The 2100 cable car descends a 200-m-high hill at a 30 degree slope. In addition to its brakes, the cable car controls its speed by pulling an 1700 counterweight up the other side of the hill at a 20 degree slope. 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? The answer i have is correct but I'm looking for an explanation to why i did what i did. Im having a hard time wrapping my head around it completely.
bf(brake force)=mg(sin theta)-mg(sin theta)
=(2100kg)(9.81)(sin 30)-(1700kg)(9.8)(sin 20)=4592 N
The only reason i know the answer is because i've seen a similar problem done but like i said i dont understand why i need to use the equation i used. I appreciate the help.
How much braking force does the cable car need to descend at constant speed? The answer i have is correct but I'm looking for an explanation to why i did what i did. Im having a hard time wrapping my head around it completely.
bf(brake force)=mg(sin theta)-mg(sin theta)
=(2100kg)(9.81)(sin 30)-(1700kg)(9.8)(sin 20)=4592 N
The only reason i know the answer is because i've seen a similar problem done but like i said i dont understand why i need to use the equation i used. I appreciate the help.