Given T<(2/3)mg
Tension of rope
T+ma=mg =>
T
≤mg-ma
=m(g-a)
Solve for a in
(2/3)mg=m(g-a)
with what minimum acc can a fireman slide down a rope whose breaking strenght is two third of his weight ?
1 answer
1 answer
3 answers
1 answer
1 answer
3 answers