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Multiple Choice Questions 1. All the vectors quantities obey: A) Parallelogram law of addition B) Parallelogram law of multiplication C) Parallelogram law of addition of square root of their magnitudes D) Parallelogram law of addition of square of their magnitudes 2. Express the force in the figure as a Cartesian vector. F-S00N A) F=(250i-354j+250k) N B) F=(-250i-354j+250k) N C) F=(354i-250j+250k) N D) F=(-354i+250j+250k) N 3. Determine the magnitude of the x and y components of the 2 kN force. 2AN A) Fx = -4.000 kN, Fy = -2.312 kN B) Fx = -1.000 kN, Fy = -1.732 kN C) Fx = -1.414 kN, Fy = -1.414 kN D) Fx = -1.732 kN, Fy = -1.000 kN 4. Express the force F2 in Cartesian vector form. F,-400 Ib 45 45% F;-600 lb

Multiple Choice Questions 1. All the vectors quantities obey: A) Parallelogram law of addition B) Parallelogram law of multiplication C) Parallelogram law of addition of square root of their magnitudes D) Parallelogram law of addition of square of their magnitudes 2. Express the force in the figure as a Cartesian vector. F-S00N A) F=(250i-354j+250k) N B) F=(-250i-354j+250k) N C) F=(354i-250j+250k) N D) F=(-354i+250j+250k) N 3. Determine the magnitude of the x and y components of the 2 kN force. 2AN A) Fx = -4.000 kN, Fy = -2.312 kN B) Fx = -1.000 kN, Fy = -1.732 kN C) Fx = -1.414 kN, Fy = -1.414 kN D) Fx = -1.732 kN, Fy = -1.000 kN 4. Express the force F2 in Cartesian vector form. F,-400 Ib 45 45% F;-600 lb

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4. Express the force F2 in Cartesian vector form. F,-400 Ib F,- 600 Ib 1 ENGR 2301 A) F2 = (212 i + 212 j – 519 k) Ib B) F2 = (155 i + 155 j + 300 k) Ib C) F2 = (367 i + 367 j – 300 k) Ib D) F2 = (155 i + 155 j – 300 k) Ib 5. Force F acts on A such that one of its components, lying in the x-y plane, has a magnitude of 50 lb. Express F as a Cartesian vector. 50 Ib A) F = (43.3 i – 25.0 j + 28.9 k) Ib B) F = (43.3 i + 25.0 j + 28.9 k) Ib C) F = (43.3 i + 25.0 j + 25.0 k) Ib D) F = (43.3 i - 25.0 j + 25.0 k) Ib 6. What is the correct expression of position vector rcB (in feet)? 2 ft Fc=490 lb F = 600 lb 4 ft 6 ft 3 ft 4 ft 2 f 4 ft A) -4i - j+ 2k B) 4i + j - 2k C) 4j + 2k D) -4j - 2k E) -4i + 3j + 4k
Transcribed Image Text:4. Express the force F2 in Cartesian vector form. F,-400 Ib F,- 600 Ib 1 ENGR 2301 A) F2 = (212 i + 212 j – 519 k) Ib B) F2 = (155 i + 155 j + 300 k) Ib C) F2 = (367 i + 367 j – 300 k) Ib D) F2 = (155 i + 155 j – 300 k) Ib 5. Force F acts on A such that one of its components, lying in the x-y plane, has a magnitude of 50 lb. Express F as a Cartesian vector. 50 Ib A) F = (43.3 i – 25.0 j + 28.9 k) Ib B) F = (43.3 i + 25.0 j + 28.9 k) Ib C) F = (43.3 i + 25.0 j + 25.0 k) Ib D) F = (43.3 i - 25.0 j + 25.0 k) Ib 6. What is the correct expression of position vector rcB (in feet)? 2 ft Fc=490 lb F = 600 lb 4 ft 6 ft 3 ft 4 ft 2 f 4 ft A) -4i - j+ 2k B) 4i + j - 2k C) 4j + 2k D) -4j - 2k E) -4i + 3j + 4k
ENGR 2301 CH 2 Exam Review Problems Multiple Choice Questions 1. All the vectors quantities obey: A) Parallelogram law of addition B) Parallelogram law of multiplication C) Parallelogram law of addition of square root of their magnitudes D) Parallelogram law of addition of square of their magnitudes 2. Express the force in the figure as a Cartesian vector. F- S00N A) F=(250i-354j+250k) N B) F=(-250i-354j+250k) N C) F=(354i-250j+250k) N D) F=(-354i+250j+250k) N 3. Determine the magnitude of the x and y components of the 2 kN force. 2 KN A) Fx = -4.000 kN, Fy = -2.312 kN B) Fx = -1.000 kN, Fy = -1.732 kN C) Fx = -1.414 kN, Fy = -1.414 kN D) Fx = -1.732 kN, Fy = -1.000 kN 4. Express the force F2 in Cartesian vector form. F,- 400 Ib F,- 600 Ib 1
Transcribed Image Text:ENGR 2301 CH 2 Exam Review Problems Multiple Choice Questions 1. All the vectors quantities obey: A) Parallelogram law of addition B) Parallelogram law of multiplication C) Parallelogram law of addition of square root of their magnitudes D) Parallelogram law of addition of square of their magnitudes 2. Express the force in the figure as a Cartesian vector. F- S00N A) F=(250i-354j+250k) N B) F=(-250i-354j+250k) N C) F=(354i-250j+250k) N D) F=(-354i+250j+250k) N 3. Determine the magnitude of the x and y components of the 2 kN force. 2 KN A) Fx = -4.000 kN, Fy = -2.312 kN B) Fx = -1.000 kN, Fy = -1.732 kN C) Fx = -1.414 kN, Fy = -1.414 kN D) Fx = -1.732 kN, Fy = -1.000 kN 4. Express the force F2 in Cartesian vector form. F,- 400 Ib F,- 600 Ib 1
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