Question

A135° 1. 25° Problem #4 Counterweight The 3000 kg cable car shown in figure descends a 200 m-high hill. In addition to its brakes, the cable car controls its speed by pulling a 2500 kg 200 m counterweight up the other side of the hill. The rolling friction of both the cable car and the counterweight are negligible. A. Draw the free body diagrams of cable car (call it m ) and the counterweight (call it m2) mass. Draw component diagrams. B. Choose an appropriate set of axes and write Newton's laws for both masses in x and y directions. C. Solve the equations and find out how much braking force does the cable car need to descend at a constant speed?

Answer 1

Given m, 3000 kg mg = 2500 kg T = Tension is the Streina e = 350 coso, a = acceleration - 0 m/s2 F = Brecaking force masine 19 US cos _ Giver) Constant Speed means accel Seat speed means acceleration eys 2030. 6= acceleration due to creavity = 9,8 m/s² mssa Fore mass'm F&T - mosina = M, a - T - -F + missing + m2 - Fore mass ma massina - Tama u Petting T value foom equation @ Ento equation am sine – (F & magsine, tm, a) - ma y mas sinea +F - mis sing, - ma mga since a so >> mag sinea +F - mig sine, -0-0 fa mig sine, - mas sinen F = (3000kg) & (9,8 m²) x sin (35) - (2500 k) & g. 8 m/ex sinlar, F = 16863.142 - 10354.144 » F = 6508.9998 N $ 6509 N - 6.509 xloºn