#$&* course Phy 231 7/4 4 Is it correct to state that for a ball flying through the air in a parabolic path, that force resistance will always be 180 degrees opposite of motion throughout?
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Given Solution: `a** Gravity exerts a downward force equal to the weight of the ball. While in contact with the ball, and only while i contact, the bat exerts a normal force, which pushes outward along a line originating from the central axis of the bat. This force is perpendicular to the surface of the bat at the point of contact. Unless the direction of the ball is directly toward the center of the bat, which will not be the case if the ball is hit at an upward angle by a nearly level swing, there will also be a frictional force between bat and ball. This frictional force will be parallel to the surface of the bat and will act on the ball in the 'forward' direction. COMMON STUDENT ERROR: The gravitational force and the force exerted by the ball on the bat are equal and opposite. The force of the bat on the ball and the gravitational force are not equal and opposite, since this is not an equilibrium situation--the ball is definitely being accelerated by the net force, so the net force is not zero. ** COMMON STUDENT ERROR: Confusing motion in a direction with force in that direction. There is no force associated with the motion of the ball. The velocity of the ball in will remain unchanged if there is no net force on the ball. Furthermore, if the is net force has zero component in the x direction, the x velocity remains unchanged; the analogous statement holds for the y direction. STUDENT QUESTION I got confused about the motion in the direction with the force in that direction. I think I understand. INSTRUCTOR RESPONSE The force tells you the direction of the acceleration, not the direction of the velocity. From the force you can therefore tell the direction of the change in velocity, not the direction of the velocity itself. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Just to clarify Force by gravity for the ball is massball * gravity and is down towards earth. The bat applies a normal force on the ball that is 90 degrees up on the point of contact between the ball and bat.
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Given Solution: To clarify, there are three frictional forces operating here. One is the static frictional force between the tires and the road. The force exerted by the tires on the road is 2100 N backward, and the frictional force exerted by the road on the tires is 2100 N forward. The 'force of friction including air resistance' includes the rolling friction of between the tires and the road, and the air resistance. This force totals 250 N, and is in the direction opposite motion. Taking the forward direction as positive, the net force exerted on the car is therefore 2100 N - 250 N = 1850 N. Rearranging the formulation F_net = m a of Newton's Second Law, we obtain m = F_net / a = 1800 N / (1.8 m/s^2) = 1000 N / (m/s^2) = 1000 kg m/s^2 / (m/s^2) = 1000 kg. Your solution should have included a description of your free-body diagram. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Force tires on road = -2100 N with respect to positive x axis. Force static Road on tires = + 2100 N with respect to positive x axis. These forces are equal and opposite. ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: Openstax: Two children pull a third child on a snow saucer sled exerting forces F1 and F2 as shown from above in Figure 4.36. Find the acceleration of the 49.00-kg sled and child system. Note that the direction of the frictional force is unspecified; it will be in the opposite direction of the sum of F1 and F2 . YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: F1x = 10 cos (45) N f1y = 10 sin(45) f2x = 8 cos (-30) N f2y = 8 sin (-30) N Fresultant = f1 + f2 = (F1x + F2x) + (f1y + f2y) = (10 cos (45) + 8 cos (-30))i + (10 sin (45) + 9 sin (-30)) j = (13.9993 i + 3.07107j) N FRX = 13.9993N Fry = 3.07107N theta Resultant = arctan (Fry/Frx) = 12.3732 degrees above positive x axis. Force friction will be directed 180 degrees + 12.3732 degrees = 192.4 degrees Fresultant = 'sqrt ( Frx^2 + fry^2) = 14.3322N Fnet = FR + FFric = 14.3322N + (-7.5N) = 6.83217 N Fnet = mass * accel Acceleration = Fnet /mass = 6.83217 N / 49.00 kg = 0.1394 m/s^2 ^2 The sled will accelerate until it hits a final velocity at a direction of 12.4 degrees above the positive x axis with a value of 0.1394 m/s^2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Taking the rightward direction as that of the positive x axis, we have forces F_1 of 10 N at 45 deg and F_2 of 8 N at 330 deg. The components of these forces are F_1_x = 10 N cos(45 deg) = 7 N, approx. and F_1_y = 10 N sin(45 deg) = 7 N, approx. F_2_x = 8 N cos(330 deg) = 4 N F_2_y = 8 N sin(330 deg) = -7 N, approx. If F = F_1 + F_2 then F_x = 7 N + 4 N = 11 N, approx. F_y = 7 N - 7 N = 0 N, approx. Thus F = 11 N at 0 deg, i.e., 11 N directed approximately along the positive x axis. The frictional force will therefore be directed approximatly along the negative x axis, and the net force will be F_net = 11 N - 7.5 N = 3.5 N along the positive x axis. The acceleration of the sled will be a = F_net / m, where m is the 49 kg mass of the system. We get a = 3.5 N / 49 kg = .07 m/s^2 in the positive x direction. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ********************************************* Question: Openstax: What force is exerted on the tooth in Figure 4.38 if the tension in the wire is 25.0 N? Note that the force applied to the tooth is smaller than the tension in the wire, but this is necessitated by practical considerations of how force can be applied in the mouth. Explicitly show how you follow steps in the Problem-Solving Strategy for Newton’s laws of motion. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: T1 = Ti cos(-15)i + T1Sin(-15)j T2 = T2 cos (180 + 15) i + t2 sin (180 + 15)j Sum of fx = t1 cos (-15) + t2 cos (195) = 0 N Sum of fy= t1 sin (-15) + t2 sin (195) = - 12.9 N The wire exerts a force of 12.9 N on the tooth into the mouth. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The tension acts toward the right and down, and toward the left and down. Taking the rightward direction as the positive x direction, the directions of the two tension forces are 345 deg and 195 deg. The magnitude of each tension force is 25 N. So, using T_1 and T_2 for two two tension forces, the x and y components are T_1_x = 25 N cos(345 deg) = 24 N, approx.. T_1_y = 25 N sin(345 deg) = -6 N, approx.. T_2_x = 25 N cos(195 deg) = -24 N, approx.. T_2_y = 25 N sin(195 deg) = -6 N, approx.. The x component of the net force is therefore T_1_x + T_2_x = 24 N + (- 24 N) = 0, and the y component is -6 N + (- 6 N) = -12 N. The net force is thus 12 N in the downward y direction (toward the back of the mouth). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qgen phy list the forces on the ball while flying toward the outfield, and describe the directions of these forces YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: As the ball flies toward the out field, gravity pulls down the ball slowing it down in the y direction, the x component of the initial force is decreased slightly by friction forces which are 180 degrees opposite of motion, and the ball follows a parabolic path through the field. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a**After impact the forces are gravity, which is constant and in the y direction, and air resistance. The direction of the force of air resistance is opposite to the direction of motion. The direction of motion is of course constantly changing, and the magnitude of the force of air resistance depends on the speed of the ball with respect to the air, which is also changing. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qgen phy give the source of each force you have described YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Source of motion is the after collision with the bat giving the ball momentum on its trajectory. Force friction by air pushing back on the ball always 180 degrees opposite of direction of motion. Gravity constantly pulls the ball back down to earth. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The gravitational force is the result of the gravitational attraction between the ball and the Earth. The normal force is the result of the elastic compression of bat and ball. The frictional force is due to a variety of phenomena related to the tendency of the surfaces to interlock (electromagnetic forces are involved) and to encounter small 'bumps' in the surfaces. ** ERRONEOUS STUDENT ANSWER: the air, the pitcher, the bat/ batter. friction. gravity INSTRUCTOR RESPONSE: All these are sources of force in one or both situations (bat striking ball, ball flying toward outfield) except the pitcher. The pitcher exerted a force previously, and that force was instrumental in delivering the ball to the batter, but that force ended well before any of these events occurred. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I think i got it, i assumed just as the ball was flying through the air. I can visualize the ball compressing on impact with the bat though. ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qgen phy what is the direction of the net force on the ball while in contact with the bat? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: My drawing has a big circle representing the bat, and a small circle representing the ball. FbaT is directed above the positive x axis of the ball at some angle between 0, 90 degrees. F gravity is drawn at the point of contact and is straight down at 270 degrees with respect to the ball. I have a small arrow that is opposite the Fbat and represents the compression of the ball on the bat.
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Given Solution: `a** We assume that the y axis is directed vertically upward, and the x axis is horizontal. The normal force will vary from 0 at the instant of first contact, to a maximum at the instant of greatest compression, and back to 0 at the instant contact ceases. So there is no single normal force. However we can represent 'the' normal force as the average normal force. The gravitational force will remain constant. The frictional force will vary with the changing normal force, and we will speak here of the average frictional force. The average normal force will be the greatest of these forces, much greater than friction or gravity. The frictional force will likely also exceed the gravitational force. The y component of the normal force will overwhelm the y components of the frictional force and the gravitational force, both of which are downward, giving us a net y component slightly less than the y component of the normal force. The x component of the normal force will be reinforced by the x component of the frictional force, making the x component of the net force a bit greater than the x component of the normal force. This will result in a net force that is 'tilted' forward and slightly down from the normal force (see the figure at the end for a vector diagram showing normal, frictional, gravitational and resultant forces). Note that the frictional and gravitational forces will tend to 'spin' the baseball as well as contributing to its translational acceleration. The spinning effect is a topic for a later chapter. ** IMPORTANT NOTE: It is essential that you sketch a diagram showing these forces. You are very unlikely to understand the explanation given here without a picture. Even with a picture this might be challenging. If you are not sure you understand, you should submit a copy of this question and solution, along your questions and/or commentary (mark insertions with ****). STUDENT COMMENT: Not sure about the frictional force. Why is it down? How do we calculate it? INSTRUCTOR RESPONSE: The frictional force exerted on the ball by the bat is perpendicular to the normal force, so the frictional force is exerted in the plane tangent to both the ball and the bat (imagine a flat piece of cardboard sandwiched between the ball and the bat; it lies in this tangent plane. If you have a line segment connecting the middle of the ball with the middle of the bat, it is perpendicular to the tangent plane (this line segment would cut through the piece of carboard at a right angle). Note that the direction of the normal force on the ball is along this line.). It is clear that the x component of the frictional force on the ball is in the 'forward' direction of motion. It is also clear that the in the tangent plane, the 'forward' direction is also downward. So the frictional force has a positive x component, and a negative y component. Assuming the ball does not 'slip' in contact with the bat, the frictional force is the force of static friction. The force of static friction cannot exceed the product of the coefficient of friction and the normal force: f_static < = mu * N, where f_static is the force of static friction, mu is the coefficient of static friction and N is the normal force. STUDENT QUESTION So the net force is tilted slightly, what does this mean, is it parallel to the ball and then slightly perpendicular at the same time so it tends to curve????????????????????????????????????????????????????? Is there a ready made sketch in our notes so that I can see if what I drew is correct? INSTRUCTOR RESPONSE The ball is represented in the figure below by the light pink circle, the bat by the green circle. The ball has arrived from the left, the bat is being swung to the right. The normal force acts perpendicular to the surface where the ball and bat make contact; the normal force is represented by the vector pointing toward upper right. The frictional force acts parallel to the surface of contact, and is represented by the shorter vector pointing down and to the left. (The ball and bat actually compress significantly, the ball more than the bat, and that compression is the source of the normal force. However the compression is not depicted in the figure.) The figure does not represent the gravitational force on the ball, which would be depicted as a downward force acting at the center of the ball. In a typical 'hit', the gravitational force would be much less than either the frictional or the normal force. The three forces are shown in the figure below, head-to-tail, along with the resultant force (the resultant is in red; the gravitational force is in the downward vertical direction and would likely be much less than depicted here). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Interesting visualization of the forces. So the weight is straight down, but the normal force is perpendicular to the contact point between the ball and bat at some angle. The net force of the bat is at some angle above the x axis, and the normal force is 90 degrees less than this net force. ------------------------------------------------ Self-critique Rating: 2 ********************************************* Question: `qgen phy what is the net force on the ball while flying toward the outfield? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Net force at any instant is the force forward + force by resistance opposite of motion. So as the ball approaches the arch of its path. FResistance is directed 180 degrees from its motion. On its way down it still has forward motion, gravity pulls it down to earth, and f resistance pushes it up at an angle 180 degrees from it motion. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The net force will consist of the downward gravitational force and the force of air resistance opposing the motion. If the ball is rising the y component of the air resistance will be in the downward direction, reinforcing the gravitational force and giving a net downward y component slightly exceeding that of gravity. If the ball is falling the y component will be in the upward direction, opposing the gravitational force and giving a net downward y component slightly less than that of gravity. In either case the x component will be in the direction opposite to the 'forward' motion of the ball, so the net force will be directed mostly downward but also a bit 'backward'. There are also air pressure forces related to the spinning of the ball; the net force exerted by air pressure causes the path of the ball to curve a bit, but these forces won't be considered here. ** STUDENT QUESTION What about as the ball is moving forward, is ther no air resistance being pushed against the ball horizontally as it flies to the outfield? INSTRUCTOR RESPONSE The ball typically experiences air resistance with components in both the x and the y direction. If it's rising the y component of the air resistance is downward, if it's falling the y component is upward. If it's at the very top of its arc, then for an instant it is neither rising nor falling and there is no air resistance in the y direction. The x component of the air resistance is in the direction opposite the 'forward' motion of the ball. I believe this is the force you asked about in your question. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qUniv. 5.88 (5.84 10th edition). Elevator accel upward 1.90 m/s^2; 28 kg box; coeff kin frict 0.32. How much force to push at const speed? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: W =m*gravity. Since acceleration is up, the box feels more weight, this force pushes the box down towards the elevator floor. W = mass *(gravity + acceleration) = 28 kg * (-9.80 m/s^2 + (-1.90 m/s^2) = -327.6 N Normal force will be toward the ceiling of the elevator at 327.6 N Force forward = Normal * uk = 327.6 N * 0.32 = 105 N confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aSTUDENT SOLUTION AND INSTRUCTOR COMMENT: The magnitude of kinetic friction force is fk = mu-sub k * N. First we add the 1.9 to 9.8 and get 11.7 as the acceleration and times that by the 28 kg and get 327.6 as the force so plugging in we get fk = 0.32 * 327.6 = 104.8 N. ** Good. The net force Fnet on the box is Fnet = m a = 1.90 m/s^2 * 28 kg. The net force is equal to the sum of the forces acting on the box, which include the weight mg acting downward and the force of the floor on the box acting upward. So we have Fnet = Ffloor - m g = m a. Thus Ffloor = m g + m a = 28 kg * 9.8 m/s^2 + 28 kg * 1.90 m/s^2 = 28 kg * 11.7 m/s^2 = 330 N, approx. Being pushed at constant speed the frictional force is f = `mu * N, where N is the normal force between the box and the floor. So we have f = .32 * 330 Newtons = 100 N, approx. ** STUDENT QUESTION: I don't understand why the net force is the weight of the box + the upward force of the elevator. Since the weight is directed downwards, and the elevator is going upwards, shouldn't it be the force of the elevator - weight? ???? INSTRUCTOR RESPONSE There are two forces acting on the box in the vertical direction, the weight (acting downward) and the normal force exerted by floor of the the elevator on the box (which acts upward). Using Ffloor for the normal force, the net vertical force is therefore net vertical force = Ffloor - weight. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: