course Phy 231 7/11 9 025. `query 25*********************************************
.............................................
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. Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* OK ********************************************* Question: `qgen phy list the forces on the ball while flying toward the outfield, and describe the directions of these forces YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Air resistance on the ball will be opposite to the direction of motion and the amount depends on the velocity of the ball. Gravity acting on the ball will be downward and equal to the weight of the ball. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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): OK ------------------------------------------------ Self-critique rating #$&* OK ********************************************* Question: `qgen phy give the source of each force you have described YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Air resistance comes from air hitting the surface of the ball and making friction, which is a negative force (in the direction opposite to motion) and causes the ball to slow down. Gravity comes from the gravitational attraction between the earth and the ball. Normal force is from the compression between the ball and the bat when they hit each other. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* OK ********************************************* Question: `qgen phy what is the direction of the net force on the ball while in contact with the bat? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: You have to look at the average frictional and normal forces because they change over time. The normal force is the greatest followed by friction and gravity. Because the y component of the normal force is the greatest, the y component will by going up the positive y axis. The x component is along the positive x axis, so the net force will be going forward, but with a smaller angle than the original normal force. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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 ****). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* OK ********************************************* Question: `qgen phy what is the net force on the ball while flying toward the outfield? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The y component of the force will be downward because gravity is the biggest force. Air resistance will be going against the direction of motion, but depending on the way the ball is travelling it will either be helping or going against the force of gravity. Therefore, the net force on the ball will be going down and in the negative x direction a little bet, but mostly down. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* OK ********************************************* Question: `qUniv. 5.88 (5.84 10th edition). Elevator accel upward 1.90 m/s^2; 28 kg box; coeff kin frict 0.32. How much force to push at const speed? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To push at a constant speed you have to have enough force to go exactly against friction. Normal force = 1.9m/s^2(28kg) + 9.8m/s^2(28kg) = 53.2N + 274.4N = 327.6N Friction = 327.6N * .32 = 105 N You need equal and opposite of the 105N frictional force to push the box at a constant speed. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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. 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* OK "