Exam 1, Fall 2014 CE 2200

Section 1…. FORCE VECTORS

1.1 F2 = 70 lb. What is the magnitude of the resultant force acting on the bracket? (5 pts) A. 111 lb B. 121 lb C. 131 lb D. 141 lb E. 151 lb

1.2 F1 = 675 N. What is the i component of F1 ? (5 pts) A. 416 N B. 421 N C. 433 N D. 450 N E. 468 N (4/5)(675)cos30 = 468 N 1.3 What is the k component of F1 ? (5 pts) A. 405 N B. 390 N C. 382 N D. 375 N E. 360 N (3/5)(675) = 405 N

1.4 A force is expressed as a Cartesian vector F = {35 i – 62 j + 87 k} N. What is the coordinate direction angle beta β of the force? (5 pts) A. 109º B. 114º C. 119º D. 123º E. 128º β = cos-1 =123°

ϴ = cos-1 [F ∙ r / F(r) ] = cos-1 4/34.08 = 83.3° 1.5 F = {2i + 5j + 10k} kN. What is the angle θ between the force F and the pole? (5 pts) A. 77.6° B. 80.3° C. 83.3° D. 86.5° E. 88.8° F = = 11.36 kN rOA = (2i + 2j – 1k) m rOA = = 3 m ϴ = cos-1 [F ∙ r / F(r) ] = cos-1 4/34.08 = 83.3°

Section 2…. Particle Equilibrium

FED cos30 - 3/5 (FEB) = 0 FED sin30 + 4/5 (FEB) -90 = 0 2.1 (5 pts) If the bucket weighs 90 lb, determine the tension developed in wire ED. A. 42.3 lb B. 30.2 lb C. 54.4 lb D. 66.5 lb E. 75.8 lb FED cos30 - 3/5 (FEB) = 0 EB ED FBD of E FED sin30 + 4/5 (FEB) -90 = 0 90 lb FED = 54.39 lb FEB = 78.5 lb

ƩFx and ƩFy @ E ƩFx and ƩFy @ B 4 Equations… then 2.2 (5 pts) Knowing the weight of the bucket and assuming no other unknowns have been solved for previously, what is the minimum number of scalar equations that are required to solve for the tension developed in cable BA? (Fill in the blank on the cover answer sheet) 4 Equations… ƩFx and ƩFy @ E then ƩFx and ƩFy @ B

2.3 (5 pts) If the particle shown is in equilibrium and F = 1000 N, determine the magnitude of F1. B. 1200 N C. 1000 N D. 1100 N E. 800 N ƩFz =0 -1000cos30+F1sin60 = 0 F1 = 1000 N 2.4 (5 pts) What is the magnitude of the resultant force of the particle system shown above? (Fill in the blank on the cover answer sheet) ZERO – Particle is at EQUILBRIUM

Sum forces ƩFx = 0 ƩFY = 0 ƩFZ = 0 2.5 (5 pts) For the same system shown above, how many scalar equations can be written and what is the maximum number of unknowns that can be solved? A. 3 unknowns, 3 equations B. 2 unknowns, 4 equations C. 9 unknowns, 6 equations D. 4 unknowns, 4 equations E. none of the above 3 Equations… 3 Unknowns Sum forces ƩFx = 0 ƩFY = 0 ƩFZ = 0

Section 3…. Moments

+ MR = – 50(2) +20(3sin30) – 60(4+3cos30) MR = – 465.885 N-m 3.1 (5 pts) If F = 60 N, determine the magnitude of the equivalent resultant moment acting at point O. A. 465.9 N∙m B. 531.9 N∙m C. 399.9 N∙m D. 333.9 N∙m E. 672.2 N∙m + MR = – 50(2) +20(3sin30) – 60(4+3cos30) MR = – 465.885 N-m 3.2 (5 pts) The direction of the resultant moment calculated above is: A. clockwise B. counterclockwise C. unable to be determined

3.3 (5 pts) The x-component of F2 shown causes a moment about: A. the x-axis only B. the y-axis only C. the z-axis only D. the x and y axes E. the x and z axes F. the y and z axes G. all axes H. none of the above F2Y F2X F2X : is parallel to the x-axis and cannot create a moment ABOUT the x-axis. is not parallel to either the y or z axes. has a line of action which does not pass through the y or z axes. F2Z

rOB rOC The position vector used MUST: 3.4 (5 pts) It is desired to calculate the moment due to the force about point O. Determine the position vector(s) that could be used in the moment cross product calculation (r x F). A. rOB G. rAC B. rBO H. rCA C. rOC I. A. and C. only D. rCO J. E. and G. only E. rAB K. B. and D. only F. rBA L. F. and H. only rOB rOC The position vector used MUST: Go FROM the point of interest TO a point on the line of action of the force.

MO = r x F (lies perpendicular to the plane incl. r & F) MX = MO u 3.5 (5 pts) The cutting tool on the lathe exerts a force F on the shaft as shown. When finding the moment due to F about the x-axis using the triple scalar product u∙(r x F), determine the vector u. A. [6, -4, -7] B. [1, 0, 0] C. [0.597, -0.398, -0.697] D. [0, 1, 0] E. [23.0, 0, 19.3] F. [0, 0, 1] r u MO MX MO = r x F (lies perpendicular to the plane incl. r & F) MX = MO u u represents the DIRECTION of the vector component

Section 4 … Couples, Equivalent Systems, Distributed Loads

MRY = (30cos60)(4) – (30cos60)(8) = – 60 N-m 4.1 Determine the sum of the moments exerted on the pipe by the two couples shown. The magnitude of F = 20 N and the magnitude of P = 30 N. (5 pts) A. {40 i + 60 j – 103.9 k} N∙m B. {60 i + 40 j – 69.3 k} N∙m C. {– 40 i – 60 j + 103.9 k} N∙m D. {– 60 i – 40 j + 69.3 k} N∙m MRX = – (20)(2) = – 40 N-m MRY = (30cos60)(4) – (30cos60)(8) = – 60 N-m MRZ = – (30sin60)(4) + (30sin60)(8) = 103.9 N-m MR = { – 40i – 60j + 103.9k } N-m

FRX = 40cos30 = 34.64 kN FRY = 30 + 40sin30 = 50.0 kN Replace the force and couple system acting on the beam by an equivalent force and couple system at point O. (8 pts) F1 = 30 kN a = 3 m F2 = 40 kN b = 2 m M = 210 kN-m c = 5 m   4.2 Express the resultant equivalent force in Cartesian vector format. A. {34.6 i + 50.0 j} kN B. {50.0 i + 34.6 j} kN C. {– 34.6 i – 50.0 j} kN D. {– 50.0 i – 34.6 j} kN FRX = 40cos30 = 34.64 kN FRY = 30 + 40sin30 = 50.0 kN

+ MR = (30)(3) + (40sin30)(5) + 210 = 400 kN-m Replace the force and couple system acting on the beam by an equivalent force and couple system at point O. (8 pts) F1 = 30 kN a = 3 m F2 = 40 kN b = 2 m M = 210 kN-m c = 5 m   4.3 Express the resultant moment in Cartesian vector format. A. {400 k} kN-m B. {2290 k} kN-m C. {– 400 k} kN-m D. {– 2290 k} kN-m + MR = (30)(3) + (40sin30)(5) + 210 = 400 kN-m

4.4 Replace the force and couple system acting on the beam by only a single equivalent force and find the location, x, where this force acts on the beam measured (in meters) from point O. (Fill in the blank on the cover answer sheet) (5 pts) 400 kN-m = (50.0 kN)(x) x = 8.0 m

FR = (1/2)(480)(3) + (1/2)(600)(6) + (600)(2) FR = 720 + 1800 + 1200 4.5 Replace the distributed loading with an equivalent resultant force. If a = 3 ft, b = 2 ft and z = 600 lb/ft determine the magnitude of that force. (5 pts) A. 11,280 lb B. 4,720 lb C. 3,720 lb D. 6,240 lb E. 1,480 lb F. 2,560 lb FR = (1/2)(480)(3) + (1/2)(600)(6) + (600)(2) FR = 720 + 1800 + 1200 FR = 3720 lb