I want to know the technical reason of different dye uptake of the same fibre and yarn whether it is cotton?
This is a Statics problem, but simple trigonometry can be used to solve itYou'll want to start by drawing all the forces acting on the objectIn this case, there is the object's weight of 20,240 lbf (lbf is an abbreviation for pound-force, or simply pounds), and the unknown tensions in both cablesDraw them at their respective angles from the horizontalYou'll need two equations to solve; the sum of the forces in the vertical direction, and the sum of the forces in the horizontal directionBecause we're given the angles, this is relatively straightforwardLet T1 be the force in the cable at 44.5 degrees and T2 be the force in the cable at 42.3 degreesFor the forces in the vertical direction: T1sin(44.5) + T2sin(24.3) 20240 For the horizontal direction: T1cos(44.5) - T2cos(24.3) 0 Solve both of these equations for T1 and T2 to find the tension in each cable T1 19,785.8 lbf T2 15,484.1 lbf This answer makes sense because both of those forces summed are larger than the weight of the object.
Not sure of your question butthe permeability of materials (ability to take and hold dye for example) can vary due to many different factors: density, porosity, composition, temperature, presence of impurities/surface contamination, water content, etc There are different types of dyes which work different chemicallyIf the primary cause of the variation you are seeing is due to variation in the incoming material, you may be able to improve your results by using a different dye chemistry good luck good luck
Tension T1 20/sin44.5 28.53 poundsT2 240/sin24.5 578.74 pounds
defining variables: a angle one 44,5 b angle two 24.3 T1 tension in rope one T2 tension in rope two you can solve this by using vector algebra or newtonian mechanicsboth will lead to the same answer, i will use newtonian mechanicsthis is a 2nd law problem i.eFmA we need to break this problem up into its componentsthe net force in the x direction is; Fnet,x T1cos(a) - T2cos(b) since the object is not moving in the x direction the forces must cancel out and equal zerotherefore; 0 T1cos(a) - T2cos(b) T2 [cos(a)/cos(b)](T1) the net force in the y direction is; Fnet,y T1sin(a) + T2sin(b) - Fg where Fg is the objects weight; 20240 lbs your problem suggests that the object is moving, but since there is not an acceleration the net forces must equal zerotherefore; Fg T1sin(a) + T2sin(b) now we have 2 equations and two unknownsso, we can solve this system using any method of choice ) Fg T1sin(a) + T2sin(b) Fg T1sin(a) +sin(b)[cos(a)/cos(b)](T1) T1{sin(a) +[cos(a)/cos(b)]} Fg T1 [Fg]/{sin(a) +[cos(a)/cos(b)]} - answer T2 [Fgcos(a)]/[sin(a)cos(b) + cos(a)] - answer sorry i don't have a calculator so i only can give you a general solution to this problemhope it was helpful none the less.
