course PHY201 ????~??F??C??assignment #025
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18:57:50 principles of physics and gen phy 4.26 free-body diagram of baseball at moment hit, flying toward outfield gen phy list the forces on the ball while in contact with the bat, and describe the directions of these forces
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RESPONSE --> At the moment the bat and the ball make contact, there are four forces exerted. There is the force that is on the ball pushing it in the direction of the bat, and there is the force of gravity pulling on its weight. Then there is also the force of the bat appling force in the direction of the ball perpendicular to the bat. Also, there are frictional forces that is exerted on the ball in the parallel direction of the bat. After contact, while the ball is flying toward outfield, there is the force of gravity applied on the ball. Next is the pushing force in the angled direction of the ball's motion. Also, there are the forces applied in the x and y direction of the balls motion.
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18:58:47 ** 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. **
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19:01:19 gen phy list the forces on the ball while flying toward the outfield, and describe the directions of these forces
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RESPONSE --> While the ball is flying toward the outfield, there is the force of gravity pulling on the ball in the downward direction. Next is the force pushing the ball in the angled direction of its motion. Then there are the forces applied in the x and y directions relevant to the pushing force in the angled direction.
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19:01:49 **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. **
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RESPONSE --> I forgot about the air resistance.
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19:10:47 gen phy give the source of each force you have described
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RESPONSE --> The force of gravity in the force the earth pulls on the objects on the surface of each. The pushing force behind the ball as it heads in the direction of the batter is from the projectile force applied from the pitcher. The force applied on the ball perpendicular to the bat is from the batter swinging the bat along its angular path of motion. There is also the force of friction between the two surfaces of the ball and bat.
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19:11:52 ** 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. **
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19:17:03 gen phy what is the direction of the net force on the ball while in contact with the bat?
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RESPONSE --> The direction of net force applied on the ball is the sum of all forces exerted on the ball, which should result in the direction perpendicular to the bat, as the ball seems to follow the path in which it was hit.
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19:18:29 ** The normal force will vary from 0 at the instant contact begins 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 along with the normal force, and we will speak here of the average frictional force.The average normal force will be the greatest force, 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. Note that the frictional force will tend to 'spin' the baseball but won't contribute much to the translational acceleration of the ball. This part is a topic for another chapter. **
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RESPONSE --> A bit more in depth than my answer. And I guess I missed the part of the force of gravity remaining constant.
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19:26:57 gen phy what is the net force on the ball while flying toward the outfield?
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RESPONSE --> The net force will be the sum of the forces pushing the ball in its direction of motion(in referance to the x and y vectors), the effects of the air resistance in the opposite direction of motion, and the force of gravity pulling the ball downward toward the surface of Earth.
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19:40:35 ** 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 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. **
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RESPONSE --> I have a question or thought. According to the First Law of Motion, objects at rest stay at rest unless an external force is applied. Objects in motion remain in motion unless an external force is applied. Well, we know that gravity and air resistance are the external forces in this example; but what about the force in the direction of the objects motion. Is there a force continually pushing the object? I was thinking that there would be, because the object has a mass and a rate of acceleration in its direction motion; and therefore, F = m * a, so there would a force pushing it. I am thinking that it is only the external forces that slow down the rate of acceleration and cause the ball to fall. The force is pushing the ball until the air resistance and gravity slowing reduce the momentum of the force pushing the ball until another force eventually becomes the primary affecting force driving the balls motion. Is this correct or am I missing something? Or am I just nuking it?
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20:20:46 Univ. 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?
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20:20:55 STUDENT 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. **
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