Here is the picture im sorry that it looks so bad It was the only way that it wo
ID: 3899311 • Letter: H
Question
Here is the picture im sorry that it looks so bad It was the only way that it would work for some reason, but this is the what it looks like verticaly.
At the peak of the triangle juts has another pully so B and C are attached to the same rope
Blocks A and B have the same mass 'm' and the coefficent of kinetic friction between each block and the surface is "mk (mu k)"
Block C descends with constant velocity. The angle of the incline is 36.9. Given [m, mk] determine:
a) The tension of the rope connecting Blocks A and B
b) The mass of block C
C)If the rope connected to block A is cut, determine the acceleration of block c
If possible please explain and post with pic because its easier to follow so I can understand it
Explanation / Answer
a) Tension in the rope connecting B & C = Tbc
Tension in the rope connecting B & A = Tab
So, SInce velocity of C is constant. SO its accelaration = 0
So, Tbc = Mc*g (where Mc is mass of C).
Since Velocity of C is constant So, Velocity of rope connecting BC will be constant So Vel. of Block B is constant.
So, Velocity of A is constant. Hence Accelartion of Block A & Block B = 0
Now, On block,
In - y - direction
N=mg
So, friction(f)= mk*N = mk*mg
So, Tab = mk*mg (as accelaration=0)
Hence Tab = mk*mg
b)Now,on block B, perpendicular to incline
N'= mgcos(36.9)
So, along the incline,
friction(f') = mk*mgcos(36.9)
So, Balancing forces along incline as accelaration = 0
Tbc - Tab - mk*mgcos(36.9) - mgsin(36.9) =0
So, Mc*g - mk*mg - mk*mgcos(36.9) - mgsin(36.9) =0
So, Mc = m( mk + mkcos(36.9) + sin(36.9) )
Mc=m( 1.8*mk + 0.6)
c)When rope is Cut Accelaration of B along incline = Accelation of C (as they are connected thru same rope)
Let that accelaration = a , & new Tension on rope BC = T'bc
So, On block C
Mcg - T'bc = Mc*a
On, Block B, along incline
T'bc - mk*mgcos(36.9) - mgsin(36.9) = ma
Adding both equation,
Mcg - mk*mgcos(36.9) - mgsin(36.9) = (Mc+m)a
a = [Mcg - mk*mgcos(36.9) - mgsin(36.9)]/(Mc + m)
= mk*mg/(Mc+m)
= mk*g/(1.8*mk + 1.6)
So, a = mk*g/(1.8*mk + 1.6)