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http://www.saburchill.com/physics/chapters/0017.htm l Forces in Two Dimensions http://www.physics-net.com/force/sf000.htm 1 Objectives re Vectors • Evaluate the sum of two or more vectors both mathematically and graphically • Determine the components of vectors • Utilize basic trigonometry: sine, cosine, tangent, Law of Cosines, Law of Sines 2 What is a vector? • A quantity with both magnitude (number with units) and direction. Examples are: • Displacement 340 km W • Force 24.0 n S • Velocity 125.6 m/s E 3 Formulae to use • Trig functions: – Sine q = opp / hyp – Cos q = adj / hyp – Tan q = opp / adj – Law of Cosines: R2 = A2 + B2 - 2AB(cosq) – Law of Sines: R = A = B Sin q Sin a Sin b • The first three only work with 90o between vectors • (Get handout if you have not had trigonometry) 4 When vectors are at right angles to one another: • Draw out vectors (use protractor with metric rule) set to some scale (i.e. 1cm = 25km) using heel-to-toe • Draw the resultant (measure direction and magnitude) • Find the angle between the resultant and one of the given vectors • Find the magnitude of the resultant • So… let’s try one. Problem 1 in Ch5 • Protractor & Calculator needed! 5 Car driven 125.0 km W then 65.0 km S . Find mag & dir of resultant. • Let scale be 1cm:10km (125xcm/10=12.5cm) and 65xcm/10=6.5cm) 125.0 km W • Draw to scale 65.0 km S q • Measure – Magnitude of resultant(convert cm back to km) – Direction of resultant 6 Now for the math. Your answers should be close to what you measured. • This time we are going to find direction Adj, 125 first. q Opp 65 • Tan q = opp = 65 • adj 125 q = tan-1 65 = 27.5o S of E • 125 7 Find the magnitude of the resultant • Let’s use sin relationship • sin q = opp/hyp = 65/R • R = 65/sin q = 65/sin 27.5o • R = 140.77 = 1.41x102 km • You could also use the Law of cosines and get the same magnitude. 8 Another problem • Find the displacement of two shoppers who walk 250.0 m then turn 90o to the right and walk 60.0 m. (Using Pythagorean theorem…) 250.0 m • R = \ (A2 + B2) 60.0 m • R = \ (250.0m)2+60.0m2) • R = 257 m • (Law of Cosines would work, too) 9 Force & Motion in Two Dimensions • In this section you will • Determine the force that produces equilibrium when three forces act on an object • Determine the components of vectors • Analyze the motion of an object on an inclined plane with and without friction. 10 • Now you will use your skill in adding vectors to analyze situations in which the forces acting on an object are at angles other than 90°. • Recall that when the net force on an object is zero, the object is in equilibrium. • According to Newton’s laws, the object will not accelerate because there is no net force acting on it; an object in equilibrium is motionless or moves with constant velocity. 11 • Suppose that two forces are exerted on an object and the sum is not zero. • How could you find a third force that, when added to the other two, would add up to zero, and therefore cause the object to be in equilibrium? • To find this force, first find the sum of the two forces already being exerted on the object. 12 • This single force that produces the same effect as the two individual forces added together, is called the resultant force. • The force that you need to find is one with the same magnitude as the resultant force, but in the opposite direction. • A force that puts an object in equilibrium is called an equilibrant. 13 Finding the equilibrant of A and B Asgn: 5/ 14 Components of vectors • Now we are going to go backwards and resolve a vector into the vectors that caused it (vector resolution) +y Ry q +x Rx 15 To find the x component, Rx: cos q = adj/hyp = Rx / 5.0 km cos q = Rx/5.0km Rx = 5.0km cos q = 5.0km (cos 40) Rx = 3.8 km To find the y component, Ry: sin q = opp/hyp = Ry / 5.0 km sin q = Ry/5.0km Ry = 5.0km sin q = 5.0 km (sin 40) Ry = 3.2 km +y opp = Ry Ry q = 40. o +x Rx 16 Section 5.1 assignment • 3, 4, 7, 12-14 & 80 (angle for both E and F is 45o) • Section check frames 32-40 17 Review What is the normal force, FN? The perpendicular contact force exerted by a surface on another object. FN Fw 18 Friction • For objects at rest to move, they must overcome inertia. To move a couch, you must apply force to get it going then less force to keep it going. • Kinetic Friction Force: Ff kinetic = mk F N • Static friction force: Ff static ≤ ms FN • Sketch and analyze the problem. 19 #17. A girl exerts a 36 n horizontal force while pulling a 52 n sled across a sidewalk at constant speed. Find the coefficient of kinetic friction between the sidewalk and the sled. • Fn = mg = 52n • Friction force equals girl’s pulling force. • Ff kinetic = mkFN so mk = Ff/Fn=36n/52n mk = 0.69 20 #19. Corey is dragging a box full of books (combined weight of 134n) from his house to the car. How hard must Corey pull on the box in order to start it moving if the coefficient of static friction between the pavement and the box is 0.12? 21 Solution • F Corey on Box = F friction • F Corey = msFN = msmg • F Corey = 0.12(134n) = 16.1n 22 #5/22. A 1.4 kg block slides across a rough surface in such a way that it slows down with an acceleration of 1.25 m/s 2. Find the coefficient of kinetic friction between the block and the surface. 23 FN a = 1.25 m/s2 Fforward Ffriction Fw = mg Is there a net force? 24 Fnet = mk FN ma = mk mg a = mkg mk = a/g = 1.25 m/s2 / 9.80 m/s2 = 0.128 25 • New problem: 5/23 • You help your mom move a 41 kg bookcase in the living room. You push with a force of 65n and the bookcase accelerates at 0.12 m/s2. What is the coefficient of kinetic friction between the bookcase and the carpet? 26 FN Ff = mkFN F Fw Fnet = F - mkFN = F - mkmg = ma mk = (F - ma)/mg working equation! = [65n - (41kg)(0.12m/s2)] / 41kg(9.80m/s2) = 0.15 27 • Your assignment for 5.2 is 18&20, 24,25 • Section check frames 67-74 of Publisher’s PPt. 28 The drawing below shows the vectors acting on a skier going down a hill. Acceleration is parallel to the slope, on one axis, usually the x-axis The y-axis is perpendicular to the x-axis and perpendicular to the surface of the slope. There are two forces—normal and frictional forces. These forces are in the direction of the coordinate axes. However, the weight is not. When an object is placed on an inclined plane, the magnitude of the normal force between the object and the plane will usually not be equal to the object’s weight. 29 Apply Newton’s laws once in the x-direction and once in the y-direction. The weight does not point in either of these directions, you will need to break this vector into its x- and y-components before you can sum your forces in these two directions 30 Let’s look at Ex. 6 on page 134 • A 62-kg person on skis is going down a hill sloped at 37o. The coefficient of kinetic friction between the skis and the snow is 0.15. How fast is the skier going 5.0 s after starting from rest? 31 33. An ant climbs at a steady speed up the side of its anthill, which is inclined 30.0° from the vertical. Sketch a free-body diagram for the ant. 32 • Assignment for 5.3 is 36 and 95, 97-99 • Section check 85-90 33 Multi Vector Problems 34 1 5 4 2 3 35 36 Magnitude 23.3 N 25 N • Sin q = a / h h = O / Sin q • H = 23.3 N / Sin 43o • 34.2 N • Add 180o and you get the equilibrant! 37

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posted: | 3/24/2011 |

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