1. The problem statement, all variables and given/known data If particles at B and C of equal mass m are connected by strings to each and to points A and D as shown, all points remaining in the same horizontal plane. If points A and D move with the same acceleration a along parallel paths, solve for the tensile force in each of the strings. Assume that all points retain their initial relative positions. PLEASE SEE ATTACHMENT 2. Relevant equations Kinematics- total acceleration: A = 0(i) + g(-j) + a(j) + 0(k) = Ax(i) + Ay(j) + Az(k) Mechanics: ∑F - mA = 0; Magnitude of acceleration due to gravity: g =9.81 Labeling: Tension between A & B: Tab, etc Coordinate System: x to the right y upwards z out of the page 3. The attempt at a solution Freebody Diagram on left point mass (EQ1) ∑F1 - mA = 0 = Tab*cosd(45)(-i) + Tbc(i) + Tab*sind(45)(j) - mA Freebody Diagram on right point mass (EQ2) ∑F2 - mA = 0 = Tdc*cosd(45)(i) + Tbc(-i) + Tdc*sind(45)(j) - mA SOLVING: x-componenets of (EQ1) and (EQ2): Tab = Tdc Substituting Tab for Tdc into (EQ2) y-component: m( g(-j) + a(j) ) = Tab*sind(45)(j) Showing y-component of (EQ1): m( g(-j) + a(j) ) = Tab*sind(45)(j) RESULTING: => Tab = sqrt(2)mAy = Tdc => Tbc = mAy note: A = 0(i) + g(-j) + a(j) + 0(k) = Ax(i) + Ay(j) + Az(k) Ay = -g + a My Questions: 1) The sign of acceleration due to gravity (g) is negative while on the external forces side of the equation, when usually it is negative. Which tells me that there is something wrong with my acceleration term. 2) But, how else can we solve this problem without including the acceleration due to gravity in the total acceleration? Else, I would need to totally neglect it from the solution. 3) If neglected is that the same thing as considering only the dynamic case? i.e. forget the weight = mg. Why? Is there a more intuitive approach? Thank you.