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Can I have a more detailed explanation for number 2 please. Thank you. I dont get where the distances of the 7200 lb force (8-5) and (5-4) came from.

Can I have a more detailed explanation for number 2 please. Thank you. I dont get where the distances of the 7200 lb force (8-5) and (5-4) came from.

Structural Analysis
Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Can I have a more detailed explanation for number 2 please. Thank you. I dont get where the distances of the 7200 lb force (8-5) and (5-4) came from. 

8:11 M O ll 95%1 CE 2111 Module 3 Equil... Z 12' Px Py Fy 3600 Ib 1200 Ib CD= 24' CE= 12' 87 Property of and for the exclusive use of SLU. Reproduction, storing in a retrieval system, distributing, uploading or posting oline, or transmitting in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise of any part of this document, without the prior written permission of SLU, is strictly prohibited. To solve for the forces independently, extend the action line of two unknown forces and take the moment of forces at the point of intersection. (Recall the moment of a force about a point along its line of action is zero) +u £ Mg = 0 forces P and F are concurrent at B (3600)(12) – (T)(8) = 0 T = 5400 lb +u £Mp -(3600)(12) + (1200)(24) +(F)(36) = 0 = 0 forces P and T are concurrent at D. Resolve force F at E into horizontal and vertical V13 F = 721.11 lb components. The moment of Fx at D is zero. +U £Mg = 0 forces F and T are concurrent E. Resolve (3600)(24) – (1200)(12) – '곯(P) (36)%3D0 force P at D into horizontal and vertical P = 6324.56 lb components. The moment of Px at E is zero. 2. The weight of the trapezoidal block is 7200 lb acting where shown the figure. The ground reaction varies uniformly from an intensity of pa Ib/ft at A to ps Ib/ft Determine pA and pB. В. Given: 5' 7" 7200 Ib Pa Ib/ft PA Ib/ft Required: Pa and pB Solution: R1 = Aq = (12)(PA) = 6pA FBD: R2 = A2 = (12)(PB) = 6pB +u £M@ = 0 R1(4) – 7200(8 – 5) = 0 5' 7' 6Pa(4) – 7200(3) = 0 7200 Ib PA = 900 lb/ft +U £M@ = 0 -R2(4) + 7200(5 – 4) = 0 PB Ib/ft -6pB (4) + 7200(1) = 0 PA Ib/ft Ri R2 4' 4' PB = 300 lb/ft Property of and for the exclusive use of SLU. Reproduction, storing in a retrieval system, distributing, uploading or posting online, or transmitting in any form or by any 88 means, electronic, mechanical, photocopying, recording, or otherwise of any part of this document, without the prior written permission of SLU, is strictly prohibited. NIVERS 3. A pulley of 1-ft radius, supporting a load of 500 lb, is mounted at B on a horizontal beam as shown in the figure. If the beam weighs 20O Ib and the pulley Can't connect. Check whether you're online or not. II
Transcribed Image Text:8:11 M O ll 95%1 CE 2111 Module 3 Equil... Z 12' Px Py Fy 3600 Ib 1200 Ib CD= 24' CE= 12' 87 Property of and for the exclusive use of SLU. Reproduction, storing in a retrieval system, distributing, uploading or posting oline, or transmitting in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise of any part of this document, without the prior written permission of SLU, is strictly prohibited. To solve for the forces independently, extend the action line of two unknown forces and take the moment of forces at the point of intersection. (Recall the moment of a force about a point along its line of action is zero) +u £ Mg = 0 forces P and F are concurrent at B (3600)(12) – (T)(8) = 0 T = 5400 lb +u £Mp -(3600)(12) + (1200)(24) +(F)(36) = 0 = 0 forces P and T are concurrent at D. Resolve force F at E into horizontal and vertical V13 F = 721.11 lb components. The moment of Fx at D is zero. +U £Mg = 0 forces F and T are concurrent E. Resolve (3600)(24) – (1200)(12) – '곯(P) (36)%3D0 force P at D into horizontal and vertical P = 6324.56 lb components. The moment of Px at E is zero. 2. The weight of the trapezoidal block is 7200 lb acting where shown the figure. The ground reaction varies uniformly from an intensity of pa Ib/ft at A to ps Ib/ft Determine pA and pB. В. Given: 5' 7" 7200 Ib Pa Ib/ft PA Ib/ft Required: Pa and pB Solution: R1 = Aq = (12)(PA) = 6pA FBD: R2 = A2 = (12)(PB) = 6pB +u £M@ = 0 R1(4) – 7200(8 – 5) = 0 5' 7' 6Pa(4) – 7200(3) = 0 7200 Ib PA = 900 lb/ft +U £M@ = 0 -R2(4) + 7200(5 – 4) = 0 PB Ib/ft -6pB (4) + 7200(1) = 0 PA Ib/ft Ri R2 4' 4' PB = 300 lb/ft Property of and for the exclusive use of SLU. Reproduction, storing in a retrieval system, distributing, uploading or posting online, or transmitting in any form or by any 88 means, electronic, mechanical, photocopying, recording, or otherwise of any part of this document, without the prior written permission of SLU, is strictly prohibited. NIVERS 3. A pulley of 1-ft radius, supporting a load of 500 lb, is mounted at B on a horizontal beam as shown in the figure. If the beam weighs 20O Ib and the pulley Can't connect. Check whether you're online or not. II
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