Unformatted text preview: T in the cable? The angle between the cable and the strut is θ – φ = 15º. Since the strut is in equilibrium, 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 θ – MgLcos θ + Tsin( θ – φ )L = 0 ⇒ T = (M + m/2)gcos θ /sin( θ – φ ) = 1.38 × 10 4 N b) What are the horizontal and vertical components of the force on the strut from the hinge? The net force exerted on the strut is zero. The balance of forces equations yield x: F x – Tcos φ = 0 F x = Tcos φ = 1.25 × 10 4 N y: F y – mg – Mg – Tsin φ = 0 F y = mg + Mg + Tsin φ = 1.12 × 10 4 N...
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- Spring '08
- Force, strut, uniform strut, Tomoyuki Nakayama, Tsinφ