An 800N painter stands 3.00m from the left end of a scaffold that is 4.00m long. The uniform wooden scaffold weighs 100N and is hung by a chain at each end. What is the tension in each chain?A uniform meter stick weighs 25 N. An 80 N weight is hung at the 20.0cm mark. A 70.0N weight is hung at the 90cm mark. what is the magnitude of the upward force needed to balance the meter stick?Please help! I have a huge exam on these and my teacher wouldn't help me! I am so stumped please help!
Required conditions for equilibrium (Other wise the scaffold is moving) Sum of forces = 0 Sum of Torques = 0 Sum of Forces Painter: 800 N DOWN Scaffold: 100 N Down Sum of tensions in chains = 800 + 100 = 900 N UP Sum of Torques (Pick one end, it does not matter which. I choose the Left end) Torque = Force * Lever arm Painter is 3.0 meters from Left end Tp = 800 N * 3.00 m = 2,400 N-m CW (Clockwise) Center of gravity is at the center of the scaffold or 4/2 = 2 meters from the Left enf Ts = 100 * 2 = 200 N-m CW This means that the chain on the Right end must supply 2400 + 200 = 2600 N-m CCW Tc = Fc * 4.00 m 2600 = Fc * 4 Fc= 650 N UP Total need up is 900 N====> 900 - 650 = 250 N in Left chain QED
