The figure shows a 2000 kg cable car descending a high hill. A counterweight of mass on the other side of the hill aids the brakes in controlling the cable car's speed. The rolling friction of both the cable car and the counterweight are negligible. How much braking force does the cable car need to descend at constant speed?
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mcablecar=2000 kg mcounterweight=18000 kg g= 9.8 m/s2 h=200 m θcc= 30o θcw= 20o For the braking force: (2000kg)(9.8 m/s2)(sin30o) - (1800kg)(9.8m/s2)(sin20o)=...
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Answer:
3766.76
Explanation:
So there is no picture, so I am going to assume your picture is the same as mine, so adjust accordingly.
If the traincar is moving at a constant speed, acceleration is 0. Because of this:
Tension+Breaking Force-Force of gravity=0
=T1+F(b)-m1gsin(X)=0.
We don't know tension either, so we will solve for that.
On the counterweight:
Tension-Force of Gravity=0, or T1=m2gsin(Y)
plug this T1 into the other equation to get
m2gsin(Y)+F(b)-m1gsin(X)=0, or F(b)=m1gsin(X)-m2gsin(Y)=3766.7 N
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