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Vectors • Vector Operations • Components • Inclined Planes • Equilibrium • 2-D Motion & Force Problems • Trig Applications • Relative Velocities • Free Body Diagrams Vector Addition Suppose 3 forces act on an object at the same time. Fnet is not 15 N because these forces aren’t working together. But they’re not completely opposing each either. So how do find Fnet ? The answer is to add the vectors ... not their magnitudes, but the vectors themselves. There are two basic ways to add vectors: • Tip to tail method • Parallelogram method 8N 4N 3N Tip to Tail Method 12 N in-line examples 9N 21 N 9N 12 N 20 N Place the tail of one vector at 16 N the tip of the other. The vector sum (also called the resultant) is shown in red. It 20 N starts where the black vector 16 N began and goes to the tip of 4N the blue one. In these cases, the vector sum represents the net force. You can only add or subtract magnitudes when the vectors are in-line! Tip to Tail – 2 Vectors To add the red and blue displacement vectors first note: • Vectors can only be added if they are of the same quantity—in this case, displacement. • The magnitude of the resultant must be less than 7 m (5 + 2 = 7) and greater than 3 m (5 - 2 = 3). 5m 2m Interpretation: Walking 5 m in the direction of the blue vector and then 2 m in the direction of the red one is equivalent to walking in the direction of the black vector. The distance walked this way is the black vector’s magnitude. Commutative Property As with scalars (ordinary numbers), the order of addition is irrelevant with vectors. Note that the resultant (black vector) is the same magnitude and direction in each case. (We’ll learn how to find the resultant’s magnitude soon.) Tip to Tail – 3 Vectors We can add 3 or more vectors by placing them tip to tail in any order, so long as they are of the same type (force, velocity, displacement, etc.). Parallelogram Method This time we’ll add red & blue by placing the tails together and drawing a parallelogram with dotted lines. The resultant’s tail is at the same point as the other tails. It’s tip is at the intersection of the dotted lines. Note: Opposite sides of a parallelogram are congruent. Comparison of Methods Tip to tail method Parallelogram method The resultant has the same magnitude and direction regardless of the method used. Opposite of a Vector v -v If v is 17 m/s up and to the right, then -v is 17 m/s down and to the left. The directions are opposite; the magnitudes are the same. Scalar Multiplication x 3x -2x ½ x Scalar multiplication means multiplying a vector by a real number, such as 8.6. The result is a parallel vector of a different length. If the scalar is positive, the direction doesn’t change. If it’s negative, the direction is exactly opposite. Blue is 3 times longer than red in the same direction. Black is half as long as red. Green is twice as long as red in the opposite direction. Vector Subtraction red - blue Put vector tails together and complete the triangle, pointing to the vector that “comes first in the subtraction.” Why it works: In the first diagram, blue + black = red blue + (red – blue) = red blue - red red = red Note that red - blue is the opposite of blue - red. Other Operations • Vectors cannot be multiplied but there are two types of vector products. – Cross product – Dot product – We’ll use these products in Advanced Physics • There is no such thing as division of vectors – Vectors can be divided by scalars. – Dividing by a scalar is the same as multiplying by its reciprocal. Comparison of Vectors Which vector is bigger? 43 m 15 N 27 m/s 0.056 km The question of size here doesn’t make sense. It’s like asking, “What’s bigger, an hour or a gallon?” You can only compare vectors if they are of the same quantity. Here, red’s magnitude is greater than blue’s, since 0.056 km = 56 m > 43 m, so red must be drawn longer than blue, but these are the only two we can compare. 150 N Horizontal component Vertical component Vector Components A 150 N force is exerted up and to the right. This force can be thought of as two separate forces working together, one to the right, and the other up. These components are perpendicular to each other. Note that the vector sum of the components is the original vector (green + red = black). Components can also be drawn like this: Finding Components with Trig v sin Multiply the magnitude of the original vector by the sine & cosine of the angle made with the horizontal. The units of the components are the same as the units for the original vector. Here’s the correspondence: v v cos cosine adjacent side sine opposite side Component Example 30.814 m/s 14.369 m/s 25 34 m/s A helicopter is flying at 34 m/s at 25 S of W (south of west). The magnitude of the horizontal component is 34 cos 25 30.814 m/s. This is how fast the copter is traveling to the west. The magnitude of the vertical component is 34 sin 25 14.369 m/s. This is how fast it’s moving to the south. Note that 30.814 + 14.369 > 34. Adding up vector components gives the original vector (green + red = black), but adding up the magnitudes of the components is meaningless. Pythagorean Theorem 30.814 m/s 14.369 m/s 25 34 m/s Since components always form a right triangle, the Pythagorean theorem holds: (14.369)2 + (30.814)2 = (34)2. Note that a component can be as long, but no longer, than the vector itself. This is because the sides of a right triangle can’t be longer than the hypotenuse. Other component pairs v cos v v sin v v There are an infinite number of component pairs into which a vector can be split. Note that green + red = black in all 3 diagrams, and that green and red are always perpendicular. The angle is different in each diagram, as well as the lengths of the components, but the Pythagorean theorem holds for each. The pair of components used depends on the geometry of the problem. Inclined Plane A crate of chop suey of mass m is setting on a ramp with angle of inclination . The weight vector is straight down. The parallel component (blue) acts parallel to the ramp and is the component of the weight pulling the crate down the ramp. The perpendicular component (red) acts perpendicular to the ramp and is the component of the weight that tries to crush the ramp. parallel component Note: red + blue = black perpendicular component mg continued on next slide Inclined Plane (continued) The diagram contains 2 right triangles. is the angle between black and blue. + = 90 since they are both angles of the right triangle on the right. Since blue and red are perpendicular, the angle between red and black must also be . Imagine the parallel component sliding down (dotted blue) to form a right triangle. Being opposite , we use sine. Red is adjacent to , so we use cosine. mg sin mg cos mg continued on next slide Inclined Plane (continued) mg sin mg cos mg The diagram does not represent 3 different forces are acting on the chop suey at the same time. All 3 acting together at one time would double the weight, since the components add up to another weight vector. Either work with mg alone or work with both components together. How the incline affects the components mg sin mg sin mg cos mg mg The steeper the incline, the greater is, and the greater sin is. Thus, a steep incline means a large parallel component and a small horizontal one. Conversely, a gradual incline means a large horizontal component and a small vertical one. Extreme cases: When = 0, the ramp is flat; red = mg; blue = 0. When = 90, the ramp is vertical; red = 0; blue = mg. Inclined Plane - Pythagorean Theorem mg sin mg cos mg Let’s show that the Pythagorean theorem holds for components on the inclined plane: (mg sin )2 + (mg cos )2 = (mg)2 (sin2 + cos2 ) = (mg)2 (1) = (mg)2 Inclined Plane: Normal Force N = mg cos mg sin mg cos mg (continued) Recall normal force is perpendicular to the contact surface. As long as the ramp itself isn’t accelerating and no other forces are lifting the box off the ramp or pushing it into the ramp, N matches the perpendicular component of the weight. This be the case, otherwise the must box would be accelerating in the direction of red or green. That is, N > mg cos would mean the box is jumping off the ramp. N < mg cos would mean that the ramp is being crushed. Net Force on a Frictionless Inclined Plane ) With no friction, Fnet = mg + N = mg cos + mg sin + N = mg sin . N = mg cos mg sin (mg cos + N = 0 since their magnitudes are equal but the directions opposite. That is, the perpendicular component of the weight and the normal cancel out.) mg cos mg Therefore, the net force is the parallel force in this case. Acceleration on a Frictionless Ramp Here Fnet = mg sin = ma. So, a = g sin . Since sin has no units, a has the same units as g, as they should. Both the net force and the acceleration are down the ramp. mg sin mg cos mg Incline with friction at equilibrium At equilibrium Fnet = 0, so all forces must cancel out. Here, the normal force cancels the perpendicular component of the weight, and the static frictional force cancels the parallel component of the weight. mg sin N = mg cos fs = mg sin mg cos continued on next slide mg Incline with friction at equilibrium fs s N = s mg cos . Also, fs = mg sin (since we have equilibrium). So, mg sin s mg cos . Since the mg’s cancel out and tan = sin / cos , we have s tan . (cont.) N = mg cos fs = mg sin mg sin mg cos continued on next slide mg Incline with friction at equilibrium (cont.) Suppose we slowly crank up the angle, gradually making the ramp steeper and steeper, until the box is just about to budge. At this angle, N = mg cos fs = fs, max = s N = s mg cos . So now we have mg sin = s mg cos , and fs = mg sin mg sin s = tan . (Neither of these quantities have units.) An adjustable ramp is a convenient way to find the coefficient of static friction between two materials. mg cos mg Acceleration on a ramp with friction In order for the box to budge, mg sin must be greater than N = mg cos fs, max which means tan must be greater than s. If this is the case, forget about s and use k. fk = kN = kmg cos . Fnet = mg sin - fk = ma. So, mg sin - kmg cos = ma. fk = kmg cos mg sin The m’s cancel, which means a is independent of the size of the box. Solving for a we get: a = g sin - kg cos . Once again, the units work out right. mg cos mg Parallel applied force on ramp In this case FA and mg sin are working together against friction. Assuming FA + mg sin > fs, max the box budges and the 2nd Law tells us FA + mg sin - fk = ma. Mass does not cancel out this time. If FA were directed up the ramp, we’d have acceleration up or down the ramp depending on the size of FA compared to mg sin . If FA were bigger, friction acts down the ramp and a is up the ramp. N fk FA mg sin mg mg cos Non-parallel applied force on ramp Suppose the applied force acts on the box, at an angle above the horizontal, rather than parallel to the FA sin ( + ) ramp. This considerably complicates FA the problem. The first thing we have N to do, after making the diagram, is to fk resolve FA into parallel and FA cos ( + ) perpendicular components mg sin (orange and gray) using the angle + . FA serves to increase acceleration directly and indirectly: directly by orange pulling the box down the ramp, and indirectly by gray lightening the contact force with the ramp (thereby reducing friction). mg mg cos continued on next slide Non-parallel applied force on ramp FA sin ( + ) (cont.) Because of the perp. comp. of FA FA, N < mg cos . Assuming N FA sin ( + ) is not big enough f k to lift the box off the ramp, there is no acceleration in the FA cos ( + ) perpendicular direction. So, mg sin FA sin ( + ) + N = mg cos . Remember, N is what a scale would read if placed under the box, and a scale reads less if a force lifts up on the object. So, N = mg cos - FA sin ( + ), mg mg cos which means f = N k k = k [mg cos - FA sin ( + )]. continued on next slide Non-parallel applied force on ramp Assuming the combined force of orange and blue is enough to budge the box, we have Fnet = orange + blue - brown = ma. FA sin ( + ) FA FA cos ( + ) mg sin Substituting, we have (cont.) N fk FA cos ( + ) + mg sin - k [mg cos - FA sin ( + )] = ma. mg mg cos Hanging Sign Problem Support Beam T2 T1 1 2 mg continued on next slide Hanging sign f.b.d. Free Body Diagram T2 T1 1 2 mg Since the sign is not accelerating in any direction, it’s in equilibrium. Since it’s not moving either, we call it Static Equilibrium. Thus, red + green + black = 0. continued on next slide Hanging sign force triangle F = 0 means a closed vector polygon ! net As long as Fnet = 0, this is true no matter many forces are involved. T2 mg Vector Equation: T1 + T2 + mg = 0 T1 continued on next slide Components & Scalar Equations Hanging sign equations T2 T1 T1 sin 1 T2 sin 2 1 T1 cos 1 2 T2 cos 2 We use Newton’s 2nd Law twice, once in each dimension: Vertical: T1 sin 1 + T2 sin 2 = mg mg Horizontal: T1 cos 1 = T2 cos 2 Hanging sign sample Support Beam T1 T2 35 62 75 kg Answers: T1 = 347.65 N T2 = 606.60 N Accurately draw all vectors and find T1 & T2. Vector Force Lab Simulation Go to the link below. This is not exactly the same as the hanging sign problem, but it is static equilibrium with three forces. Equilibrium link 1. Change the strengths of the three forces (left, right, and below) to any values you choose. (The program won’t allow a change that is physically impossible.) 2. Record the angles that are displayed below the forces. They are measured from the vertical. 3. Using the angles given and the blue and red tensions, do the math to prove that the computer program really is displaying a system in equilibrium. 4. Now click on the Parallelogram of Forces box and write a clear explanation of what is being displayed and why. 3 - Way Tug-o-War Bugs Bunny, Yosemite Sam, and the Tweety Bird are fighting over a giant 450 g Acme super ball. If their forces remain constant, how far, and in what direction, will the ball move in 3 s, assuming the super ball is initially at rest ? Sam: 111 N Tweety: 64 N 38° To answer this question, we must find a, so we can do kinematics. But in order to find a, we must first find Fnet. 43° Bugs: 95 N continued on next slide 3 - Way Tug-o-War Sam: 111 N 68.3384 N 38° 87.4692 N (continued) 43.6479 N Tweety: 64 N 43° 46.8066 N continued on next slide Bugs: 95 N First, all vectors are split into horiz. & vert. comps. Sam’s are purple, Tweety’s orange. Bugs is already done since he’s purely vertical. The vector sum of all components is the same as the sum of the original three vectors. Avoid much rounding until the end. 3 - Way Tug-o-War (continued) Next we combine all parallel vectors by adding or subtracting: 68.3384 + 43.6479 - 95 = 16.9863, 87.4692 N 46.8066 N and 87.4692 - 46.8066 = 40.6626. A new picture shows the net vertical and horizontal forces on the super ball. Interpretation: Sam & Tweety together slightly 95 N overpower Bugs vertically by about 17 N. But Sam & Tweety 16.9863 N oppose each other horizontally, where Sam overpowers Tweety by about 41 N. 40.6626 N 68.3384 N 43.6479 N continued on next slide 3 - Way Tug-o-War Fnet = 44.0679 N (continued) 16.9863 N 40.6626 N Find Fnet using the Pythagorean theorem. Find using trig: tan = 16.9863 N / 40.6626 N. The Newtons cancel out, so = tan-1 (16.9863 / 40.6626) = 22.6689. (tan-1 is the same as arctan.) Therefore, the superball experiences a net force of about 44 N in the direction of about 23 north of west. This is the combined effect of all three characters. continued on next slide 3 - Way Tug-o-War (final) a = Fnet / m = 44.0679 N / 0.45 kg = 97.9287 m/s2. Note the conversion from grams to kilograms, which is necessary since 1 m/s2 = 1 N / kg. As always, a is in the same direction as Fnet.. a is constant for the full 3 s, since the forces are constant. 97.9287 m/s2 22.6689 Now it’s kinematics time: Using the fact x = v0 t + 0.5 a t2 = 0 + 0.5 (97.9287)(3)2 = 440.6792 m 441 m, rounding at the end. So the super ball will move about 441 m about 23 N of W. To find out how far north or west, use trig and find the components of the displacement vector. 3 - Way Tug-o-War Practice Problem The 3 Stooges are fighting over a 10 000 g (10 thousand gram) Snickers Bar. The fight lasts 9.6 s, and their forces are constant. The floor on which they’re standing has a huge coordinate system painted on it, and the candy bar is at the origin. What are its final coordinates? Answer: Hint: Find this angle first. Larry: 150 N Curly: 1000 N 78 93 Moe: 500 N ( -203.66 , 2246.22 ) in meters How to budge a stubborn mule It would be pretty tough to budge this mule by pulling directly on his collar. But it would be relatively easy to budge him using this set-up. explanation on next slide Big Force Little Force How to budge a stubborn mule (cont.) little force tree mule overhead view Just before the mule budges, we have static equilibrium. This means the tension forces in the rope segments must cancel out the little applied force. But because of the small angle, the tension is huge, enough to budge the mule! more explanation on next slide little force tree T T mule How to budge a stubborn mule (final) Because is so small, the tensions must be large to have vertical components (orange) big enough to team up and cancel the little force. Since the tension is the same throughout the rope, the big tension forces shown acting at the middle are the same as the forces acting on the tree and mule. So the mule is pulled in the direction of the rope with a force equal to the tension. This setup magnifies your force greatly. little force tree T T mule Relative Velocities in 1 D Schmedrick and his dog, Rover, are goofing around on a train. Schmed can throw a fast ball at 23 m/s. Rover can run at 9 m/s. The train goes 15 m/s. Question 1: If Rover is sitting beside the tracks with a radar gun as the train goes by, and Schmedrick is on the train throwing a fastball in the direction of the train, how fast does Rover clock the ball? vBT = velocity of the ball with respect to the train = 23 m/s vTG = velocity of the train with respect to the ground = 15 m/s vBG = velocity of the ball with respect to ground = 38 m/s This is a simple example, but in general, to get the answer we add vectors: vBG = vBT + vTG (In this case we can simply add magnitudes since the vectors are parallel.) continued on next slide Relative Velocities in 1 D (cont.) vBG = vBT + vTG • Velocities are not absolute; they depend on the motion of the person who is doing the measuring. • Write a vector sum so that the inner subscripts match. • The outer subscripts give the subscripts for the resultant. • This trick works even when vectors don’t line up. • Vector diagrams help (especially when we move to 2-D). vBT = 23 m/s vTG = 15 m/s vBG = 38 m/s continued on next slide Relative Velocities in 1 D (cont.) Question 2: Let’s choose the positive direction to be to the right. If Schmedrick is standing still on the ground and Rover is running to the right, then the velocity of Rover with respect to Schmedrick = vRS = +9 m/s. vRS From Rover’s perspective, though, he is the one who is still and Schmedrick (and the rest of the landscape) is moving to the left at 9 m/s. This means the velocity of Schmedrick with respect to Rover = vSR = -9 m/s. vSR Therefore, vRS = - vSR The moral of the story is that you get the opposite of a vector if you reverse the subscripts. continued on next slide Relative Velocities in 1 D (cont.) Question 3: If Rover is chasing the train as Schmedrick goes by throwing a fastball, at what speed does Rover clock the ball now? Note, because Rover is chasing the train, he will measure a slower speed. (In fact, if Rover could run at 38 m/s he’s say the fastball is still from his point of view.) This time we need the velocity of the ball with respect to Rover: vBR = vBT + vTG + vGR = vBT + vTG - vRG = 23 + 15 - 9 = 29 m/s. Note how the inner subscripts match up again and the outer most give the subscripts of the resultant. Also, we make use of the fact that vGR = - vRG. vBT = 23 m/s vBG = 29 m/s vTG = 15 m/s vRG = 9 m/s River Crossing campsite Current 0.3 m/s river boat You’re directly across a 20 m wide river from your buddies’ campsite. Your only means of crossing is your trusty rowboat, which you can row at 0.5 m/s in still water. If you “aim” your boat directly at the camp, you’ll end up to the right of it because of the current. At what angle should you row in order to trying to land right at the campsite, and how long will it take you to get there? continued on next slide River Crossing (cont.) campsite 0.3 m/s 0.5 m/s 0.4 m/s Current 0.3 m/s river boat Because of the current, your boat points in the direction of red but moves in the direction of green. The Pythagorean theorem tells us that green’s magnitude is 0.4 m/s. This is the speed you’re moving with respect to the campsite. Thus, t = d / v = (20 m) / (0.4 m/s) = 50 s. = tan-1 (0.3 / 0.4) 36.9. continued on next slide River Crossing: Relative Velocities The red vector is the velocity of the boat with respect to the water, vBW, which is what your speedometer would read. Blue is the velocity of the water with respect to the shore or camp, vWC. Green is the velocity of the boat with respect to the camp, vBC. The only thing that could vary in our problem was . It had to be determined so that red + blue gave a vector pointing directly across the river, which is the way you wanted to go. continued on next slide campsite 0.3 m/s 0.5 m/s 0.4 m/s river Current 0.3 m/s River Crossing: Relative Velocities vWC vBW (cont.) vBW = vel. of boat w/ respect to water vBC vWC = vel. of water w/ respect to camp vBC = vel. of boat w/ respect to camp Look how they add up: vBW + vWC = vBC The inner subscripts match; the out ones give subscripts of the resultant. This technique works in 1, 2, or 3 dimensions w/ any number or vectors. Law of Sines The river problem involved a right triangle. If it hadn’t we would have had to use either component techniques or the two laws you’ll also do in trig class: Law of Sines & Law of Cosines. C a b A Law of Sines: B c sin A a = sin B b = sin C c Side a is opposite angle A, b is opposite B, and c is opposite C. Law of Cosines C b A Law of Cosines: a c B a2 = b2 + c2 - 2 b c cos A These two sides are repeated. This side is always opposite this angle. It matters not which side is called a, b, and c, so long as the two rules above are followed. This law is like the Pythagorean theorem with a built in correction term of -2 b c cos A. This term allows us to work with non-right triangles. Note if A = 90, this term drops out (cos 90 = 0), and we have the normal Pythagorean theorem. Wonder Woman Jet Problem Suppose Wonder Woman is flying her invisible jet. Her onboard controls display a velocity of 304 mph 10 E of N. A wind blows at 195 mph in the direction of 32 N of E. What is her velocity with respect to Aqua Man, who is resting poolside down on the ground? vWA = vel. of Wonder Woman w/ resp. to the air vAG = vel. of the air w/ resp. to the ground (and Aqua Man) vWG = vel. of Wonder Woman w/ resp. to the ground (and Aqua Man) We know the first two vectors; we need to find the third. First we’ll find it using the laws of sines & cosines, then we’ll check the result using components. Either way, we need to make a vector diagram. continued on next slide Wonder Woman Jet Problem 32 80 (cont.) 32 100 vWG vWG 10 vWA + vAG = vWG 80 The 80 angle at the lower right is the complement of the 10 angle. The two 80 angles are alternate interior. The 100 angle is the supplement of the 80 angle. Now we know the angle between red and blue is 132. continued on next slide Wonder Woman Jet Problem (cont.) The law of cosines says: v2 = (304)2 + (195)2 - 2 (304) (195) cos 132 So, v = 458 mph. Note that the last term above appears negative, but it’s really positive, since cos 132 < 0. The law of sines says: sin 132 v = sin 195 So, sin = 195 sin 132 / 458, and 18.45 132 v This mean the angle between green and the horizontal is 80 - 18.45 61.6 Therefore, from Aqua Man’s perspective, Wonder Woman is flying at 458 mph at 61.6 N of E. 80 Wonder Woman Problem: Component Method This time we’ll add vectors via components as we’ve done before. Note that because of the angles given here, we use cosine for the vertical comp. of red but sine vertical comp. of blue. All units are mph. 103.3343 32 165.3694 299.3816 10 continued on next slide 52.789 Wonder Woman: Component Method (cont.) Combine vertical & horiz. comps. separately and use Pythag. theorem. = tan-1 (218.1584 / 402.7159) = 28.4452. is measured from the vertical, which is why it’s 10 more than was. 165.3694 103.3343 218.1584 mph 52.789 103.3343 52.789 165.3694 Comparison of Methods We ended up with same result for Wonder Woman doing it in two different ways. Each way requires some work. You can only use the laws of sines & cosines if: • you’re dealing with exactly 3 vectors. (If you’re adding three vectors, the resultant makes 4, and this method won’t work • the vectors form a triangle. Regardless of the method, draw a vector diagram! To determine which two vectors add to the third, use the subscript trick. Free body diagrams #1 For the next several slides, draw a free body diagram for each mass in the set-up and find a (or write a system of 2nd Law equations from which you could find a.) F2 v F1 m floor Two applied forces; F2 < mg; coef. of kinetic friction = k answer: F2 N F1 fk mg ma = F1 - fk = F1 - kN = F1 - k(mg – F2) (to the right). There is not enough info to determine whether or not N is bigger than F2. Free body diagrams #2 Bodies start at rest; m3 > m1 + m2; frictionless pulley with negligible mass. answer: Let’s choose clockwise as the + direction. T1 T1 m1 m1g m3 m2g m2 T1 - m1g -T2 = m1a m2: T2 - m2g = m2a m3: m3g - T1 = m3a system: m3g - m1g - m2g = (m1 + m2 + m3)a (Tensions are internal and cancel out.) T2 T2 m1: So, a = (m3 - m1 - m2)g / (m1 + m2 + m3) m3g If masses are given, find a first with last equation and substitute above to find T’s. Free body diagrams #3 answer: Note: T1 must be > T2 otherwise m2 couldn’t accelerate. T2 - m3g = m3a T1 - T2 - km2g = m2a m1g - T1 = m1a system: m1g - km2g - m3g = (m1 + m2 + m3) a m1 > m3 v m2 T1 T2 m3 m3g k N T2 T1 m1 fk m2g m1g Free body diagrams #4 answer: D v m Rock falling down in a pool of water m mg mg - D = ma. So, a = (mg - D)/m. Note: the longer the rock falls, the faster it goes and the greater D becomes, which is proportional to v. Eventually D = mg and a becomes zero, as our equation shows, and the rock reaches terminal velocity. Free body diagrams #5 A large crate of cotton candy and a small iron block of the same mass are falling in air at the same speed, accelerating down. R R cotton candy mg Fe mg answer: Since the masses are the same, a = (mg - R) / m for each one, but R is bigger for the cotton candy since it has more surface area and they’re moving at the same speed (just for now). So the iron has a greater acceleration and will be moving faster than the candy hereafter. The cotton candy will reach terminal vel. sooner and its terminal vel. will be less than the iron’s. Free body diagrams The boxes are not sliding; coefficients of static friction are given. 2 N1 f1 m2g m1 1 #6 a answer: There is no friction acting on m2. It would not be in equilibrium otherwise. f1 1N1 = 1(m1 + m2)g = T = m3g. f1’s reaction pair acting on table is not shown. 2 is extraneous m2 info in this problem, but not in the next one. m1 T T N2 m1g m3 m2 m2g m3g Free body diagrams Boxes accelerating (clockwise); m1 & m2 are sliding; coef’s of kinetic friction given. 2 N1 f2 m1 1 answer: There is friction acting on m2 now. It would not be accelerating otherwise. m3g - T = m3a; f2 = m2a; T - f1 - f2 = m1a, where f1 = 1N1 = 1(m1 + m2)g and f2 = 2N2 = 2m2g. m2 Note: f2 appears twice; they’re reaction pairs. v m1 T f1 m2g #6 b T N1 m1g m2 m2g m3 f2 m3g Free body diagrams #7 Boxes moving clockwise answer: Constant velocity is the same as no at a constant speed. velocity when it comes to the 2nd Law. Since a = 0, m2g = T = m1g sin + fk = m1g sin + km1g cos m2 = m1 (sin + k cos ). This is the relationship between the masses that must exist for equilibrium. N T v fk k m1g T m2g Note: sin , cos , and k are all dimensionless quantities, so we have kg as units on both sides of the last equation. Free body diagrams #8 Mr. Stickman is out for a walk. He’s moseying along but picking up speed with each step. The coef. of static friction between the grass and answer: his stick sneakers is s. Here’s a case where friction is a good thing. Without it we couldn’t walk. (It’s difficult to walk on ice since s is so small.) We use fs here since we assume he’s not slipping. Note: friction is in the direction of motion in this case. His pushing force does not appear in the free body diag. since it acts on the ground, not him. The reaction to his push is friction. Fnet = ma fs = ma smg ma N v fs mg a sg Since fs fs,max = sN Free body diagrams answer: Note: is measured with respect to the vertical here. F N fk m F sin v k ground mg #9 Box does not get lifted up off the ground as long as F cos mg. If F cos mg, then N = 0. Box budges if F sin > fs, max = sN = s(mg - F cos ). While sliding, F sin - k(mg - F cos ) = ma.