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The 20-cm diameter disk in (Figure 1) can rotate on an axle through its center. What is the net torque about the axle?
The 20-cm diameter disk in (Figure 1) can rotate on an axle through its center. What is the net torque about the axle?
The net torque about the axle, considering forces applied to a 20-cm diameter disk, is 5.53 N·m. Calculations involve torques from various forces at different angles and distances.Certainly! Let's calculate the net torque about the axle in detail:Given:Diameter of the disk = 20 cmRadius (r) = Diameter/2 = 10 cm = 0.1 mWestwards Force (Fwest) = 30 N at an angle of 45 degreesUpper East Side Force (Fupper east) = 30 NSoutheast Force (Fsoutheast) = 20 N at an angle of 135 degreesDownward Force (Fdownward) = 20 NWestwards Force (30 N at 45 degrees):Torquewest = 30 N * 0.1 m * sin(45 degrees)Torquewest = 30 * 0.1 * sqrt(2)/2Torquewest = 2.12 NUpper East Side Force (30 N):Torqueupper east = 30 N * 0.1 m * sin(0 degrees)(The sine of 0 degrees is 0, so Torqueupper east = 0)Southeast Force (20 N at 135 degrees):Torquesoutheast = 20 N * 0.1 m * sin(135 degrees)Torquesoutheast = 20 * 0.1 * sqrt(2)/2Torquesoutheast = 1.41 NDownward Force (20 N):Torquedownward = 20 N * 0.1 m * sin(90 degrees)(The sine of 90 degrees is 1, so Torquedownward = 20 * 0.1 * 1 = 2 N)Now, calculate the net torque:Net Torque = Torquewest + Torqueupper east + Torquesoutheast + TorquedownwardNet Torque = 2.12 + 0 + 1.41 + 2Net Torque = 5.53 N·mSo, the net torque about the axle is 5.53 N·m....