Factor of friction strange behavior

AI Thread Summary
The discussion revolves around measuring the factor of friction between steel-brass and steel-aluminum surfaces under varying applied forces. It is noted that the friction coefficient (μ) decreases with increased force for steel-brass, while it increases for steel-aluminum. Participants suggest that the experimental setup may not align with ideal conditions, impacting the results. Key details include the use of a steel rotating disk, smoothened surfaces, and specific pressure levels generating forces of 106 N, 212 N, and 318 N. The conversation emphasizes the importance of understanding experimental variables that could influence friction behavior.
Achileas7
Messages
2
Reaction score
0
Hello everyone, I have a really interesting question here. I am measuring the factor of friction between two different metals. My results are a beat cloudy. When I use Steel - Brass surfaces the factor of friction, μ is being reduced while I am increasing the applied force. In other side, when I use Steel - Aluminum surfaces, when I am increasing the applied force on the two materials the factor of friction, μ goes up. Anyone can explain what is happening? There is no measurement mistake, that' s guaranteed.
 
Physics news on Phys.org
Achileas7 said:
Hello everyone, I have a really interesting question here. I am measuring the factor of friction between two different metals. My results are a beat cloudy. When I use Steel - Brass surfaces the factor of friction, μ is being reduced while I am increasing the applied force. In other side, when I use Steel - Aluminum surfaces, when I am increasing the applied force on the two materials the factor of friction, μ goes up. Anyone can explain what is happening? There is no measurement mistake, that' s guaranteed.

You'll have to tell us a bit more: What are the magnitudes and ranges of normal forces that you're working with, how are the friction surfaces prepared, what pressure is involved, how is the normal force applied and how is the coefficient of friction measured? There are a lot of ways that your experimental setup could deviate from the ideal conditions in which the coefficient of friction is close to independent of the normal force.

(Also "no measurement mistake, that's guaranteed" is something that we've all regretted saying at least once. Just sayin' :smile:)
 
Nugatory said:
(Also "no measurement mistake, that's guaranteed" is something that we've all regretted saying at least once. Just sayin' :smile:)

One of our lab tutor told as so as, we aren' t supposed to focus on such a thing and focus on what' s going on :smile: I can understand you objections on that.

Now with the experiment part:
Our experimental setup includes a steel rotation disk, and to measure the friction factor we use some aluminum and brass rods of the same radius. We use pressures of 1, 2 and 3 bars which creates forces of about 106 N, 212 N and 318 N in our experimental setup.
The surfaces are being threw a high standard rectifier machine, which makes them really smooth. The forces are vertically applied on the rod. The coefficient of friction is measured by a electronical device which gives as the force of the friction. We use the type μ=T/(2*F) as we used two rods at the time, at the opposite side of the disk.

F: Applied Force
T: Friction Force
μ: Friction Factor
 
Last edited:
The rope is tied into the person (the load of 200 pounds) and the rope goes up from the person to a fixed pulley and back down to his hands. He hauls the rope to suspend himself in the air. What is the mechanical advantage of the system? The person will indeed only have to lift half of his body weight (roughly 100 pounds) because he now lessened the load by that same amount. This APPEARS to be a 2:1 because he can hold himself with half the force, but my question is: is that mechanical...
Hello everyone, Consider the problem in which a car is told to travel at 30 km/h for L kilometers and then at 60 km/h for another L kilometers. Next, you are asked to determine the average speed. My question is: although we know that the average speed in this case is the harmonic mean of the two speeds, is it also possible to state that the average speed over this 2L-kilometer stretch can be obtained as a weighted average of the two speeds? Best regards, DaTario
Some physics textbook writer told me that Newton's first law applies only on bodies that feel no interactions at all. He said that if a body is on rest or moves in constant velocity, there is no external force acting on it. But I have heard another form of the law that says the net force acting on a body must be zero. This means there is interactions involved after all. So which one is correct?
Back
Top