Question

The 2040 kg cable car shown in the figure descends a 200-m-high hill. In addition to its brakes, the cable car controls its speed by pulling an 1860 kg counterweight up the other side of the hill. The rolling friction of both the cable car and the counterweight are negligible.

Part A: How much braking force does the cable car need to descend at constant speed?

Part B: One day the brakes fail just as the cable car leaves the top on its downward journey. What is the runaway car's speed at the bottom of the hill?

Answer 1

m1 = 2040 kg                  m2 = 1860 kg

for m2

T - m2*g*sintheta2 = m2*a

as the car is moving with constant speed , a = 0

T = m2*g*sin20.......(1

for m1

m1*g*sintheta1 -F - T = m1*a

m1*g*sintheta1 = F + T

2040*9.8*sin30 = F + (1860*9.8*sin20)

F = 3761.6 N <<-------answer

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B)

F = 0

for m2

T - m2*g*sin20 = m2*a

for m1

m1*g*sin30 - T = m1*a

m1*g*sin30 - m2*sin20 = (m1+m2)*a

(2040*9.8*sin30)-(1860*9.8*sin20) = (2040+1860)*a

a = 0.96 m/s^2

v = sqrt(2*a*L)

sin30 = h/L

L = h/sin30

v = sqrt(2*0.96*200/sin30)

v = 27.7 m/s