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.
Constraint: Total rope length ( L = \underbrace{y_B} {\text{horizontal top left to B}} + \underbrace{\sqrt{y_B^2 + H^2}} {\text{diagonal from B up to fixed pulley?}} ) — This gets messy. Let's do the : Two movable pulleys.
(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} ).
This example focuses on a common but subtle topic: and relative velocity , which often trips students up. Sample Problem (Inspired by Bedford & Fowler, Ch. 2-3) Problem: Block A is pulled down the inclined plane at a constant speed ( v_A = 2 \text{ m/s} ). The rope system shown (a single continuous rope, fixed at the top left, passing through a movable pulley attached to block B, and then down to block A) causes block B to move horizontally. Determine the velocity of block B when the rope segment between the fixed pulley and block B makes an angle ( \theta = 30^\circ ) with the horizontal. The rope is always taut and inextensible. Better: Known result — for a 2:1 mechanical
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:
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.
For ( \theta = 30^\circ ), ( \cos 30^\circ = 0.866 ): (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 ).
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 )). Let ( x_A ) = distance of block
[ v_B = \frac{v_A}{\cos\theta} ]
[ v_B = \frac{v_A}{\cos\theta} ]
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).