PART A:
When being taught about Newton's Second Law, I learned many things. The majority of this is comprised of the acceleration a system experiences when acted upon by an unbalanced force. This is includes the friction that an object experiences when moving (neglecting air resistance, usually). I learned about using my FBDs to calculate the sum of the forces, ∑Fx and ∑Fy, and setting them equal to ma (mass*acceleration). This includes pulley systems and Atwood's machines. I also learned about the coefficient of friction, called mu. This can be obtained from the equation Ff=mu*Fn. Mu is actually a legal naked number because it is obtained by dividing newtons by newtons.
What I have found particularly difficult is solving for the acceleration with very limited information, such as not knowing the mass or weight of the object. I also find it very difficult to find mu in complex situations, such as when the object is accelerating or under certain circumstances and restrictions.
As far as my problem solving skills go, I am satisfied. I feel as though I have mastered FBDs and finding the sum of the forces in one axis. I am good at solving equations for what is needed. I also particularly enjoy the problems relating to Atwood's machines. I am still new to finding mu, which poses difficulties. I also worry when I am given limited information. I always do work to my fullest effort though, so success is soon to follow.
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Your description as perfect, not too long and not too brief! And good confidence at the end of your reflection!
ReplyDeleteGood Job Cyrus! The lenght of your paragraph is easy to understand and very infromative. I liked how you gave the equations and clarifed what the mu is and why it was a legal naked number. Over all you did a very good job and explained everything we have learned so far very well and specific
ReplyDeleteI am so thankful for your kind words about my reflection. Thanks for taking the time to read my thoughts on what we have learned.
ReplyDeleteJust so you know the sum of the forces is only equal to mass*acceleration in special cases for complete accuracy it should really be change in momentum over change in time(dp/dt). You also at one point made it seem like mass and weight are the same thing when weight is actually mass being acted upon by gravity. As well as from the outside I have no clue what you mean by FBDs.
ReplyDeleteWell, we learned that the sum of the forces was equal to ma, because we hadn't yet learned of momentum. What I was implying with the mass and weight was that they essentially are the same piece of data, because with one, I can automatically find the other, I was not at all referring to them as the same thing. FBDs are free-body diagrams.
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