Module 7 -phys_13___11a_application_of__newtons_law (1)
- 1. PHYSICS 13 & 11A GENERAL PHYSICS 1 MARLON FLORES SACEDON Dynamics of Motion: Application of Newton’s Laws of Motion
- 2. 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. APPLICATION OF NEWTON’S LAWS OF MOTION Friction 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 Where: N is the normal force exerted by the contact surface to the object. Normal force is always acting against and perpendicular to the surface in contact. m 𝑁 𝑃 𝑤 𝑓 𝜃 𝑦 𝑥 𝜃 𝑓 = 𝜇𝑁
- 3. Free-Body Diagram (FBD) m 𝑁 𝑃 𝑤 𝑓 𝜃 𝑦 𝑥 𝜃 FBD is a vectors diagram showing all forces acting on the body. APPLICATION OF NEWTON’S LAWS OF MOTION
- 4. Free-Body Diagram (FBD) FBD of Furniture 𝑦 𝑥 𝜮𝑭 𝒂 𝑷 𝒘 = 𝒎𝒈 N 𝒇 = 𝝁𝑵 Σ 𝐹𝑦 = 𝑁 − 𝑤 = 0 (for static equilibrium) Σ 𝐹𝑥 = 𝑃 − 𝑓 (Net External force) 𝜮𝑭 = 𝑃𝑥 − 𝑓 Net external force Note: If pulling force 𝑃 is less than friction 𝑓 then furniture is at rest otherwise it will move to the right.. APPLICATION OF NEWTON’S LAWS OF MOTION FBD is a vectors diagram showing all forces acting on the body.
- 5. Free-Body Diagram (FBD) N P 𝒘 = 𝒎𝒈 𝑦 𝑥 𝑷 𝒙 𝑷 𝒚 FBD of box Σ 𝐹𝑦 = 𝑁 + 𝑃𝑦 − 𝑤 = 0 (for static equilibrium) 𝒇 = 𝝁𝑵 Σ 𝐹𝑥 = 𝑃𝑥 − 𝑓 (Net External force) 𝜮𝑭 = 𝑃𝑥 − 𝑓 Note: If Σ 𝐹𝑦 is not equal to zero then box didn’t touch the ground surface. Therefore Net external force is not equal to Σ 𝐹𝑥. Net external force 𝜮𝑭 𝒂 APPLICATION OF NEWTON’S LAWS OF MOTION FBD is a vectors diagram showing all forces acting on the body.
- 6. Free-Body Diagram (FBD) Exercise Problem: Construct the free-body diagram (FBD) 𝒘 = 𝒎𝒈 FBD of box 𝒘 𝒚 𝒘 𝒙 20 𝑜 APPLICATION OF NEWTON’S LAWS OF MOTION
- 7. 𝑠 𝑡 = 2𝑠 𝑚 𝐴 𝑚 𝐵 Figure: 𝑚 𝐴 𝑚 𝐵 𝑚 𝐵 𝑚 𝐴 𝑠 𝑡 = 2𝑠 APPLICATION OF NEWTON’S LAWS OF MOTIONProblem: A horizontal cord is attached to a 6.0-kg body in a horizontal table whose coefficient of kinetic friction 𝜇 𝑘 = 0.25. 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 acceleration of the system and tension in the cord? Find the distance the two bodies will travel after 2s, if they start from rest.
- 8. Problem: A horizontal cord is attached to a 6.0-kg body in a horizontal table whose coefficient of kinetic friction 𝜇 𝑘 = 0.25. 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 acceleration of the system and tension in the cord? Find the distance the two bodies will travel after 2s, if they start from rest. 𝑚 𝐵 𝑚 𝐴 𝑠 𝑡 = 2𝑠 𝑚 𝐴 𝑚 𝐵 Figure: 𝑚 𝐴 𝑚 𝐵 𝑠 𝑡 = 2𝑠 𝑦 𝑥 𝑦 𝑥𝑚 𝐴 𝑦 𝑥 𝑚 𝐵 +𝑇𝑐𝑜𝑟𝑑 −𝑇𝑐𝑜𝑟𝑑 −𝑤 𝐴= 𝑚 𝐴 𝑔 𝑤 𝐵 = 𝑚 𝐵 𝑔 𝑁 Σ 𝐹𝐴 Σ 𝐹𝐵 𝑎 𝑎 FBD of 𝑚 𝐴 FBD of 𝑚 𝐵 Σ 𝐹𝑥 = +𝑇𝑐𝑜𝑟𝑑 − 𝑓 Σ 𝐹𝑦 = 𝑁 − 𝑤 𝐴 = 0 (static equilibrium) Σ 𝐹𝐴 = 𝑚 𝐴 𝑎 From Second law 𝑇𝑐𝑜𝑟𝑑 − 𝑓 = 𝑚 𝐴 𝑎 Σ 𝐹𝑦 = 0 (static equilibrium) Σ 𝐹𝑥 = −𝑇𝑐𝑜𝑟𝑑+𝑚 𝐵 𝑔 Σ 𝐹𝐵 = 𝑚 𝐵 𝑎From Second law −𝑇𝑐𝑜𝑟𝑑+𝑚 𝐵 𝑔 = 𝑚 𝐵 𝑎 Substi. 𝐸𝑞 2 in 𝐸𝑞. 1 to eliminate 𝑇𝑐𝑜𝑟𝑑 then solve Acceleration 𝑎. 𝑎 = 𝑔(𝑚 𝐵 − 𝜇 𝑘 𝑚 𝐴) 𝑚 𝐴 + 𝑚 𝐵 𝑎 = 9.81(8 − 0.25(6) 6 + 8 = 4.55 𝑚/𝑠2 Net external force Net external force APPLICATION OF NEWTON’S LAWS OF MOTION −𝑓 𝑇𝑐𝑜𝑟𝑑 = 𝑚 𝐵 𝑔 − 𝑚 𝐵 𝑎 Eq.2 = 𝑇𝑐𝑜𝑟𝑑 − 𝜇 𝑘 𝑚 𝐴 𝑔 𝑇𝑐𝑜𝑟𝑑 − 𝜇 𝑘 𝑚 𝐴 𝑔 = 𝑚 𝐴 𝑎 Eq.1
- 9. 𝑚 𝐵 𝑚 𝐴 𝑠 𝑡 = 2𝑠 𝑚 𝐴 𝑚 𝐵 Figure: 𝑚 𝐴 𝑚 𝐵 𝑠 𝑡 = 2𝑠 𝑦 𝑥 𝑦 𝑥 a) Solve for tension in the cord 𝑇𝑐𝑜𝑟𝑑 using Eq.1 𝑇𝑐𝑜𝑟𝑑 = 6 4.55 + 0.25(6)(9.81)= 42.08 N ANSWER b) Solve for the distance 𝑠 of two bodies after travelling 2 Given: 𝑣𝑖 = 0 𝑡 = 2 𝑠 𝑎 = 4.55 𝑚/𝑠2 From kinematics: 𝑠 = 𝑣1 𝑡 + 1 2 𝑎𝑡2 𝑠 = 0 2 + 1 2 4.55 22 = 9.10 𝑚 ANSWER APPLICATION OF NEWTON’S LAWS OF MOTION 𝑇𝑐𝑜𝑟𝑑 − 𝑓 = 𝑚 𝐴 𝑎 Eq.1 𝑇𝑐𝑜𝑟𝑑 = 𝑚 𝐴 𝑎 + 𝜇 𝑘 𝑚 𝐴 𝑔 Problem: A horizontal cord is attached to a 6.0-kg body in a horizontal table whose coefficient of kinetic friction 𝜇 𝑘 = 0.25. 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 acceleration of the system and tension in the cord? Find the distance the two bodies will travel after 2s, if they start from rest.
- 10. 𝒎 𝟏 Atwood Machine 𝑤1 = +𝑚1 𝑔 −𝑇𝑐𝑜𝑟𝑑 Σ 𝐹1 𝑎 FBD of 𝑚1 𝑦 +𝑥 Σ 𝐹𝑥 = +𝑤1 − 𝑇𝑐𝑜𝑟𝑑 Net external force Σ 𝐹1 = 𝑚1 𝑎 From Second law +𝑤1 − 𝑇𝑐𝑜𝑟𝑑 = 𝑚1 𝑎 𝒎 𝟐 −𝑤2= 𝑚2 𝑔 +𝑇𝑐𝑜𝑟𝑑 Σ 𝐹2 𝑎 FBD of 𝑚2 𝑦 +𝑥 Σ 𝐹𝑥 = −𝑤2 + 𝑇𝑐𝑜𝑟𝑑 Net external force Σ 𝐹2 = 𝑚2 𝑎 From Second law −𝑤2 + 𝑇𝑐𝑜𝑟𝑑 = 𝑚2 𝑎 𝑇𝑐𝑜𝑟𝑑 = 𝑚2 𝑎 +𝑤2 𝑚1 𝑔 − 𝑇𝑐𝑜𝑟𝑑 = 𝑚1 𝑎 Eq.1 Substi Eq.2 in eq.1 𝑚1 𝑔 − (𝑚2 𝑎 +𝑚2 𝑔) = 𝑚1 𝑎 𝑇𝑐𝑜𝑟𝑑 = 𝑚2 𝑎 +𝑚2 𝑔 Eq.2 𝑎 = 𝑔(𝑚1 − 𝑚2) 𝑚1 + 𝑚2
- 11. Problem: Two blocks connected by a cord passing over a small, frictionless pulley rest on a double inclined plane. Block A whose mass is 𝑚 𝐴 = 100 𝑘𝑔 is on left side and Block B with mass 𝑚 𝐵 = 50 𝑘𝑔 is on the right side inclined plane.. The coefficient of kinetic frictions between block and inclined plane on the right and left side are 𝜇 𝐴 = 0.30 and 𝜇 𝐵 = 0.25 respectively. The angles of the inclined plane are 𝜃 = 30 𝑜 (left side) and 𝛽 = 53.1 𝑜 (right side). (a) Calculate the acceleration "𝑎" of the system, (b) calculate also the tension in the cord 𝑇𝐶. 𝜃 𝛽
- 12. . 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. ASSIGNMENT Figure:
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