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Physics 13 General Physics 1
Dynamics of Motion: Application of Newtonโs Laws of Motion MARLON FLORES SACEDON
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Application of Newtonโs Laws of Motion
Friction Friction ๐ - 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 ๐ - 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. ๐= ๐ ๐ ๐=๐๐ Where: N is the normal force exerted by the contact surface to the object. ๐ฆ ๐ฅ ๐ฆ ๐ฅ ๐ฆ ๐ฅ ๐ ๐ ๐ ๐ ๐ ๐ m m ๐ m ๐ ๐ค ๐ค ๐ ๐ ๐ค
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Application of Newtonโs Laws of Motion
Free-Body Diagram (FBD) FBD is a vectors diagram showing all forces acting on the body. ๐ฆ ๐ฅ m ๐ ๐ ๐ค ๐ ๐
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Application of Newtonโs Laws of Motion
Free-Body Diagram (FBD) FBD is a vectors diagram showing all forces acting on the body. ฮฃ ๐น ๐ฆ =๐โ๐ค=0 (for static equilibrium) ๐ฆ ๐ฅ ๐ฎ ๐ญ ฮฃ ๐น ๐ฅ =๐โ๐ (Net External force) ๐ N FBD of Furniture ๐ฎ ๐ญ = ๐ ๐ฅ โ๐ Net external force ๐=๐๐ต Note: If pulling force ๐ is less than friction ๐ then furniture is at rest otherwise it will move to the right.. ๐ท ๐=๐๐
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Application of Newtonโs Laws of Motion
Free-Body Diagram (FBD) FBD is a vectors diagram showing all forces acting on the body. FBD of box ๐ฆ ๐ฅ ๐ฎ ๐ญ ฮฃ ๐น ๐ฆ =๐+ ๐ ๐ฆ โ๐ค=0 (for static equilibrium) N ๐ ฮฃ ๐น ๐ฅ = ๐ ๐ฅ โ๐ (Net External force) P ๐ท ๐ ๐ฎ ๐ญ = ๐ ๐ฅ โ๐ ๐=๐๐ต Net external force ๐=๐๐ ๐ท ๐ Note: If ฮฃ ๐น ๐ฆ is not equal to zero then box didnโt touch the ground surface. Therefore Net external force is not equal to ฮฃ ๐น ๐ฅ .
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Application of Newtonโs Laws of Motion
Free-Body Diagram (FBD) Exercise Problem: Construct the free-body diagram (FBD) FBD of box ๐ฎ ๐ญ ๐ ๐ฆ ๐ฅ N ๐=๐๐ต ๐=๐๐ ๐ ๐ ๐ ๐ 20 ๐
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Application of Newtonโs Laws of Motion
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? ๐ ๐ก=2๐ Figure: ๐ ๐ด ๐ ๐ต ๐ ๐ด ๐ ๐ต ๐ ๐ต ๐ ๐ด ๐ ๐ก=2๐
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Application of Newtonโs Laws of Motion
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? ๐ฆ ๐ฅ ฮฃ ๐น ๐ด ๐ โ๐ ๐๐๐๐ ๐ฆ ๐ฅ FBD of ๐ ๐ด FBD of ๐ ๐ต ๐ ๐ฆ ๐ฅ ๐ ๐ก=2๐ Figure: +๐ ๐๐๐๐ ฮฃ ๐น ๐ต ๐ ๐ ๐ด ๐ ๐ต ๐ ๐ด ๐ ๐ต ๐ ๐ด ๐ ๐ต ๐ ๐ต ๐ ๐ด โ๐ค ๐ด = ๐ ๐ด ๐ ๐ค ๐ต = ๐ ๐ต ๐ ๐ฆ ๐ฅ ฮฃ ๐น ๐ฆ =๐โ ๐ค ๐ด =0 (static equilibrium) ๐ ๐ก=2๐ ฮฃ ๐น ๐ฅ = +๐ ๐๐๐๐ ฮฃ ๐น ๐ฆ =0 (static equilibrium) Net external force ฮฃ ๐น ๐ฅ = โ๐ ๐๐๐๐ + ๐ ๐ต ๐ Net external force ฮฃ ๐น ๐ด = ๐ ๐ด ๐ From Second law ฮฃ ๐น ๐ต = ๐ ๐ต ๐ From Second law Add ๐ธ๐ 1and ๐ธ๐.2 to eliminate ๐ ๐๐๐๐ then solve Acceleration ๐. ๐ ๐๐๐๐ = ๐ ๐ด ๐ Eq.1 โ๐ ๐๐๐๐ + ๐ ๐ต ๐= ๐ ๐ต ๐ Eq.2 ๐= ๐ ๐ต ๐ ๐ ๐ด + ๐ ๐ต ๐= 8(9.81) 6+8 =5.61๐/ ๐ 2 Contโn
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Application of Newtonโs Laws of Motion
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. ๐ฆ ๐ฅ a) Solve for tension in the cord ๐ ๐๐๐๐ using Eq.1 ๐ ๐ก=2๐ Figure: ๐ ๐๐๐๐ =6(5.61)=33.66 N ANSWER ๐ ๐ด ๐ ๐ต ๐ ๐ด ๐ ๐ต ๐ ๐ต ๐ ๐ด ๐ฆ ๐ฅ b) Solve for the distance ๐ of two bodies after travelling 2 sec. Given: ๐ ๐ก=2๐ ๐ฃ ๐ =0 ๐ก=2 ๐ ๐=5.61 ๐/ ๐ 2 From kinematics: ๐ = ๐ฃ 1 ๐ก ๐ ๐ก 2 ๐ = =11.22๐ ANSWER
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ASSIGNMENT Figure: . Given: m = 10 kg h = 5 m L = 10m S1 = 2m ร =44o
๐=0.06 coef. of kinetic friction Note: Particle โmโ 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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eNd
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