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lab 4 forces of friction 83.5 out of 85.docx

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Department
Physics
Course
PHY 122
Professor
Dolenko
Semester
Spring

Description
μ μ Lab 4: Forces of Friction and Determining s and k T.A: Monday at 2:00 p.m Abstract: This experiment was performed in order to calculate the total frictional forces and to determine how the coefficients of friction were related and what the total forces of friction depended on. For part one, two masses were added to a block and the total friction over fifty trial runs was recorded by a force sensor. For part two, only one mass was added and only 30 trials were done. The calculated results for the coefficient of static friction were nearly identical for part one and part two. The result for part one determined the coefficient of static friction to be .41 +/- .07 and part two determined to be .41 +/- .08. The total coefficients for kinetic friction for both parts were very similar as well with part one determining the kinetic friction coefficient to be .28 +/- . 05 Newtons and part two .28 +/- .05 Newtons. The results for the total friction on the block were the greatest when the applied force was the greatest and decrease in order of the highest force to the least. This suggests that the coefficients of friction are independent of the force being applied to the object. It also was determined that the total force acting on the object depends on both the force and on the coefficient of friction. The results obtained from this experiment were expected. The histograms of the total friction did show some outliers which most likely were caused by flaws in the system (such as the table and with varying forces applied). Objective The objective of this lab was to calculate the total frictional forces on the block and from there determine the coefficients of kinetic and static friction for the system. As a result of this lab, it will be determined if the force required to keeping an object moving is greater or less than the force required to set it in motion. Equipment The equipment used in this lab was a wooden block with weights attached to a force sensor by  string.  Graphical analysis was used to record the forces on the block.  Procedure For part one, two weights were added to a wooden block. Agroup member pulled on the force sensor and the total tension was recorded for fifty runs. For part two, the same procedure was done except that only one weight was added to the wooden block and only 30 runs were done. Results Part 1: μs=μ saverage−∆μ s = .41 +/- .07 μk=μ kaverage−∆ μ k = .28 +/- .05 Part 2: μs=μ saverage−∆μ s = .41 +/- .08 μ =μ +¿−∆ μ k kaverage k = .26 +/- .05 Kinetic Friction Figure 1: Frequency of recorded kinetic friction in Newtons Static Friction Figure 2: Frequency of recorded static friction in Newtons Data Analysis: Part 1: μs=μ saverage−∆ μ s F sf average μsaverage: Normal Force(totalmass∗gravity) Example Calculation for μsaverage: .497+.497+.1236 (¿)∗9.81 ¿ ¿ = .412 4.52 ¿ ∆ μ = ∆F sf s NormalForce(totalmass∗gravity) Example Calculation for ∆μ s : .497+.497+.1236 (¿)∗9.81 ¿ ¿ = .065 .712 ¿ Therefore the total static friction is: .41 +/- .07 μk=μ kaverage∆μ k μk average : F kf average Normal Force(totalmass∗gravity) μkaverage Example Calculation for : .497+.497+.1236 (¿)∗9.81 ¿ ¿ = .284 3.11 ¿ ∆F kf ∆ μk= Normal Force(totalmass∗gravity) Example Calculation for ∆μ k : .497+.497+.1236 (¿)∗9.81 ¿ = .052 ∆ μk= .566 ¿ Therefore the total kinetic friction is .28 +/- .05 F sf sf averagef sf f sf averag: slope on figure 1 Slope= 4.52 f ∆ fsf sf average : √n * n= number of trials ∆ f Example Calculation for sf = ∆ f 4.52 sf : √ 50 = .639 Therefore, F sf.5+¿−.6 Newtons F kf kf average∆f kf f kf averag: Slope on figure 2 Slope= 3.11 f ∆ fkf kf average : √ n *n= number of trials ∆ fkf Example Calculation for = 3.11 ∆ fkf : = .44 √ 50 Therefore, F sf.11+¿−.4 Newtons Figure 3: Graph of static friction vs. itself in Newtons Figure 4: Graph of kinetic friction plotted against itself in Newton’s Part 2: μs=μ saverage−∆μ s F sf average μsaverage: Normal Force(totalmass∗gravity) Example Calculation for μsaverage: .497+.1236 (¿)∗9.81 ¿ = .406 ¿ 2.47 ¿ ∆ μ = ∆F sf s N ∆μ Example Calculation for s : .497+.1236 (¿)∗9.81 ¿ ¿ = .083 .505 ¿ Therefore the total static friction is: .41 +/- .08 μk=μ kaverage−∆μ k μk averag: F kf average Normal Force(totalmass∗gravity) μ Example Calculation for kaverag: .497+.1236 (¿)∗9.81 ¿ ¿ = .263 1.6 ¿ ∆F ∆ μk= kf N Example Calculation for ∆μ k : .497+.1236 (¿)∗9.81
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