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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This note was uploaded on 12/10/2011 for the course PHY 2048 taught by Professor Field during the Spring '08 term at University of Florida.
 Spring '08
 Field

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