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Resultant Forces: Analyzing Vertical and Horizontal Forces in Different Scenarios

This resource presents practice problems focused on calculating resultant forces from different scenarios involving ground reaction forces, therapist-patient interactions, and muscle contractions. Specific questions detail vertical and horizontal forces, allowing students to apply vector addition methods and trigonometric functions to derive resultant magnitudes and directions. Each question includes step-by-step solutions for clear understanding and mastery of force resolution concepts. Perfect for students learning mechanics and dynamics in physics.

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Resultant Forces: Analyzing Vertical and Horizontal Forces in Different Scenarios

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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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