A window washer is standing on a scaffold supported by a vertical rope at each end. The scaffold weighs 190 N and is 3.30 m long. What is the tension in each rope when the 710 N worker stands 2.30 m from one end?smaller tensionN?larger tensionN?
think of statics (in terms of forces and torques): because of fact the gadget is in equilibrium, the sum of the forces = 0 and the sum of the torques = 0 on the grounds that sum of forces = 0, then: ft(left) + ft(suitable) - Fg(scaffold) - Fg(employee) = 0 on the grounds that sum of torques = 0, then: -Set some element as your axis of rotation. for my area, I set between the ft as my axis. to that end, my torque equation will become ft(suitable)(2.ninety m) – Fg(scaffold)(a million.40 5 m) – Fg(employee)(a million.ninety m) = 0 you already know Fg(scaffold), Fg(employee), so resolve for ft(suitable). Plug this value into the tension equation and resolve for ft(left).
If the scaffold weight is uniformly distributed, each rope will support one half of the weight. Left reaction = 95N Right reaction = 95N The 710N worker is standing closer to one end (assume left). Sum moments about the left end = 0 1m * 710N - 3.3m * Right reaction = 0 Right reaction = 710/3.3 = 215.2N Sum forces in Vertical direction = 0 Left reaction = 710 - 215.2 = 494.8N Sum all forces Left reaction = 95N + 494.8N = 589.8N larger tension force Right reaction = 95N + 215.2N = 310.2N smaller tension force
