Engineering Mechanics Dynamics Fifth Edition Bedford Fowler Solutions Manual Link

[ v_B = \frac{v_A}{\cos\theta} ]

Therefore:

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: Since total rope length constant: ( v_A = v_B \cos\theta )

For ( \theta = 30^\circ ), ( \cos 30^\circ = 0.866 ): That diagonal length change rate = ( v_B

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 ). 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 )).