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Practice Problems (Forces)

Practice Problems (Forces). Question (1). If the ground reaction forces include 1200N of vertical force & 240N of horizontal force, as pictured What is the resultant of these combined forces?. Answer. F hx = 240N F hy = zero F vx = zero F vy = 1200N

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Practice Problems (Forces)

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  1. Practice Problems(Forces)

  2. Question (1) • If the ground reaction forces include 1200N of vertical force & 240N of horizontal force, as pictured • What is the resultant of these combined forces?

  3. Answer • Fhx = 240N Fhy= zero • Fvx= zero Fvy= 1200N Fx = Fhx + Fvx = 240 N + O = 240 N Fy = Fhy + Fvy = 0 + 1200 N = 1200 N R = [ (Fx)2 + (Fy)2 ]½ = [ 2402 + 12002]1/2 R = 1223.76 N tan = Fy/Fx = 1200/240 = 5  = 78.69°

  4. Question (2) • Therapist is helping a patient stand up. The patient gathers all strength & pushes straight up from the chair with a force of 600N. The therapist, using proper body mechanics, pulls up & out at an angle of 65° with the vertical & with a force of 1280N. • What is the magnitude & direction of the resultant force moving the patient? ( refer to the pcicture in your handout)

  5. Answer • Fpx= zero Fpy= 600N • Ftx = Ft cos (65°+90°) = 1280 cos 155° = -1160 N  Fty= Ft sin (65°+90°) = 1280 sin 155° = 540.9 N Fx = Fpx + Ftx = 0 + -1160 N = -1160 N Fy = Fpy + Fty = 600N + 540.9 N = 1140.9 N

  6. Answer R = [ (Fx)2 + (Fy)2 ]½ = [-1160 2 + 1140.9 2]1/2 R = 1627.03 N tan = Fy/Fx = 1140.9 /-1160 = -0.98  = -44.52°

  7. Question (3) • The vastus medialis (VM) muscle pulls on the patella at 40° from shaft of the femur with a magnitude of 240N. The vastus lateralis (VL) muscle pulls on the patella at 28 ° from the shaft of the femur with a magnitude of 200N. The rectus femoris (RF) muscle pulls on the patella at 10° from shaft of the femur with a magnitude of 100° • What is the magnitude & direction of the quadriceps muscle force on the patella? ( refer to the pcicture in your handout)

  8. Answer • FxVM= FVM sin 40° = 240 sin 40° = 154.2 N • FyVM= FVM cos 40° = 240 cos 40° = 183.8 N • FxVL= FVL cos (28°+ 90°) = 200 cos (28°+ 90°) = -93.9N • FyVL= FVL sin (28°+ 90°) = 200 sin (28°+ 90°) = 176.5 N • FxRF= FRF cos (10°+ 90°) = 100 cos (10°+ 90°) = -17.36N • FyRF= FRF sin (10°+ 90°) = 100 sin (10°+ 90°) = 98.48 N

  9. Answer Fx = FxVM + FxVL+ FxRF = 154.2 N + -93.9N + -17.36N = 42.94N Fy = FyVM + FyVL+ FyRF = 183.8N + 176.5N + 98.48 N = 458.78N R = [ (Fx)2 + (Fy)2 ]½ = [42.942 + 458.782]1/2 R = 460.78 N tan = Fy/Fx = 458.78 / 42.94= 10.68  = 84.65°

  10.  Study Hard & Good Luck 

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