Applications of Newton’s Laws of Motion Chapter Opener. Caption: Newton’s laws are fundamental in physics. These photos show two situations of using Newton’s laws which involve some new elements in addition to those discussed in the previous Chapter. The downhill skier illustrates friction on an incline, although at this moment she is not touching the snow, and so is retarded only by air resistance which is a velocity-dependent force (an optional topic in this Chapter). The people on the rotating amusement park ride below illustrate the dynamics of circular motion.

 We need VECTOR addition to add forces in the 2nd Law! Remember that forces are VECTORS! Newton’s 2nd Law: ∑F = ma ∑F = VECTOR SUM of all forces on mass m  We need VECTOR addition to add forces in the 2nd Law! Forces add according to the rules of VECTOR ADDITION! Get Force Components from trig, as discussed earlier.

Problem Solving Procedures Make a Sketch. For each object separately, sketch a free-body diagram, showing all the forces acting on that object. Make the magnitudes & directions as accurate as you can. Label each force. Resolve Vectors into Components. Apply Newton’s 2nd Law separately to each object & for each vector component. Solve for the unknowns. Note: This often requires algebra, such as solving 2 linear equations in 2 unknowns!

Example FR = [(F1)2 + (F2)2](½) = 141 N tanθ = (F2/F1) = 1  θ = 45º Square Root! FR = [(F1)2 + (F2)2](½) = 141 N tanθ = (F2/F1) = 1  θ = 45º

Example If the boat moves with acceleration a, ∑F = = FR = ma FRx = max, FRy = may

Example FRCx = FRCcosθ FRCy = -FRCsinθ FRBx = -FRBcosθ FRBy = -FRBsinθ Newton’s 3rd Law  FBR = -FRB, FCR = -FRC FRCx = FRCcosθ FRCy = -FRCsinθ FRBx = -FRBcosθ FRBy = -FRBsinθ

Example A box of mass m = 10 kg is pulled by an attached cord along a horizontal smooth (frictionless!) surface of a table. The forc exerted is FP = 40.0 N at a 30.0° angle as shown. Calculate: a. The acceleration of the box. b. The magnitude of the upward normal force FN exerted by the table on the box. Figure 4-21. Caption: (a) Pulling the box, Example 4–11; (b) is the free-body diagram for the box, and (c) is the free-body diagram considering all the forces to act at a point (translational motion only, which is what we have here). Answer: (a) The free-body diagram is shown in part (b); the forces are gravity, the normal force, and the force exerted by the person. The forces in the vertical direction cancel – that is, the weight equals the normal force plus the vertical component of the external force. The horizontal component of the force, 34.6 N, accelerates the box at 3.46 m/s2. (b) The vertical component of the external force is 20.0 N; the weight is 98.0 N, so the normal force is 78.0 N.

The normal force, FN is NOT equal & opposite to the weight!! A box of mass m = 10 kg is pulled by an attached cord along a horizontal smooth (frictionless!) surface of a table. The force exerted is FP = 40.0 N at a 30.0° angle as shown. Calculate: a. The acceleration of the box. b. The magnitude of the upward normal force FN exerted by the table on the box. Free Body Diagram Figure 4-21. Caption: (a) Pulling the box, Example 4–11; (b) is the free-body diagram for the box, and (c) is the free-body diagram considering all the forces to act at a point (translational motion only, which is what we have here). Answer: (a) The free-body diagram is shown in part (b); the forces are gravity, the normal force, and the force exerted by the person. The forces in the vertical direction cancel – that is, the weight equals the normal force plus the vertical component of the external force. The horizontal component of the force, 34.6 N, accelerates the box at 3.46 m/s2. (b) The vertical component of the external force is 20.0 N; the weight is 98.0 N, so the normal force is 78.0 N. The normal force, FN is NOT equal & opposite to the weight!!

Components of FP FPy = FP sin() = 40sin(30.0°) = 20 N A box of mass m = 10 kg is pulled by an attached cord along a horizontal smooth (frictionless!) surface of a table. The force is FP = 40.0 N at a 30.0° angle as shown. Calculate: a. The acceleration. b. The normal force FN Components of FP FPx = FP cos() = 40cos(30.0°) = 34.6 N FPy = FP sin() = 40sin(30.0°) = 20 N Figure 4-21. Caption: (a) Pulling the box, Example 4–11; (b) is the free-body diagram for the box, and (c) is the free-body diagram considering all the forces to act at a point (translational motion only, which is what we have here). Answer: (a) The free-body diagram is shown in part (b); the forces are gravity, the normal force, and the force exerted by the person. The forces in the vertical direction cancel – that is, the weight equals the normal force plus the vertical component of the external force. The horizontal component of the force, 34.6 N, accelerates the box at 3.46 m/s2. (b) The vertical component of the external force is 20.0 N; the weight is 98.0 N, so the normal force is 78.0 N.

FN + FPy – mg = 0; FN = mg - FPy FN = 98 – 20 = 78 N A box of mass m = 10 kg is pulled by an attached cord along a horizontal smooth (frictionless!) surface of a table. The force is FP = 40.0 N at a 30.0° angle as shown. Calculate: a. The acceleration. b. The normal force FN Components of FP FPx = FP cos() = 40cos(30.0°) = 34.6 N FPy = FP sin() = 40sin(30.0°) = 20 N Newton’s 2nd Law Vertical (y): Fy = 0 FN + FPy – mg = 0; FN = mg - FPy FN = 98 – 20 = 78 N Figure 4-21. Caption: (a) Pulling the box, Example 4–11; (b) is the free-body diagram for the box, and (c) is the free-body diagram considering all the forces to act at a point (translational motion only, which is what we have here). Answer: (a) The free-body diagram is shown in part (b); the forces are gravity, the normal force, and the force exerted by the person. The forces in the vertical direction cancel – that is, the weight equals the normal force plus the vertical component of the external force. The horizontal component of the force, 34.6 N, accelerates the box at 3.46 m/s2. (b) The vertical component of the external force is 20.0 N; the weight is 98.0 N, so the normal force is 78.0 N.

FN + FPy – mg = 0; FN = mg - FPy FN = 98 – 20 = 78 N A box of mass m = 10 kg is pulled by an attached cord along a horizontal smooth (frictionless!) surface of a table. The force is FP = 40.0 N at a 30.0° angle as shown. Calculate: a. The acceleration. b. The normal force FN Components of FP FPx = FP cos() = 40cos(30.0°) = 34.6 N FPy = FP sin() = 40sin(30.0°) = 20 N Newton’s 2nd Law Vertical (y): Fy = 0 FN + FPy – mg = 0; FN = mg - FPy FN = 98 – 20 = 78 N Horizontal (x): Fx = ma FPx = ma; a = (FPx/m) a = 3.46 m/s2 Figure 4-21. Caption: (a) Pulling the box, Example 4–11; (b) is the free-body diagram for the box, and (c) is the free-body diagram considering all the forces to act at a point (translational motion only, which is what we have here). Answer: (a) The free-body diagram is shown in part (b); the forces are gravity, the normal force, and the force exerted by the person. The forces in the vertical direction cancel – that is, the weight equals the normal force plus the vertical component of the external force. The horizontal component of the force, 34.6 N, accelerates the box at 3.46 m/s2. (b) The vertical component of the external force is 20.0 N; the weight is 98.0 N, so the normal force is 78.0 N.

Example: Pulling against friction A box, mass m = 10 kg, is pulled along a horizontal surface by a force FP = 40.0 N applied at a 30.0° angle above horizontal. The coefficient of kinetic friction is k = 0.3. Calculate a. the acceleration, b. the normal force FN. Figure 5-5. Solution: Lack of vertical motion gives us the normal force (remembering to include the y component of FP), which is 78 N. The frictional force is 23.4 N, and the horizontal component of FP is 34.6 N, so the acceleration is 1.1 m/s2.

A box, mass m = 10 kg, is pulled along a horizontal surface by a force FP = 40.0 N applied at a 30.0° angle above horizontal. The coefficient of kinetic friction is k = 0.3. Calculate a. the acceleration, b. the normal force FN. Components of FP FPx = FP cos() = 40cos(30.0°) = 34.6 N FPy = FP sin() = 40sin(30.0°) = 20 N Figure 5-5. Solution: Lack of vertical motion gives us the normal force (remembering to include the y component of FP), which is 78 N. The frictional force is 23.4 N, and the horizontal component of FP is 34.6 N, so the acceleration is 1.1 m/s2.

FN + FPy – mg = 0; FN = mg - FPy FN = 98 – 20 = 78 N A box, mass m = 10 kg, is pulled along a horizontal surface by a force FP = 40.0 N applied at a 30.0° angle above horizontal. The coefficient of kinetic friction is k = 0.3. Calculate a. the acceleration, b. the normal force FN. Components of FP FPx = FP cos() = 40cos(30.0°) = 34.6 N FPy = FP sin() = 40sin(30.0°) = 20 N Newton’s 2nd Law Vertical (y): Fy = 0 FN + FPy – mg = 0; FN = mg - FPy FN = 98 – 20 = 78 N Figure 5-5. Solution: Lack of vertical motion gives us the normal force (remembering to include the y component of FP), which is 78 N. The frictional force is 23.4 N, and the horizontal component of FP is 34.6 N, so the acceleration is 1.1 m/s2.

FN + FPy – mg = 0; FN = mg - FPy FN = 98 – 20 = 78 N A box, mass m = 10 kg, is pulled along a horizontal surface by a force FP = 40.0 N applied at a 30.0° angle above horizontal. The coefficient of kinetic friction is k = 0.3. Calculate a. the acceleration, b. the normal force FN. Components of FP FPx = FP cos() = 40cos(30.0°) = 34.6 N FPy = FP sin() = 40sin(30.0°) = 20 N Newton’s 2nd Law Vertical (y): Fy = 0 FN + FPy – mg = 0; FN = mg - FPy FN = 98 – 20 = 78 N Figure 5-5. Solution: Lack of vertical motion gives us the normal force (remembering to include the y component of FP), which is 78 N. The frictional force is 23.4 N, and the horizontal component of FP is 34.6 N, so the acceleration is 1.1 m/s2. Friction: Ffr = kFN = (0.3)(78) = 23.4 N

FN + FPy – mg = 0; FN = mg - FPy FN = 98 – 20 = 78 N A box, mass m = 10 kg, is pulled along a horizontal surface by a force FP = 40.0 N applied at a 30.0° angle above horizontal. The coefficient of kinetic friction is k = 0.3. Calculate a. the acceleration, b. the normal force FN. Components of FP FPx = FP cos() = 40cos(30.0°) = 34.6 N FPy = FP sin() = 40sin(30.0°) = 20 N Newton’s 2nd Law Vertical (y): Fy = 0 FN + FPy – mg = 0; FN = mg - FPy FN = 98 – 20 = 78 N Figure 5-5. Solution: Lack of vertical motion gives us the normal force (remembering to include the y component of FP), which is 78 N. The frictional force is 23.4 N, and the horizontal component of FP is 34.6 N, so the acceleration is 1.1 m/s2. Friction: Ffr = kFN = (0.3)(78) = 23.4 N Horizontal (x): Fx = ma FPx - Ffr = ma; a = (FPx - Ffr)/m) = 1.14 m/s2

Example: Two boxes and a pulley 2 boxes are connected by a cord running over a pulley. The coefficient of kinetic friction between box A & the table is k = 0.2. Ignore mass of cord & pulley & friction in the pulley, which means a force applied to one end of the cord has the same magnitude at the other end. Find the acceleration, a, of the system, which has the same magnitude for both boxes if the cord doesn’t stretch. As box B moves down, box A moves to right. Find the tension FT in the cord. a  a  ∑F = ma For EACH mass separately! x & y components plus friction Ffr = μkFN

Example: Two boxes & a pulley Coefficient of kinetic friction is k = 0.2. Find the acceleration, a, of the system & the tension FT in the cord. ∑F = ma For EACH mass separately! x & y components plus friction Ffr = μkFN mA: ∑Fx = mAa = FT – Ffr ∑Fy = 0 = FN – mAg FN = mg, Ffr = μkFN = μkmAg So: mAa = FT – μkmAg (1) mB: ∑Fy = mBa = mBg – FT So: mBa = mBg – FT (2) Solve (1) & (2) to find a & FT a  a 

Example: Two boxes and a pulley Solutions are: a = (mBg – Ffr)/(mA + mB) a = 1.4 m/s2 FT = Ffr + mAa FT = 17 N a  a  ∑F = ma For EACH mass separately! x & y components plus friction Ffr = μkFN

Conceptual Example: To push or to pull a sled? Your little sister wants a ride on her sled. If you are on flat ground, will you exert less force if you push her or pull her? Assume the same angle θ in each case. ∑F = ma Pulling Pushing Figure 5-6. Answer: Pulling decreases the normal force, while pushing increases it. Better pull. y forces: ∑Fy = 0 FN – mg –Fy = 0 FN = mg + Fy Ffr (max) = μsFN x forces: ∑Fx = ma y forces: ∑Fy = 0 FN - mg + Fy = 0 FN = mg - Fy Ffr (max) = μsFN x forces: ∑Fx = ma