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File Edit View History Bookmarks Iools Help DCCCD Care. DCCCD Sea DCCCD Det... DCCCD Det DCCCD Det Staff Richla Search Vibiscms/mod/ibis/ php?id 3029018 sa plinglearning.com p Most visited Getting started Financial Aid Office :t.. Physical Quan es an Periodic Table Calcul 00 Question 7 of 9 Map 95 Sapling Learning 90 After an unfortunate accident at a local warehouse you have been contracted to determine the cause. A jib crane collapsed and injured a worker. An image of this type of crane is shown in the figure below. The horizontal steel beam had a mass of 81.70 kg per meter of length and the tension in the cable was 00 T 30 N. The crane was rated for a maximum load of 454.5 kg. If d 6.160 m, s 0.450 m x 1.800 m and h 2.250 m, what was the magnitude of WL (the load on the crane) before the collapse? 00 What was the magnitude of force at the attachment point P? The acceleration due to gravity is g 9.810 m/s2 00 Numbe 00 00 Next Previous 2 Check Answer x H how to scree Events for De, Richland Available From: /11/2016 10:50 AM 11/16/2016 1 55 PM Due Date: Points Possible 00 Grade Category: Graded Description: Policies: Homework Li You can check your answers You can view solutions after the due date n keep trying to answer each question u you get it right or give up. You lose 5% of the points available to each answer in your question for each incorrect attempt at that answer O eTextbook O Help With This Topic O Web Help & Videos O Technical Support and Bug Reports 5.35 PM /14/2016

Explanation / Answer

Sum the moments about P

T(d - s) sin theta - W(d - x) - F(d/2) = 0

theta = tan-1(h/(d-s)) = tan-1(2.25 / 5.71) = 21.5 deg

11130 * 5.71 * sin 21.5º - W * 4.36 - 81.7 * 6.16 * 9.8 * 6.16/2 = 0

4.36 W = 8133 N

W = 1865 N

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sum the vertical forces

Fv + T sin theta - W - 81.7 * 6.16 * 9.8 = 0

Fv + 11130 * sin 21.5 - 1865 - 4932 = 0

Fv = 2712 N

sum the horizontal forces Fh - T cos theta= 0

Fh - 11130 * cos 21.5º = 0

Fh = 10350 N

magnitude P = sqrt (103502 + 27122) = 10700 N

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