This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: = 25. The strut is in equilibrium. Therefore, the net torque exerted on the strut is zero. We choose our rotational axis at the hinge. The balance of torques equation yields -mg(L/2)cos MgLcos + Tsin( )L = 0 T = (M + m/2)gcos /sin( ) = 5.15 10 3 N b) What are the horizontal and vertical components of the force on the strut from the hinge? Since the strut is in equilibrium, the net force exerted on the strut is zero. The balance of forces equation yields x: F x Tcos = 0 F x = Tcos = 4.97 10 3 N y: F y mg Mg Tsin = 0 F y = mg + Mg + Tsin = 4.57 10 3 N...
View Full Document
- Summer '08