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NFPA 1006 Office Of The Fire Marshal (OFM) Practice Test

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With a 400-lb load going through a directional pulley with an angle of 120°, what is the force exerted on the directional anchor?
Correct Answer:
400 lbs
Explanation:
The force on a directional anchor comes from the vector sum of the tensions in the two rope segments that meet at the redirect. If the load is 400 lb and the rope on both sides carries that same tension, the two tension vectors are equal in magnitude and separated by the given angle. Use the vector add-on approach: F^2 = T^2 + T^2 + 2 T^2 cos θ, where θ is the angle between the rope segments. With θ = 120° and T = 400 lb, cos 120° = -1/2. So F^2 = 2(400^2) + 2(400^2)(-1/2) = 320,000 - 160,000 = 160,000, and F = 400 lb. Intuitively, as the angle between the rope legs grows, the anchor force decreases from 800 lb (when the rope angles are the same) toward 0 lb (when the legs oppose each other); at 120°, the anchor bears 400 lb.

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