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 lbB. 121 lbC. 131 lbD. 141 lbE. 151 lb

1.3 What is the k component of F1 ? (5 pts)A. 405 NB. 390 NC. 382 ND. 375 NE. 360 N

1.2 F1 = 675 N. What is the i component of F1 ? (5 pts)A. 416 NB. 421 NC. 433 ND. 450 NE. 468 N

(4/5)(675)cos30 = 468 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. 114ºC. 119ºD. 123º E. 128º

β = cos-1

=123°

2 2 235

62

62 87

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 kN2 2 22 5 10

rOA = = 3 m 2 2 22 2 1 rOA = (2i + 2j – 1k) m

ϴ = cos-1 [F ∙ r / F(r) ] = cos-1 4/34.08 = 83.3°

Section 2…. Particle Equilibrium

2.1 (5 pts) If the bucket weighs 90 lb, determine the tension developed in wire ED.A. 42.3 lbB. 30.2 lbC. 54.4 lbD. 66.5 lbE. 75.8 lb

+ 0xF

+ 0yF

FED cos30 - 3/5 (FEB) = 0

FED sin30 + 4/5 (FEB) -90 = 0

EB ED

FBD of E

90 lb

FED = 54.39 lb FEB = 78.5 lb

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 thenƩFx and ƩFy @ B

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

2.3 (5 pts) If the particle shown is in equilibrium and F = 1000 N, determine the magnitude of F1.A. 900 NB. 1200 NC. 1000 ND. 1100 NE. 800 N

ƩFz =0 -1000cos30+F1sin60 = 0 F1 = 1000 N

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 equationsB. 2 unknowns, 4 equationsC. 9 unknowns, 6 equationsD. 4 unknowns, 4 equationsE. none of the above

3 Equations… 3 Unknowns

Sum forces ƩFx = 0 ƩFY = 0 ƩFZ = 0

Section 3…. Moments

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. clockwiseB. counterclockwiseC. unable to be determined

+

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.

3.3 (5 pts) The x-component of F2 shown causes a moment about:

A. the x-axis onlyB. the y-axis onlyC. the z-axis onlyD. the x and y axesE. the x and z axesF. the y and z axesG. all axesH. none of the above

F2Z

F2Y

F2X

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. onlyD. rCO J. E. and G. onlyE. rAB K. B. and D. onlyF. rBA L. F. and H. only

The position vector used MUST:Go FROM the point of interest TO a point on the line of action of the force.

rOC

rOB

1) MO = r x F (lies perpendicular to the plane incl. r & F)

2) MX = MO u u represents the DIRECTION of the vector component

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

uMOMX

Section 4 … Couples, Equivalent Systems, Distributed Loads

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

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 mF2 = 40 kN b = 2 mM = 210 kN-m c = 5 m 4.2 Express the resultant equivalent force in Cartesian vector format.A. {34.6 i + 50.0 j} kNB. {50.0 i + 34.6 j} kNC. {– 34.6 i – 50.0 j} kND. {– 50.0 i – 34.6 j} kN

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 mF2 = 40 kN b = 2 mM = 210 kN-m c = 5 m 4.3 Express the resultant moment in Cartesian vector format.A. {400 k} kN-mB. {2290 k} kN-mC. {– 400 k} kN-mD. {– 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

FR = 3720 lb

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 lbB. 4,720 lbC. 3,720 lbD. 6,240 lb E. 1,480 lbF. 2,560 lb