Problem:
Is there a relationship between the mass of an object and the force of static friction and force of kinetic friction of that object? If so, what type of relationship is there?
Background:
Friction is a type of force that is created when two surfaces move across one another. Static friction is friction between two or more solid objects that are not moving relative to one another. Kinetic friction is the amount of friction between two objects that are moving relative to each other. All of these types can be determined by the “FUN” equation (f = μn) if you know the coefficient of each type and the normal force.
Mass may influence friction by also influencing a dependent relationship towards friction and the normal force. …show more content…
It was increasing at almost a constant rate, as seen in Figures 4, 5 and 6.
Block Mass (g) Trial 1 (N) Trial 2 (N) Trial 3 (N) Trial 4 (N) Trial 5 (N) Average Force (Static Friction, N)
304.6 0.634 0.619 0.630 0.554 0.632 0.614
404.6 0.937 0.866 0.766 0.835 0.820 0.845
504.6 1.023 0.969 1.043 1.078 1.093 1.041
604.6 1.191 1.263 1.410 1.319 1.319 1.300
704.6 1.489 1.241 1.652 1.434 1.478 1.459
804.6 1.762 1.757 1.723 1.726 1.770 1.726
Block Mass (g) Trial 1 (N) Trial 2 (N) Trial 3 (N) Trial 4 (N) Trial 5 (N) Average Force (Kinetic Friction, N)
304.6 0.473 0.464 0.467 0.450 0.465 0.464
404.6 0.593 0.603 0.598 0.613 0.625 0.606
504.6 0.753 0.751 0.751 0.750 0.787 0.758
604.6 0.902 0.897 0.894 0.903 0.912 0.902
704.6 1.068 1.060 1.068 1.076 1.072 1.069
804.6 1.208 1.203 1.181 1.144 1.168 1.175
Calculations Normal Force/Weight F_W= mg F_W=(0.3046g)(9.81m/s^2 ) F_W=2.0988126 N= 2.10 N
Average Force
F_AVG=(Trials 1 to 5)/5
F_AVG=(0.473N+0.464N+0.467N+0.450N+0.465N)/5
F_AVG=0.4638 N=0.464 N
Coefficient of Friction (Kinetic and Static) f=μn → μ=f/n
μ=(0.4638N/2.0988126N) …show more content…
To begin with, in order to get rid of the error caused by a non-zeroed Force Sensor, be sure to calibrate it every time you start a new trial and keep checking to make sure it is zeroed. Also, make sure that there is no draft in the room to be certain that there will be no air resistance on the wooden block as you pull it. And finally, you can use an instrument that will help pull the block at a constant velocity in order to get an accurate measurement for kinetic friction. If these are taken into consideration, then the lab will be more
Is there a relationship between the mass of an object and the force of static friction and force of kinetic friction of that object? If so, what type of relationship is there?
Background:
Friction is a type of force that is created when two surfaces move across one another. Static friction is friction between two or more solid objects that are not moving relative to one another. Kinetic friction is the amount of friction between two objects that are moving relative to each other. All of these types can be determined by the “FUN” equation (f = μn) if you know the coefficient of each type and the normal force.
Mass may influence friction by also influencing a dependent relationship towards friction and the normal force. …show more content…
It was increasing at almost a constant rate, as seen in Figures 4, 5 and 6.
Block Mass (g) Trial 1 (N) Trial 2 (N) Trial 3 (N) Trial 4 (N) Trial 5 (N) Average Force (Static Friction, N)
304.6 0.634 0.619 0.630 0.554 0.632 0.614
404.6 0.937 0.866 0.766 0.835 0.820 0.845
504.6 1.023 0.969 1.043 1.078 1.093 1.041
604.6 1.191 1.263 1.410 1.319 1.319 1.300
704.6 1.489 1.241 1.652 1.434 1.478 1.459
804.6 1.762 1.757 1.723 1.726 1.770 1.726
Block Mass (g) Trial 1 (N) Trial 2 (N) Trial 3 (N) Trial 4 (N) Trial 5 (N) Average Force (Kinetic Friction, N)
304.6 0.473 0.464 0.467 0.450 0.465 0.464
404.6 0.593 0.603 0.598 0.613 0.625 0.606
504.6 0.753 0.751 0.751 0.750 0.787 0.758
604.6 0.902 0.897 0.894 0.903 0.912 0.902
704.6 1.068 1.060 1.068 1.076 1.072 1.069
804.6 1.208 1.203 1.181 1.144 1.168 1.175
Calculations Normal Force/Weight F_W= mg F_W=(0.3046g)(9.81m/s^2 ) F_W=2.0988126 N= 2.10 N
Average Force
F_AVG=(Trials 1 to 5)/5
F_AVG=(0.473N+0.464N+0.467N+0.450N+0.465N)/5
F_AVG=0.4638 N=0.464 N
Coefficient of Friction (Kinetic and Static) f=μn → μ=f/n
μ=(0.4638N/2.0988126N) …show more content…
To begin with, in order to get rid of the error caused by a non-zeroed Force Sensor, be sure to calibrate it every time you start a new trial and keep checking to make sure it is zeroed. Also, make sure that there is no draft in the room to be certain that there will be no air resistance on the wooden block as you pull it. And finally, you can use an instrument that will help pull the block at a constant velocity in order to get an accurate measurement for kinetic friction. If these are taken into consideration, then the lab will be more