Question

V- 70 Question 2 (30 points) 2 A 60 kg passenger sits in an amusement park ride. The arm of the ride CB is 12 m long. The passenger remains in an upright position as the arm rotates about point C. The arm rotates back and forth, remaining in the plane of the page. At the instant 0 350, the passenger has a speed of 10 m/s which is decreasing at 2 m/s a) Determine the forces that the chair exerts on the passenger in order to produce this motion. These forces can be expressed in any convenient coordinate system -90 Draw a separate diagram of the passenger that shows the directions of their velocity and acceleration b) c) If the ride were altered such that the arm changed length as it rotated, what, if anything, would change about your approach for solving for the forces? What additional information, if any, would you need about the ride? ATe AIB Don't solve with moments, use equations of motiowm ae -190 300

Answer 1

Given that, The arm of the ride C B is 12 m long theta = 35 degree Passenger has a speed of 10 m/s which is decreasing at 2 m/s^2 To Calculate R_B rightarrow force that due chair exerts an passenger (R_C)_V = vertical reaction force at point �C� (R_C)_H = horizontal reaction force at point �C� R_B = force exerted by chair on passenger R_B cos theta & R_B sin theta = Component of R_B i e: vertical & horizontal component m g = weight of due passenger acting vertical downward. \r\n rightarrow from Delta C O B cos 35 = C O/C B = C O/12 (we know C B = 12) cos 35 = C O/12 C O = (cos 35) 12 C O = 0.819 times 12 C O