Lecture 10 - Friction 2

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  • เผยแพร่เมื่อ 13 ม.ค. 2025

ความคิดเห็น • 6

  • @m.rajakumarreddy7425
    @m.rajakumarreddy7425 2 ปีที่แล้ว

    Still sign conventions are confusing

  • @priyanksharma1518
    @priyanksharma1518 7 ปีที่แล้ว +1

    FBD is wrong. there should be Fcy,though it will be zero.Also Fbd shown in FBD is also wrong as it is not the external force on the frame but on a particular joint.While analysing joint B showing Fbd is correct but not during the FBD of entire frame.

    • @narang__2309
      @narang__2309 4 ปีที่แล้ว

      Too late to reply but I feel Fbd part is correct. It is an external force on the frame, If you isolate the frame. That's the force due to which pin B holds the frame.

    • @ktk3335
      @ktk3335 4 ปีที่แล้ว

      @@narang__2309 can you please tell me why there is no fcy in the F.B.D though the support c is a pin joint.

    • @narang__2309
      @narang__2309 4 ปีที่แล้ว +4

      @@ktk3335 Even I had that doubt. This is how I approached it. If you see the resolution he has done at B. He first took the Force B along rod and then resolved in X And Y components. It means whatever the resultant of X and Y comes. It will lie along that member. ( Because it's a two force member, it can either have tension or compression along it's axis). Similarly, if you take Fcx And Fcy, it's resultant will like along the memebr CD. But since it's resultant is along Fcx. So Fcy is 0. That's my approach. Better to confirm from your teacher.

    • @mukeshkhadav3836
      @mukeshkhadav3836 4 ปีที่แล้ว

      @@narang__2309 you are correct i also felt the same problem fcy is 0