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Unformatted text preview: external forces are exerted on the system: Gravitational force at the center of the sign, tension in the cable, and x and y components of the force from the hinge. Choosing our rotational axis at the hinge, we apply the balance of torques equation to get the tension. Tsin θ d h – mg(d h – L/2) = 0 ⇒ T = mg(d h – L/2)/(sin θ d h ) = 648 N b) What are the horizontal and vertical components of the force on the rod from the wall? Take the positive directions to the right and upward respectively. We take our +x axis to the right and +y axis upward. The balance of forces equations yield x: F x – Tcos θ = 0 F x = Tcos θ = 360 N y: F y + Tsin θ – mg = 0 F y = mg – Tsin θ = 245 N...
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This note was uploaded on 01/25/2011 for the course PHY 2048 taught by Professor Field during the Summer '08 term at University of Florida.
 Summer '08
 Field
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