It slides over towards the heavier weight until the scale gets caught by the pulley, and becomes fixed in place, and then registers the smaller of the two weights.
Another physics puzzler: relative to the stationary background - does the bike travel with the train, or, stay in place (more or less)? https://www.youtube.com/watch?v=TP_0Vv5F29I 5:05 World First! Bike Flip On A Moving Train
During my engineering education I took a course in Newtonian physics in which we learned about force systems and masses in motion. Later I took a civil engineering course called statics in which we learned methods for solving tricky systems of forces and masses that were not in motion. I also took a mechanical engineering course called dynamics in which we learned methods for solving tricky systems of forces and masses that were in motion. I also taught engineering for 30 years during which I taught these techniques to my students, the primary tool is called a free body diagram (FBD). In dynamic systems force on a mass equal the mass times acceleration (ΣF=ma). In static systems forces on a mass are equal to zero because the acceleration is zero (ΣF=0). The scale problem is a trivial statics problem, the tension in each string is 100N, the scale is static and the sum of the forces on the scale is 100-100=0. The scale is in 100N of tension just like the strings. The train problem is pretty simple if the train doesn't accelerate or decelerate (meaning it is static) and the bicycle can be considered the only dynamic mass. In understanding the physics, it doesn't matter that the train is moving except it creates artificially windy conditions for the bicycle. Static means no acceleration - recall that the scale problem is hurtling through space, orbiting the sun and spinning on the earth's axis but is static because it does not accelerate in our frame of reference. As I told many students repeatedly, you can't solve the problem unless you can draw a correct FBD.
What's the YouTube video showing what happens when one of the two weights is made less than 100N, or, greater than 100N?
ReplyDeleteIt slides over towards the heavier weight until the scale gets caught by the pulley, and becomes fixed in place, and then registers the smaller of the two weights.
DeleteSimple: if it's out of balance, it falls. If it gets stopped by the pulley, that's like the bracket scenario, and it reads the lighter weight.
DeleteAnother physics puzzler: relative to the stationary background - does the bike travel with the train, or, stay in place (more or less)? https://www.youtube.com/watch?v=TP_0Vv5F29I 5:05 World First! Bike Flip On A Moving Train
ReplyDeleteDuring my engineering education I took a course in Newtonian physics in which we learned about force systems and masses in motion. Later I took a civil engineering course called statics in which we learned methods for solving tricky systems of forces and masses that were not in motion. I also took a mechanical engineering course called dynamics in which we learned methods for solving tricky systems of forces and masses that were in motion. I also taught engineering for 30 years during which I taught these techniques to my students, the primary tool is called a free body diagram (FBD). In dynamic systems force on a mass equal the mass times acceleration (ΣF=ma). In static systems forces on a mass are equal to zero because the acceleration is zero (ΣF=0). The scale problem is a trivial statics problem, the tension in each string is 100N, the scale is static and the sum of the forces on the scale is 100-100=0. The scale is in 100N of tension just like the strings. The train problem is pretty simple if the train doesn't accelerate or decelerate (meaning it is static) and the bicycle can be considered the only dynamic mass. In understanding the physics, it doesn't matter that the train is moving except it creates artificially windy conditions for the bicycle. Static means no acceleration - recall that the scale problem is hurtling through space, orbiting the sun and spinning on the earth's axis but is static because it does not accelerate in our frame of reference. As I told many students repeatedly, you can't solve the problem unless you can draw a correct FBD.
ReplyDeleteI don't care what he says, the bracket is not pulling. That can't be correct terminology.
ReplyDeleteFrom the perspective of the scale, it is. And these physics problems are always about figuring things out from the proper perspective.
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