Table 3. Total Normal Static Force Kinetic friction Friction mass Block sideway
ID: 1885020 • Letter: T
Question
Table 3. Total Normal Static Force Kinetic friction Friction mass Block sideway with 2 extra masses b%otoooor_3oto0 01 20Ht0.02 FROM TABLE 1 Rewrite the results for Trial 1 Block +2masses 4.91 ta02. To confirm the hypothesis that the area of contact is not related to the coefficient of friction, compare the forces of friction in the two conditions reported on table 3. Determine the quantity "difference" below Forcelarge areaForces (error on Forcelarge area) 2 + (err or on Forcesmall area)2 Accept or reject the hypothesis? If this ratio is smaller than 3, then your hypothesis is acceptable because it is within the margin of error. Report all your calculations on your excel file.Explanation / Answer
form the data we can see
for sideways block
Fn = 7.3+-0.0001 N
Fsn = 2.04+-0.02 N
Fkn = 1.97+-0.02 N
for normal configuration
Fn = 7.3+-0.0001 N
Fsn = 2.10+-0.02 N
Fkn = 1.84+-0.02 N
so
diff = |Fl - Fs|/sqrt((dFl^2) + (dFs)^2)
diffn = |7.3 - 7.3|/sqrt(0.0001^2 + 0.0001^2)
diffsn = |2.04 - 2.1|/sqrt(2*0.02^2) = 2.1213203435 < 3
diffkn = |1.97-1.84|/sqrt(2*0.02^2) = 4.59619407771 > 3
hence the value of kinetic firciont might actually depend on the area of contact, but the value of static friction does not
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