course Phy 121 There was only one question here for principles of physics that I saw. I hope this was correct. ??z?????????assignment #025????????????
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13:48:58 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 --> My arrow was pointing in an easterly direction toward the bat and then after it struck the bat it was pointing in a westerly direction, directly the opposite of eachother.
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13:48:58 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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13:52:25 ** 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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RESPONSE --> I didn't even think of the gravity exerting a force on the ball. I get that bat will exert a force on the ball and then does gravity again act on the ball?
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13:52:30 gen phy list the forces on the ball while flying toward the outfield, and describe the directions of these forces
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13:52:33 **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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13:52:35 gen phy give the source of each force you have described
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13:52:37 ** 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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13:52:40 gen phy what is the direction of the net force on the ball while in contact with the bat?
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13:52:42 ** 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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13:52:48 gen phy what is the net force on the ball while flying toward the outfield?
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13:52:50 ** 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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13:52:53 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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13:52:56 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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