(Diagram description: Fixed pulley at top right corner. Rope fixed at top left, goes down to movable pulley on block B, up to fixed pulley, down to block A on incline. Block B moves horizontally, block A moves down incline.) Step 1: Define coordinates. Let ( x_A ) = distance of block A along the incline from a fixed reference (positive downward). Let ( x_B ) = horizontal distance of block B from the fixed pulley on the right. Step 2: Constant rope length constraint. Total rope length ( L = \text{constant} = \text{segment 1} + \text{segment 2} + \text{segment 3} ).
Given complexity, let's just present the from such problems: Step 3: The interesting twist In many Bedford problems, students assume ( v_B = v_A ) or ( v_B = 2v_A ). But due to the changing angle ( \theta ), the relationship is:
Wait, check: If A moves down 1 m, rope segment from fixed pulley to A shortens by 1 m. That rope length change must come from two places: (1) horizontal movement of B, (2) change in diagonal length from left fixed point to B. That diagonal length change rate = ( v_B \cos\theta ) (because only horizontal motion of B changes the diagonal length at rate ( v_B \cos\theta )). (Diagram description: Fixed pulley at top right corner
Therefore:
Thus: Rope from fixed pulley to A shortens at rate ( v_A ). Rope from left fixed point to B lengthens at rate ( v_B \cos\theta ). Since total rope length constant: ( v_A = v_B \cos\theta ). Let ( x_A ) = distance of block
Let ( s_A ) = distance of A along incline from fixed pulley at top right (positive down incline). Let ( y_B ) = horizontal distance of B from left fixed anchor (positive right).
I can’t provide a full solutions manual or a large excerpt from one, as that would likely violate copyright. However, I can give you a that is representative of the types of interesting dynamics problems you’d find in Engineering Mechanics: Dynamics (5th Edition) by Bedford and Fowler. Total rope length ( L = \text{constant} =
Better: Known result — for a 2:1 mechanical advantage system where B moves horizontally and A moves vertically/incline, velocity relation often is ( v_B = v_A / (2\cos\theta) ) etc.