2.A uniform horizontal beam that weighs 150 N is attached at one end to a wall b

Question

2.A uniform horizontal beam that weighs 150 N is attached at one end to a wall by a pin connector that allows it to rotate. A cable firmly attached to the wall above the pin supports the opposite end of the beam. The beam is 4.0 m long and the cable makes an angle of 58o above the horizontal with beam. An additional mass of 50 kg is hanging down from a point that is 3.0 m from the pin. a.What is the tension in the cable necessary to keep the beam horizontal in equilibrium? b.What is the magnitude and direction of the force on the pin?

Explanation / Answer

Let's use the pin connector as a reference point. The sum of all of the torques acting to turn the beam clockwise about this pin must equal the sum of all torques acting to turn the beam in the opposite direction. torque due to weight of beam = torque due to tension in cable W(L / 2) = TL sin? W / 2 = Th / sqrt(h² + L²) (150 N) / 2 = T(3.0 m) / sqrt[(3.0 m)² + (4.0 m)²] T = 125 N All of the forces on the beam must balance, too. vertical forces: weight = (vertical component of cable tension) + (vertical component of force from pin) W = Tsin? + (Fy) 150 N = (125 N)(3 m / 5 m) + Fy Fy = 75 N horizontal forces: horizontal component of cable tension = horizontal component of force from pin Tcos? = Fx (125 N)(4 m / 5 m) = Fx Fx = 100 N magnitude of force on pin: F = sqrt[(75 N)² + (100 N)²] F = 125 N direction of force on pin: tan? = 75/100 ? = 36.9° above the beam

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