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How is free body diagram of a block on an inclined plane different than that of a vertical circle (string connected to a block doing circular motion in vertical direction) and a banked road

Because there are two types of results in both cases:- T(tension)/N(Normal Reaction) cos [some angle]= mg And Mg cos[some angle]= N/T !(https://i.sstatic.net/VIEX3.jpg)

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  • $\begingroup$ What have you tried? Show us your work, in terms of the three FBD you have drawn. $\endgroup$
    – John Alexiou
    Commented Jan 22, 2018 at 19:50
  • $\begingroup$ I couldn’t really find my problem in vertical circle FBD cause there’s a special case that might become analogue to the ones above so forgive me for that but yeah,there’s the problem $\endgroup$
    – Dilin Finn
    Commented Jan 22, 2018 at 20:53
  • $\begingroup$ Are you saying that you don't know how to draw a FBD of an object attached to a string traveling in a circle or an object on a banked turn? $\endgroup$
    – Chet Miller
    Commented Jan 22, 2018 at 23:14
  • $\begingroup$ Yeah the different results are so beyond me $\endgroup$
    – Dilin Finn
    Commented Jan 23, 2018 at 2:41
  • $\begingroup$ What you do is draw a sketch of the object upon which you want to do a force balance, and, in this sketch, show (using arrows) all the forces acting on the object. Does that sound very hard? $\endgroup$
    – Chet Miller
    Commented Jan 23, 2018 at 3:42

1 Answer 1

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This is the general inclined plane (all red arrows are vectors)

This is the general banked ramp, notice that when you solve for anything, you can use: $a = v^2/r$ and solve for various unknowns

This is the block in vertical motion, notice that:

  • when the block is at the top: $F_Total = F_T + mg$ (accel direction chosen is down)

  • When the block is at the middle, $F_Total = mg$ ($F_T$ plays no role, is perpendicular to accel)

-when block is at the bottom, $F_Total = F_T - mg$ (accels up, x+ chosen is towards center)

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  • $\begingroup$ Thanks for the great work. Now if we resolve the second diagram,we get two results in the vertical direction as I have stated in an above comment. I wanna know how $\endgroup$
    – Dilin Finn
    Commented Jan 23, 2018 at 12:03

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