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

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

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Problem Solving Procedures 1.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. 2.Resolve vectors into components. 3.Apply Newton’s 2 nd Law separately to each object & for each vector component. 4.Solve for the unknowns. Note that this often requires algebra, like solving 2 linear equations in 2 unknowns!

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Example Find the resultant force F R

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

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Example 4-9 If the boat moves with acceleration a, ∑F = F R = ma F Rx = ma x, F Ry = ma y Find the resultant force F R

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Example 4-9 Illustrates the procedures for Newton’s 2 nd 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 2 nd 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: ∑F x = ma x, ∑F y = ma y

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

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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 F P = 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 F N exerted by the table on the box. Free Body Diagram The normal force, F N is NOT always equal & opposite to the weight!!

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Two boxes are connected by a lightweight (massless!) cord & are resting on a smooth (frictionless!) table. The masses are m A = 10 kg & m B = 12 kg. A horizontal force F P = 40 N is applied to m A. Calculate: a. The acceleration of the boxes. b. The tension in the cord connecting the boxes. Example 4-12 Free Body Diagrams

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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 m C = 1000 kg. Elevator m E = 1150 kg. Calculate a. The elevator’s acceleration. b. The tension in the cable. a E = - a a C = a aa aa Free Body Diagrams

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Conceptual Example 4-14 mg = 2000 N 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? Free Body Diagram

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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/s 2 ? b. When the car moves at constant velocity, v = 90 km/h? Free Body Diagram

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

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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 Example 4-1b: Sliding Down An Incline Free Body Diagram

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Example = 300 N F T2x = F T cosθ F T2y = -F T sinθ F T1x = -F T cosθ F T1y = -F T sinθ Free Body Diagram

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Problem 32 Take up as positive! m = 65 kg mg = 637 N F T + F T - mg = ma 2F T -mg = ma F P = - F T ∑F = ma (y direction) on the woman + the bucket! FT FT FT FT FP FP mg a a Free Body Diagram Newton’s 2 nd Law Newton’s 3 rd Law!!

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Problem 32 Solution Take up as positive! m = 65 kg mg = 637 N F T + F T - mg = ma 2F T -mg = ma F P = - F T ∑F = ma (y direction) on the woman + the bucket! FT FT FT FT FP FP mg a a Free Body Diagram Newton’s 2 nd Law Newton’s 3 rd Law!!

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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 32 Solution

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Problem 33 F T1 a a m1g m1g F T2 F T2 a a m2g m2g We draw free-body diagrams for each bucket. a. Since the buckets are at rest, their acceleration is 0. Write Newton’s 2 nd Law for each bucket, calling UP the positive direction. b. Now repeat the analysis, but with a non-zero acceleration. The free-body diagrams are unchanged.

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General Approach to Problem Solving 1.Read the problem carefully; then read it again. 2.Draw a sketch, then a free-body diagram. 3.Choose a convenient coordinate system. 4.List the known & unknown quantities; find relationships between the knowns & the unknowns. 5.Estimate the answer. 6.Solve the problem without putting in any numbers (algebraically) ; once you are satisfied, put the numbers in. 7.Keep track of dimensions. 8.Make sure your answer is REASONABLE!

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Chapter 4 Summary Newton’s 1 st 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 2 nd Law: Newton’s 3 rd 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 2 nd Law along each axis.

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