SAT II Physics (Gary Graff) Episode 1 Part 4 ppsx

Peterson’s: www.petersons.com 71 CHAPTER 2 MECHANICSMECHANICS MECHANICSMECHANICS MECHANICS STATICS Objects that are not free to move are said to be in equilibrium. For an object to be in equilibrium, the following two conditions must be met: 1. All the applied forces must equal zero. 2. All the applied torques must equal zero. Let’s stop a moment to give a quick definition for force. A force is a push or a pull. That’s all there is to it, a push or a pull. All forces occur in pairs, and the force pairs act on different bodies. When a force is directed toward a point of contact, the force is called a compression. When a force is directed away from a point of contact, the force is called a tension. Using what we know about vectors, we can see that every set of forces can be resolved into just three lines of action: along the x axis, the y axis, and the z axis. We can restate the above (called the First Condition of Equilibrium) in the following manner: ΣF = 0 This can be broken down to address the three axes individually. Σ Σ Σ F F F x y z = = = 0 0 0 The symbol “Σ” is read as “the sum of.” We’ll use it to state the conditions of equilibrium mathematically. The x axis is horizontal (side to side), the y axis is vertical (up and down), and the z axis is altitude (in and out). Peterson’s SAT II Success: Physics 72 Example One illustration of the first condition of equilibrium is a book resting on a flat surface as shown above. The book rests on the surface with a force equal to its weight, say 10 N. The surface doesn’t collapse or push the book away from itself, which means that the surface is pushing back on the book with a force equal to the force the book exerts on it. The book is in equilibrium. This statement can be shown as an equation. Σ Σ Σ Σ F F F F y z =∴ = = = 0 0 0 0 x Solution Looking at the individual axes: Σ Σ Fx Fz x = = 0 (there are no forces) 0 (there are no for z cces) y ΣFFF FF book surface book surface == − = = 00 CHAPTER 2 Peterson’s: www.petersons.com 73 Example A problem involving an object hanging by a wire is solved in a similar manner. A 25N mercury vapor light is hung from the ceiling by a wire. What is the tension (T 1 ) in the wire? Don’t forget the force pairs: • The light pulls on the cable. • The cable pulls on the light. • They are equal in magnitude and opposite in direction to one another. STATICS Peterson’s SAT II Success: Physics 74 Solution Σ Σ Σ Σ ΣΣ F F F F FF x y z xz = = = = == 0 0 0 0 00 and because no forces are appplied on either of the or axes N TN 1 xz FT y Σ=− = = 1 25 0 25 Example Another problem involving objects constrained by cables requires us to use some trigonometry to find a solution. A 100N traffic light is suspended in the middle of an intersection by three different cables. The three cables meet at point A. Find the tensions in T 1 , T 2 , and T 3 . CHAPTER 2 Peterson’s: www.petersons.com 75 Solution There is no movement of point A, so point A must be in equilibrium. Summing forces about point A: Σ Σ Σ F Fnoz F z x = = = 0 0 0 (there are forces on the axis) (there aare x Fare y y forces on the axis) (there forces on the Σ=0 axis) We can solve the axis tension first N y y FT = =− =Σ 1 100 0 TT 1 100= N • T 2 is a resultant vector with components located along the x and y axis. • T 2 is an unknown value, but it does pull upward toward the +x direction. • If T 2 is both pulling up and to the right against point A, then some- thing must be pulling down and left against point A. • Pulling down is T 1 and pulling left is T 3 . STATICS Peterson’s SAT II Success: Physics 76 Having identified the equilibrant pulls as T 1 and T 3 , we can set T 2 y against the upward pull of T 2 , which is T 2y . This meets the condi- tion of equilibrium on the y axis, which is stated in the equation below. ΣFTT TT T T T yy y y =− = = = = 12 12 1 2 2 0 100N 100 N is found by substitut iing and solving 100N N 200N 2 =° = = T T T 2 2 30 100 5 (sin ) . T 3 could have been found before T 2 , but the order in which these two are found is not critical. Looking at point A again, T 3 is identified as the left pull, and we can set T 3 against the right pull of T 2 , which is T 2x . This meets the condition of equilibrium on the x axis, which we state in the following equation. ΣFTT TT T T T T xx x x y x =− = = = ° = 32 32 3 2 3 0 30 100 5 tan . Substitute for N 2 88 172 3 T = N CHAPTER 2 Peterson’s: www.petersons.com 77 Alternate method ΣFTT TT TT T T xx x x =− = = =° = 32 32 32 3 0 30 200 (cos ) (. Substitute for N 2 886 172 3 ) T = N The solution to the problem isThe solution to the problem is The solution to the problem isThe solution to the problem is The solution to the problem is T 1 = 100N T 2 = 200N T 3 = 300N TORQUES When all the forces acting on an object along the three axes sum to zero, there can be no up and down, side to side, or in and out motion. However the object can still spin or rotate. The First Condition of Equilibrium addresses straight line or concurrent forces, which do not cause objects to rotate. The Second Condition of Equilibrium relates to the turning effects that act upon an object. These turning effects are called torques. The Second Condition of Equilibrium states that the sum of all the applied torques must equal zero. ΣT = 0 Having written the second condition of equilibrium, we can restate it as: Σ (+ T) = Σ (− T) This tells us that the turning effects in one direction about a point must equal the turning effects in the other direction about the point (pivot point). These effects can be labeled clockwise/counter- clockwise, left/right, or up/down. My preference is to use the (+) and (–) signs. This way one direction can always be set opposite the other without worrying about additional perspectives. Torques are further defined as TF=  , where F is an applied force and  is the distance from the applied force to the point of rotation (pivot point). TORQUES Peterson’s SAT II Success: Physics 78 When working with torques we must also consider the weight of the object involved. Gravity pulls on each and every part of an object. These parts cause torques about the central point in the object. If all the individual torques were set against one another and canceled out, then one point, a point about which all the other points rotate, would remain. All the weight of the object could be considered to operate from this point too. That point is called the Center of Gravity (CG). A force whose action line passes through the pivot point exerts no torque. The  is the “lever arm,” which is defined as the perpen- dicular distance from the applied force to the pivot point. When an object is supported at its center of gravity, the point of rotation for torques is also located at the CG. Therefore, the weight of the object is not a factor in the identification of the applied torques. This occurs because of the definition of the lever arm. The object is supported at the CG, so the force exerted by the object’s weight does not have a lever arm; there is no distance from the applied force to the pivot point. The uniform rod above weighs 2N and is 1m long. The force F 1 is applied at the end of the rod, as shown. The lever arm  1 is the distance from the force F 1 to the pivot point P. Should values of 10N be added for the force and a length of .5m for  , the calculation of the torques becomes: TF T T = = =•  ( )(. )10 5 5 Nm Nm Note: The units of torque are a force unit multiplied by a length unit. CHAPTER 2 Peterson’s: www.petersons.com 79 The situation above can change drastically by simply allowing the rod to rotate at a different point. The new situation above requires that we remember the weight of the object in addition to the already known F 1 . The calculation of the torques is: TF F T wt wt =+()( ) 11  There are two torque-producing entities. ==•+• =•+• =• ()(.)10 1 2 5 10 11 Nm N m N m 1N m Nm T T Should we pick the other end of the rod as the pivot point, F 1 completely cancels out of the problem and the applied torque be- comes 1N • m. TORQUES [...]... final answer to the problem requires the addition of the displacement of the train during each of the three parts They are: 12 2 m 875 m + 12 7.6 m 11 24. 6 m Part 1 (acceleration) Part 2 (constant velocity) Part 3 (negative acceleration) The total distance between the two subway stations is 11 24. 6 meters FREE FALL When an object is released near the earth and nothing except the earth affects the object,... velocity) 3 The train slows down (negative acceleration) 1 2 The operation s = vo t + at is suitable for part one Remember 2 that the train started from rest, so the (vot ) term is zero The working equation is: 1 s = at 2 2 1 s = (1. 25m/s2 ) ( 14 sec)2 2 s = 12 2m 82 Peterson’s SAT II Success: Physics KINEMATICS The train coasts at velocity for the second part of the problem Once we find the velocity, we will... the opposite side 88 Peterson’s SAT II Success: Physics MOTION IN TWO DIMENSIONS After finding vy , apply the appropriate free fall equation to find the time vy = vL (cos 30°) vy = (22 m/s)(.5) = 11 m/s The equation that yields the time is vfy = voy + at (using the subscript “y” to denote the vertical axis) v fy = voy + at v fy − voy a =t (11 m/s) − ( 11 m/s) = t = 2. 24 s 9.8m/s2 The total time the lure... station and accelerates at a rate of 1. 25 m/s2 for 14 seconds, then coasts at a velocity for 50 seconds The conductor then applies the brakes to slow the train at a rate of 1. 2 m/s2 to bring it to a complete stop at the next station How far apart are the two train stations? Solution Upon analysis of the situation we find there are three parts to the problem: 1 The train speeds up (acceleration) 2 The train... no displacement (the ball started in her hand and has therefore not gone anywhere) The equation becomes: 1 0 = vo t + gt 2 2 gt 2 − vo = 2t (9.8m/s2 )(2.6s)2 − vo = 2(2.6s) − vo = 12 . 74 m/s2 Note: The negative vo lets us know that the girl is throwing the ball upward 86 Peterson’s SAT II Success: Physics MOTION IN TWO DIMENSIONS MOTION IN TWO DIMENSIONS CURVILINEAR MOTION The motion of a football observed... train when acceleration stops Remember that this is also the average velocity while the train is coasting v f = at v f = (1. 25m/s2 ) ( 14 s) v f = 17 .5m/s Now we find the displacement while the train coasts, using: s=vt s = (17 .5 m/s)(50s) s = 875 m The equation that best fits the third part of the problem is (vf2 = vo2 + 2as) Again we can drop a term from the equation The final velocity of the train is... a= (.6 kg)(9.8 m/s2 ) − (3N) 6 kg + 1. 5 kg a = 1. 37 m/s2 The 600 g (.6 kg) hanging mass is the cause of the applied force in the system Multiply this mass by g (9.8 m/s2) to convert it into weight Friction is subtracted because friction always opposes motion Both masses will be accelerated (if motion does occur), so we add them together 94 Peterson’s SAT II Success: Physics ... applied force, and Ff is the frictional force F = ma The applied force is 62.8N FA − Ff = ma The frictional force is u • n e FA − (.22) (10 0 N) = ma 62.8N − 22N = ma FA is larger than Ff The block will move mass of the block = 10 0N = 10 .2 kg 9.8 m/s2 62.8N − 22N =a 10 .2 kg 4 m/s2 = a The block accelerates in the direction of the unbalanced force Peterson’s: www.petersons.com 93 CHAPTER 2 Example The frictional... 9.8m/s2 The total time the lure is in the air (free fall) is 2. 24 seconds The x component of the lure’s velocity (vx) must also be calculated with the trigonometry vx = vL (cos 30°) vx = (22 m/s)(.866) vx = 19 m/s With both the time and x velocity known, the distance the lure travels can be found using: s=vt s = (19 m/s2) (2. 24 sec) s = 42 .56 m Peterson’s: www.petersons.com 89 CHAPTER 2 NEWTON’S LAWS... Ff is the frictional force F = ma The applied force is 20N FA − Ff = ma The frictional force is µ • n s FA − (.22) (10 0 N) = ma 20 N − 22N = ma We can stop here The block will not move! This is because the applied force is not large enough to overcome friction 92 Peterson’s SAT II Success: Physics NEWTON’S LAWS OF MOTION Let’s change the applied force to 62.8N and find the acceleration with the new conditions . three parts. They are: 12 2 m Part 1 (acceleration) 875 m Part 2 (constant velocity) + 12 7.6 m Part 3 (negative acceleration) 11 24. 6 m The total distance between the two subway stations is 11 24. 6. operation svt at o =+ 1 2 2 is suitable for part one. Remember that the train started from rest, so the (v o t ) term is zero. The working equation is: sat s s = = = 1 2 1 2 12 5 14 12 2 2 22 ( . )(. the already known F 1 . The calculation of the torques is: TF F T wt wt =+()( ) 11  There are two torque-producing entities. ==•+• =•+• =• ()(. )10 1 2 5 10 11 Nm N m N m 1N m Nm T T Should