Equilibrium of Non-Concurrent Force Systems support reactions weight of bridge deck on beam BEAM

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Concurrent and Non-Concurrent Force Systems lines of action of forces intersect at single point lines of action of forces do not intersect at single point Σ 𝐹 𝑥 =0 Σ 𝐹 𝑦 =0 Σ 𝐹 𝑥 =0 Σ 𝐹 𝑦 =0 Σ𝑀=0 FB FA x y W 45° A B C D x y qB qA FB FA W = 100 lb A D Ay Ax B C

Free Body Diagrams STEPS: pinned joint Choose bodies to include on FBD resists motion in x and y directions Choose bodies to include on FBD STEPS: 3000 lbs Draw the body of interest A Show loads exerted by interacting bodies; name the loads B C D Define a coordinate system roller support resists motion in y direction Label distances and angles Assume center of mass of car is half way between the front and rear wheels B B=1500 lbs C C=1500 lbs x y A Ax Ay 12 ft 8 ft 20 ft D

Solve for Unknown Forces x y 12 ft 8 ft 20 ft B B=1500 lbs C C=1500 lbs Ay D A Ax Σ 𝐹 𝑥 =0 Σ 𝐹 𝑦 =0 Σ𝑀=0 Strategically choosing the order in which the three equilibrium equations are applied can make the problem easier to solve. When solving for the “reaction forces” on a beam like this, it is usually easiest to sum moments about the point where there are the most unknowns. Point A has two unknowns, so let’s begin by summing moments about that point. 𝐷=1200𝑙𝑏𝑠 one equation one unknown Σ 𝑀 𝐴 =𝐷∙40𝑓𝑡−1500𝑙𝑏𝑠∙20𝑓𝑡−1500𝑙𝑏𝑠∙12𝑓𝑡=0 + Now we can sum forces in x and y. The order doesn’t matter in this case. Σ 𝐹 𝑥 = 𝐴 𝑥 =0 Σ 𝐹 𝑦 = 𝐴 𝑦 +𝐷−1500𝑙𝑏𝑠−1500𝑙𝑏𝑠=0 𝐴 𝑦 =1800𝑙𝑏𝑠 Should Ay be larger than D? Why? Think critically to evaluate the solution.

Free Body Diagram Tip Sometimes it’s helpful to strategically align your coordinate system; here, the coordinate system is aligned with the beam. In this case, it really doesn’t make that much difference in solution difficulty (compared with a horizontal / vertical alignment), but it may be a little easier since the distances used in moment calculations are clearly labeled on the beam. A B C D 20° 20° A D Ay Ax B C 20° x y 20 ft 8 ft 12 ft

Sum forces in the x‐direction to find Ax Class Problem: A man who weighs 890 N stands on the end of a diving board as he plans his dive. Draw a FBD of the beam. Sum forces in the x‐direction to find Ax Sum moments about point A to find the reaction at B (By). Sum forces in the y‐direction to find Ay. Assumptions: Ignore dynamic effects. Ignore deflection (bending) of the diving board. Assume the weight of the man can be lumped exactly 3m horizontally from point B. Solution: a. 𝑦 𝑊=890𝑁 1.5m 3m 𝐴 𝑥 𝑥 𝐴 𝑦 𝐵 𝑦

𝐹 𝑥 = 𝐴 𝑥 =0 Form good habits early! b. 1.5m 3m 𝑥 𝑦 𝐴 𝑥 𝐴 𝑦 𝐵 𝑦 𝑊=890𝑁 ALWAYS put “= 0” on the right hand side. 𝑀 𝐴 =−890𝑁∙4.5𝑚+ 𝐵 𝑦 ∙1.5𝑚=0 + ALWAYS write this when summing moments, including the point that moments are being summed about. c. 𝐵 𝑦 = 890𝑁∙4.5𝑚 1.5𝑚 =2670𝑁 𝐹 𝑦 = 𝐴 𝑦 + 𝐵 𝑦 −890𝑁=0 Tip: Always assume the reactions act in a positive coordinate reaction. The signs of the answers will set things straight in the end. d. 𝐴 𝑦 +2670𝑁−890𝑁=0 𝐴 𝑦 =−2670𝑁+890𝑁 𝐴 𝑦 =−1780𝑁 Negative means it points downward.

Class Problem: The catapult has a horizontal beam that pivots about point P. A massive boulder (mass = 720kg ) is permanently mounted at B. Before loading a smaller rock (for launch) into the scoop at M, the rope connected between the ground and R holds the system at rest. If the weight of the machine itself (including the scoop at M) is neglected, what is the tension in the rope? 𝑥 𝑦 0.6m 2.6m 0.8m 𝑃 𝑥 𝑃 𝑦 𝑅 720𝑘𝑔 𝐵=720𝑘𝑔∙9.81 𝑚 𝑠 2 𝐵=7063.2𝑁 𝑀 𝑃 =0 + =𝑅∙2.6𝑚−7063.2𝑁∙0.8𝑚 𝑅∙2.6𝑚=7063.2𝑁∙0.8𝑚 𝑅=2173.29𝑁