# Section 4-7 Solving Problems with Newton’s Laws; Free Body Diagrams

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Section 4-7 Solving Problems with Newton’s Laws; Free Body Diagrams
“It sounds like an implosion!”

∑F = ma Forces are VECTORS!! Newton’s 2nd Law:
∑F = VECTOR SUM of all forces on mass m  VECTOR addition is needed add forces in the 2nd Law! Forces add according to the rules of VECTOR ADDITION! (Ch. 3)

Problem Solving Procedures
Draw 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 that this often requires algebra, like solving 2 linear equations in 2 unknowns!

Example Find the resultant force FR

Example Find the resultant force FR FR = [(F1)2 + (F2)2](½) = 141 N
tanθ = (F2/F1) = 1, θ = 45º

Example 4-9 Find the resultant force FR If the boat moves with
acceleration a, ∑F = FR = ma FRx = max, FRy = may

Example 4-9 ∑F = ma, Note: This is the LAST step, NOT the first!
Illustrates the procedures for Newton’s 2nd Law problems: STEP 1: Sketch the situation!! Sketch a “Free Body” diagram for EACH body in problem & draw ALL forces acting on it. Part of your grade on exam & quiz problems! STEP 2: Resolve the forces on each body into components Use a convenient choice of x,y axes Use the rules for finding vector components from Ch. 3. STEP 3: Apply N’s 2nd Law to EACH OBJECT SEPARATELY ∑F = ma, Note: This is the LAST step, NOT the first! We NEED A SEPARATE equation like this for each object! Resolved into components: ∑Fx = max, ∑Fy = may

Conceptual Example 4-10 Moving at CONSTANT v, with NO friction,
which free body diagram is correct?

The normal force, FN is NOT always equal & opposite to the weight!!
Example 4-11 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 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 always equal & opposite to the weight!!

Example 4-12 Two boxes are connected by a lightweight (massless!) cord & are resting on a smooth (frictionless!) table. The masses are mA = 10 kg & mB = 12 kg. A horizontal force FP = 40 N is applied to mA. Calculate: a. The acceleration of the boxes. b. The tension in the cord connecting the boxes. Figure Caption: Example 4–12. (a) Two boxes, A and B, are connected by a cord. A person pulls horizontally on box A with force FP = 40.0 N. (b) Free-body diagram for box A. (c) Free-body diagram for box B. Answer: Free-body diagrams for both boxes are shown. The net force on box A is the external force minus the tension; the net force on box B is the tension. Both boxes have the same acceleration. The acceleration is the external force divided by the total mass, or 1.82 m/s2. The tension in the cord is the mass of box B multiplied by the acceleration, or 21.8 N. Free Body Diagrams

Example 4-13 (“Atwood’s Machine”)
Two masses suspended over a (massless frictionless) pulley by a flexible (massless) cable is an “Atwood’s machine”. Example: elevator & counterweight. Figure: Counterweight mC = 1000 kg. Elevator mE = 1150 kg. Calculate a. The elevator’s acceleration. b. The tension in the cable. Figure Caption: Example 4–13. (a) Atwood’s machine in the form of an elevator–counterweight system. (b) and (c) Free-body diagrams for the two objects. Answer: Each mass has two forces on it, gravity pulling downward and the tension in the cable pulling upward. The tension in the cable is the same for both, and both masses have the same acceleration. Writing Newton’s second law for each mass gives us two equations; there are two unknowns, the acceleration and the tension. Solving the equations for the unknowns gives a = 0.68 m/s2 and FT = 10,500 N. a a aE = - a Free Body Diagrams aC = a

Conceptual Example 4-14 Advantage of a Pulley
A mover is trying to lift a piano (slowly) up to a second-story apartment. He uses a rope looped over 2 pulleys. What force must he exert on the rope to slowly lift the piano’s mg = 2000 N weight? mg = 2000 N Free Body Diagram

Example 4-15: Accelerometer
A small mass m hangs from a thin string & can swing like a pendulum. You attach it above the window of your car as shown. What angle does the string make a. When the car accelerates at a constant a = 1.20 m/s2? b. When the car moves at constant velocity, v = 90 km/h? Figure 4-25. Answer: (a) The acceleration of the mass is given by the horizontal component of the tension; the vertical component of the tension is equal to the weight. Writing these two equations and combining them gives tan θ = a/g, or θ = 7.0°. (b) When the velocity is constant, the string is vertical. Free Body Diagram

Inclined Plane Problems
Understand ∑F = ma & how to resolve it into x,y components in the tilted coordinate system!! The tilted coordinate System is convenient, but not necessary. Engineers & scientists MUST understand these! a

Example 4-1b: Sliding Down An Incline
A box of mass m is placed on a smooth (frictionless!) incline that makes an angle θ with the horizontal. Calculate: a. The normal force on the box. b. The box’s acceleration c. Evaluate both for m = 10 kg & θ = 30º Free Body Diagram Figure Caption: Example 4–16. (a) Box sliding on inclined plane. (b) Free-body diagram of box. Answer: On an incline (or any surface), the normal force is perpendicular to the surface and any frictional forces are parallel to the surface. (a) The normal force is equal to the component of the weight perpendicular to the incline, or mg cos θ. (b) The force causing the acceleration is the component of the weight parallel to the incline; therefore the acceleration is g sin θ. (c) The normal force is 85 N and the acceleration is 4.9 m/s2.

Example = 300 N Free Body Diagram FT1x = -FTcosθ FT1y = -FTsinθ

Problem 32 Take up as positive! FP = - FT m = 65 kg mg = 637 N
FT + FT - mg = ma 2FT -mg = ma FP = - FT ∑F = ma (y direction) on the woman + the bucket! Newton’s 2nd Law FT   FT  FP  a Free Body Diagram Newton’s 3rd Law!!  mg

Problem 32 Solution Take up as positive! FP = - FT m = 65 kg
mg = 637 N FT + FT - mg = ma 2FT -mg = ma FP = - FT ∑F = ma (y direction) on the woman + the bucket! Newton’s 2nd Law FT   FT  FP  a Free Body Diagram Newton’s 3rd Law!!  mg

Problem 32 Solution The window washer pulls down on the rope with a tension force so the rope pulls up on her hands with a tension force The tension in The rope is also applied at the other end of the rope, where it attaches to the bucket. So there is another force pulling up on the bucket. The bucket-washer combination has a net force upwards. The free body diagram shows only forces on the bucket-washer, not forces exerted by them (the pull down on the rope by the person) or internal forces (normal force of bucket on person). (a) Write Newton’s second law in the vertical direction, with up as positive. The net force must be zero if the bucket and washer have constant speed. (b) Now the force is increased by 15%, so again write Newton’s second law, but with a non-zero acceleration.

Problem 33  FT1  a m1g   FT2  FT2  a  m2g
We draw free-body diagrams for each bucket. a. Since the buckets are at rest, their acceleration is 0. Write Newton’s 2nd Law for each bucket, calling UP the positive direction.  FT1  a m1g   FT2 b. Now repeat the analysis, but with a non-zero acceleration. The free-body diagrams are unchanged.  FT2  a  m2g

General Approach to Problem Solving
Read the problem carefully; then read it again. Draw a sketch, then a free-body diagram. Choose a convenient coordinate system. List the known & unknown quantities; find relationships between the knowns & the unknowns. Estimate the answer. Solve the problem without putting in any numbers (algebraically); once you are satisfied, put the numbers in. Keep track of dimensions. Make sure your answer is REASONABLE!

Chapter 4 Summary Newton’s 1st Law: If the net force on an object is zero, it will remain either at rest or moving in a straight line at constant speed. Newton’s 2nd Law: Newton’s 3rd Law: Weight is the gravitational force on an object. Free-body diagrams are essential for problem-solving. Do one object at a time, make sure you have all the forces, pick a coordinate system & find the force components, & apply Newton’s 2nd Law along each axis.

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