I am unclear as to why we are taking the components of weight and centripetal force here, have you chosen the +y axis along BC?
If I work backwards from isolating velocity I get a net force of 1.6321N, which I don't understand in terms of my beginning free body diagram using AB along the...
Sorry, just to clarify;
Whole string tension = TensBC + Tens AB
tensile force of AB = Tcos(theta)
tensile force of BC = T
(if T= 1.2255N and Tcos(53.1o) is 0.7358N, then the tension in the entire string is 2.0N)
Centripedal force = TensBCx + Tens AB
tensile force of AB = Tcos(theta)...
Sorry! I should have been more clear with my calculations straight off :)
I don't actually have the answer for this problem, so I can't say whether or not I am approaching this the right way...from my imagination of this system I think maybe that since AB is taut (horizontal) it is pulling...
Sure thing!
Ftens*sin(theta) = mg
Ftens = (0.1kg)(9.8N) / sin53.1o
Ftens = 1.2255N
So this would be the tension exerted by the portion of the rope BC
Ftens*cos(theta) = Ftensx
1.2255N*cos53.1o = Ftensx
0.7358 = Ftensx
I'm assuming where I went wrong in all this is because...
1. Homework Statement
An 100g bead is free to slide along an 80cm long piece of string ABC. The ends of the string are attached to a vertical pole at A and C, which are 40cm apart. When the pole is rotated, AB becomes horizontal.
a) find the tension in the string
b) find the speed of the...