The purpose of this experiment is to analyze the nature of the resistance of motion by static and kinetic friction. First, the maximum static friction force is attained through a system comprised of a mass hanging from a string nestled on a pulley and attached to a block and force sensor. Secondly, a block is moved at constant velocity to determine what force in the positive direction will match the force of kinetic friction in the opposite direction. Thirdly, data is collected from an accelerating block in order to calculate net force on the body, including the kinetic friction force. All of these experiments are tested with friction between cork and wood and then rubber and wood. For part one, the coefficients of static friction were determined*…show more content…*

It is present when two surfaces come into some contact with one another and it opposes the motion between the surfaces. There are two types of friction, kinetic and static, which are both studied in the following experiment. Static friction hinders motion between two surfaces so that one is not moving in relation to the other. This lack of relative motion is due to the force of static friction equating the opposite and parallel applied force. If the applied force is increased continuously, it will eventually reach some value that enables it to overcome the force of static friction. This means that force of static friction between two surfaces as a maximum value that can be derived via the following*…show more content…*

Newton’s second law explains that the sum of forces is equivalent to the product of the mass and the net acceleration. Because forces are vectors, in problems with two dimensions, the summation can be broken up into the next force in the x-direction and the net force in the y-direction. For forces that are neither parallel nor perpendicular to the x-axis, trigonometry can be used to break vectors into this parallel or perpendicular composites.

The principles of kinetic friction and Newton’s second law can be used to find the coefficient of kinetic friction from part 2 of the experiment. The summation of the forces in the sis equal to zero because the block in moving at constant velocity. Therefore, the dragging force measured in the experiment must be equivalent to the force of kinetic friction by Newton’s second law since the forces are in opposite directions. The coefficient of kinetic friction can be solved for via the following equation:

[4] μk =

It is present when two surfaces come into some contact with one another and it opposes the motion between the surfaces. There are two types of friction, kinetic and static, which are both studied in the following experiment. Static friction hinders motion between two surfaces so that one is not moving in relation to the other. This lack of relative motion is due to the force of static friction equating the opposite and parallel applied force. If the applied force is increased continuously, it will eventually reach some value that enables it to overcome the force of static friction. This means that force of static friction between two surfaces as a maximum value that can be derived via the following

Newton’s second law explains that the sum of forces is equivalent to the product of the mass and the net acceleration. Because forces are vectors, in problems with two dimensions, the summation can be broken up into the next force in the x-direction and the net force in the y-direction. For forces that are neither parallel nor perpendicular to the x-axis, trigonometry can be used to break vectors into this parallel or perpendicular composites.

The principles of kinetic friction and Newton’s second law can be used to find the coefficient of kinetic friction from part 2 of the experiment. The summation of the forces in the sis equal to zero because the block in moving at constant velocity. Therefore, the dragging force measured in the experiment must be equivalent to the force of kinetic friction by Newton’s second law since the forces are in opposite directions. The coefficient of kinetic friction can be solved for via the following equation:

[4] μk =

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