Vertical Motion question..

Hello, I have a vertical motion question which I dont seem to understand fully, though ive started it...

An acrobat of mass 'm' slides down a vertical rope of height 'h'. for the first three-quarters of her descent she grips the rope with her hands and legs so as to produce a frictional force equal to five-ninths of her weight. She then tightens her grip so that she comes to rest at the bottom of the rope. Find the frictional force she must produce in the last quarter. If the rope is 30m high, calculate
(a) Her greatest speed
(b) the time she takes to descend.

Could you at least help me with the first part...
(Finding this chapter really annoying)
Thanks a lot
From John

Reply 1

Codefusion

mg - 5/9*mg = m*a
4/9mg = m*a
a = 4/9g

a = 4/9g
u = 0
s = 3/4*30 = 22.5
v = ?

v² = 2(as) = 196
v = 14ms^-1(a)

a = ?
v = 0
u = 14
s = 1/4*30 = 7.5
196 + 15a = 0
a = -196/15^-2

mg - f = m * -196/15
f = mg + m * 196/15 = 35*9.8/15m = 7/3mg (firctional force equal to 7/3 of her weight)

time of first trip = 14/(4/9g) = 45/14s
time of second trip = -14/(-196/15) = 15/14s
time of trip = 45/14 + 15/14 = 60/14 = 4.2857 = 4.29 (3 s.f.)

Reply 2

Neo1

Thanks a lot, though I dont understand a lot how you got from

f = mg + m * 196/15 equal to that 35*9.8/15m = 7/3mg

Do u mind explaining..

Thanks
From John

Reply 3

Codefusion

f = mg + 196/15m
f = m(9.8+196/15)
f = 343/15m = 35*9.8/15m (simply divided 343 by 9.8)
f = 35/15 * mg = 7/3mg

Reply 4

dargorito

Totally wrong answer bro. It should be. 7/3mg frictional force 6s total time and greatest speed 20. Dont answer if you are going to mislead somebody that wants help. Step 1 . U make sure your values for displacement speef and acceleration are all in the same direction. Basically change the minus to a plus or plus to a minus so the direction for all s v and a are the same then properly give the correct signs ( + or minus ) . Then go for the answer i take my downwards as positive, remember resultant force or ma or fnet must be in the same direction as the acceleration you are using for your a in your ma . Then, mg - 5/9 mg=ma , m(g-5/9)=m(a) , cut all the m and you have g-5/9g=a, g(1-5/9)=a and finally a = (4/9)g Remember your final answer for fricitional force can be left as (7/3)mg you dont need the exact value. I know because i have checked my mechanics book answers for this question trust me. Then you will use g=10 since mechanics maths are outdated they dont change to 9.81 like physics and you will have a=(4/9)(10) or simply type in 4÷9×10 it is exacty the same trust me try it yourself with the calculator if you dont. So now. We use some of the general equation for mechanics, v^2=u^2+2as , assume the person starts from rest(i dont know why she starts from rest or why i should assume this but you cant get your answer if you dont assume u=0 so deal with it life is made from choices) prepare u=0, a=40/9 then substitute them into the equation u will have v^2=0^2+2(40/9)(s) now the s is mentioned in the question, 3 quarters is basically 3/4 so your s will be s=(3/4)h and h was given 60m so (3/4)(60)=s s=45 then put it in the equation again v^2=(80/9)(45) , v= sqrt.(400)

Reply 5

dargorito

Then just use s=(1/2)(u+v)(t) find time plot graph get total time . Remember the graph will be a v shape or alphabet upside down and find force using the same way you found your acceleration in the first step