑠 2 60𝑑𝑦𝑛𝑒𝑠=(15𝑔)𝑎 Formula: 𝑣 𝑓 = 𝑣 𝑖 +𝑎𝑡 Formula: s= 𝑣 𝑖 𝑡+( 1 2 )𝑎 𝑡 2 𝑎=4𝑐𝑚/ 𝑠 2 ANSWER 𝑣 𝑓 =0+4(8) s= =128𝑐𝑚 ANSWER 𝑣 𝑓 =32𝑐𝑚/𝑠 ANSWER

Newton’s laws of motion
Problems: Newton’s Second Law 20𝑘𝑔 𝑁𝑒𝑔𝑙𝑒𝑐𝑡 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑐𝑜𝑛𝑡𝑎𝑐𝑡 𝑎 =? 30 𝑜 10𝑚 𝑣 𝑓 =?

Newton’s laws of motion
𝑦 𝑥 Problems: Newton’s Second Law 20𝑘𝑔 Σ 𝐹 = 𝑤 𝑥 𝑤 𝑥 𝑤 𝑦 𝑤=𝑚𝑔 𝑎=9.81 𝑠𝑖𝑛30 =4.91𝑚/ 𝑠 2 ANSWER 30 𝑜 𝑎 =? Solving for 𝑣 𝑓 𝜃= 30 𝑜 10𝑚 𝑣 𝑓 =? 𝑣 𝑖 =0 𝑁𝑒𝑔𝑙𝑒𝑐𝑡 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑐𝑜𝑛𝑡𝑎𝑐𝑡 𝑠=10𝑚 𝑎=4.91𝑚/ 𝑠 2 𝑣 𝑓 2 = 𝑣 𝑖 2 +2𝑎𝑠 Σ 𝐹 =𝑚𝑎 Second Law of Newton 𝑣 𝑓 2 =0+2(4.91)(10) 𝑤 𝑦 𝑤 𝑤 𝑦 =𝑤𝑐𝑜𝑠𝜃 𝑣 𝑓 = 9.91 𝑚/ 𝑠 2 ANSWER 𝑤 𝑥 =𝑤𝑠𝑖𝑛𝜃 𝑤 𝑥 =𝑚𝑎 30 𝑜 𝑤𝑠𝑖𝑛𝜃=𝑚𝑎 𝑚𝑔𝑠𝑖𝑛𝜃=𝑚𝑎 𝑎= 𝑚𝑔𝑠𝑖𝑛𝜃 𝑚 𝑤 𝑥 𝑎=𝑔𝑠𝑖𝑛𝜃

Problem: A horizontal cord is attached to a 6
Problem: A horizontal cord is attached to a 6.0-kg body in a horizontal table. The cord passes over a pulley at the end of the table and to this end is hung a body of mass 8 kg. Find the distance the two bodies will travel after 2s, if they start from rest. What is the tension in the cord? 𝑠 𝑡=2𝑠 Figure: 𝑚 𝐴 𝑚 𝐵 𝑚 𝐴 𝑚 𝐵 𝑚 𝐵 𝑚 𝐴 𝑠 𝑡=2𝑠

Problem: A horizontal cord is attached to a 6.0-kg body in a horizontal table. The cord passes over a pulley at the end of the table and to this end is hung a body of mass 8 kg. Find the distance the two bodies will travel after 2s, if they start from rest. What is the tension in the cord? 𝑦 𝑥 Σ 𝐹 𝐴 𝑁 −𝑇 𝑐𝑜𝑟𝑑 𝑦 𝑥 FBD of 𝑚 𝐴 FBD of 𝑚 𝐵 𝑎 𝑦 𝑥 𝑠 𝑡=2𝑠 Figure: +𝑇 𝑐𝑜𝑟𝑑 Σ 𝐹 𝐵 𝑎 𝑚 𝐴 𝑚 𝐵 𝑚 𝐴 𝑚 𝐵 𝑚 𝐴 𝑚 𝐵 𝑚 𝐵 𝑚 𝐴 −𝑤 𝐴 = 𝑚 𝐴 𝑔 𝑤 𝐵 = 𝑚 𝐵 𝑔 𝑦 𝑥 Σ 𝐹 𝑦 =𝑁− 𝑤 𝐴 =0 (static equilibrium) 𝑠 𝑡=2𝑠 Σ 𝐹 𝑥 = +𝑇 𝑐𝑜𝑟𝑑 Σ 𝐹 𝑦 =0 (static equilibrium) Net external force Σ 𝐹 𝑥 = −𝑇 𝑐𝑜𝑟𝑑 + 𝑚 𝐵 𝑔 Net external force Σ 𝐹 𝐴 = 𝑚 𝐴 𝑎 From Second law Σ 𝐹 𝐵 = 𝑚 𝐵 𝑎 From Second law Add 𝐸𝑞 1and 𝐸𝑞.2 to eliminate 𝑇 𝑐𝑜𝑟𝑑 then solve Acceleration 𝑎. 𝑇 𝑐𝑜𝑟𝑑 = 𝑚 𝐴 𝑎 Eq.1 −𝑇 𝑐𝑜𝑟𝑑 + 𝑚 𝐵 𝑔= 𝑚 𝐵 𝑎 Eq.2 𝑎= 𝑚 𝐵 𝑔 𝑚 𝐴 + 𝑚 𝐵 𝑎= 8(9.81) =5.61𝑚/ 𝑠 2 Cont’n

Problem: A horizontal cord is attached to a 6.0-kg body in a horizontal table. The cord passes over a pulley at the end of the table and to this end is hung a body of mass 8 kg. What is the tension in the cord? Find the distance the two bodies will travel after 2s, if they start from rest. 𝑦 𝑥 a) Solve for tension in the cord 𝑇 𝑐𝑜𝑟𝑑 using Eq.1 𝑠 𝑡=2𝑠 Figure: 𝑇 𝑐𝑜𝑟𝑑 =6(5.61)=33.66 N ANSWER 𝑚 𝐴 𝑚 𝐵 𝑚 𝐴 𝑚 𝐵 𝑚 𝐵 𝑚 𝐴 𝑦 𝑥 b) Solve for the distance 𝑠 of two bodies after travelling 2 sec. Given: 𝑠 𝑡=2𝑠 𝑣 𝑖 =0 𝑡=2 𝑠 𝑎=5.61 𝑚/ 𝑠 2 From kinematics: 𝑠= 𝑣 1 𝑡 𝑎 𝑡 2 𝑠= =11.22𝑚 ANSWER

ASSIGNMENT Figure: . Given: m = 10 kg h = 5 m L = 10m S1 = 2m Ø =44o
𝜇=0.06 coef. of kinetic friction Note: Particle “m” is release from rest at pt. A and moves to pt. B, then to pt.C, and finally to pt.D. Neglect the effect of the change in velocity direction at pt.B. The same value of coef. friction, from pt.A to pt.C. Projectile motion from pt.C to pt.D. Required: a. Free-body diagram of the particle at inclined plane AB. b. Free-body diagram of the particle at horizontal plane BC. c. Unbalanced force of the particle along the inclined plane AB. d. Unbalanced force of the particle along the horizontal plane BC. e. acceleration, a1 of the particle along the inclined plane AB. f. acceleration, a2 of the particle along the horizontal plane BC. g. velocity of the particle at pt.B. h. velocity of the particle at pt.C. i. Range, (S2) j. Total time of travel of particle from pt A to pt. D. Figure:

ASSIGNMENT

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