Introduction to Engineering Mechanics Quiz Answer | 100% Correct Answers

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Introduction to Engineering Mechanics Quiz Answer

Week- 1

Forces and Particle Equilibrium

1. What is the weight in newtons of a 2500 pound vehicle?

  • a. 11,120 N
  • b. 24,525 N
  • c. 562 N
  • d. None of the above

2. The cable BCA in the figure below passes without friction through a hole at the end of the strut at C, and ties to the ground at A and to the end of the pole at B. If the tension in the cable is 800 lb, what is the force exerted by the cable onto the pole at B?

Used with permission from “Engineering Mechanics: Statics,” McGill/King, 4th Ed., 2003

a. \;-716\hat{i}  i^+ 358\hat{j} j^lb

b. \;358\hat{i}  i^+ 716\hat{j} j^​ lb

c. \;358\hat{i}  i^- 716\hat{j}  j^lb

d. \;None of the above

3. A water bucket is suspended by two cables as shown below. The maximum tension either cable can withstand is 10 lbs.

What is the maximum weight of the water bucket that can be supported?

Used with permission from “Engineering Mechanics: Statics,” McGill/King, 4th Ed., 2003

a. 18.4 lb

b. 13.7 lb

c. 15.3 lb

d. None of the above

Week- 2

Define and Calculate Moments

1. How much moment (or torque) is produced about point O by the wrench in the situation below?

  • a. 163 in-lb counterclockwise
  • b. 163 in-lb clockwise
  • c. 82.7 in-lb clockwise
  • d. None of the above

2. Determine the moment (or torque) of the 60 N force with respect to the corner C in the figure below.

  • a. 1270\hat{j}1270 j^+1020\hat{k}1020 k^N-cm
  • b. -1270\hat{j}1270 j^​ -1020\hat{k}1020 k^N-cm
  • c. -1270\hat{i}1270 i^-1020\hat{k}1020 k^N-cm
  • d. None of the above

3. Determine the moment of the couple for the system shown below. The forces shown lie in the x-y plane.

  • a. -2400\hat{k}2400 k^ft-lb
  • b. -4800\hat{k}4800 k^ft-lb
  • c. 2400\hat{k}2400 k^ft-lb
  • d. None of the above

Week- 3

Equilibrium and Equivalence of Force Systems

1. Four truss members carrying the indicated forces have their centerlines all intersecting at point P of the gusset plate as shown below. What is the resultant of the four forces?

Used with permission from “Engineering Mechanics: Statics,” McGill/King, 4th Ed., 2003

  • a. -2130\hat{i}2130 i^-65.7\hat{j}65.7 j^lb
  • b. 2130\hat{i}2130 i^+65.7\hat{j}65.7 j^​ lb
  • c. 1770\hat{i}1770 i^+300\hat{j}300 j^​ lb
  • d. None of the above

2. For the system shown below, determine an equivalent force and couple system at point C.

Used with permission from “Engineering Mechanics: Statics,” McGill/King, 4th Ed., 2003

  • a. Equivalent Force = 1730\hat{i}1730 i^+1000\hat{j}1000 j^​ lb Equivalent Moment = -6660\hat{k}6660 k^ft-lb
  • b. Equivalent Force = 1000\hat{i}1000 i^+1730\hat{j}1730 j^​ lb Equivalent Moment = -8460\hat{k}8460 k^ft-lb
  • c. Equivalent Force = 1730\hat{i}1730 i^+1000\hat{j}1000 j^​ lb Equivalent Moment = -10700\hat{k}10700 k^ft-lb
  • d. None of the above

3. Three tugboats are used to bring a large ship to a pier as shown. Each tugboat exerts a 4000 lb force pushing toward the large ship. Determine an equivalent force and couple system at point A.

  • a. Equivalent Force = 6030\hat{i}6030 i^-3570\hat{j}3570 j^​ lb Equivalent Moment = -1,150,000\hat{k}1,150,000 k^ft-lb
  • b. Equivalent Force = 5230\hat{i}5230 i^-4370\hat{j}4370 j^​ lb Equivalent Moment = -1,190,000\hat{k}1,190,000 k^ft-lb
  • c. Equivalent Force = 6030\hat{i}6030 i^-3570\hat{j}3570 j^​ lb Equivalent Moment = 1,150,000\hat{k}1,150,000 k^ ft-lb
  • d. None of the above

Week- 4

Free Body Diagrams and Equilibrium Analysis Techniques

1. Find the equivalent resultant force and the distance to the centroid in the x-direction for the system below.

Used with permission from “Engineering Mechanics: Statics,” McGill/King, 4th Ed., 2003

  • a. Equivalent Resultant Force = 5850 N down acting 4.85 meters to the right of P
  • b. Equivalent Resultant Force = 5850 N down acting 4.15 meters to the right of P
  • c. Equivalent Resultant Force = 7200 N down acting 6 meters to the right of P
  • d. None of the above

2. Draw a Free Body Diagram (FBD) for the ladder shown below. The ladder is 30-kg and the painter is 70 kg. Assume all of the weight of the painter acts down through his feet. Assume no forces are exerted by his hands on the ladder.

Used with permission from “Engineering Mechanics: Statics,” McGill/King, 4th Ed., 2003

Screenshot 333

d. None of the above

3. Find the equivalent resultant force and the distance to the centroid in the x-direction for the system below.

Used with permission from “Engineering Mechanics: Statics,” McGill/King, 4th Ed., 2003

  • a. Equivalent Resultant Force = 77.4 lb down acting 2.41 feet right of origin
  • b. Equivalent Resultant Force = 55.8 lb down acting 2.47 feet right of origin
  • c. Equivalent Resultant Force = 121 lb down acting 2.34 feet right of origin
  • d. None of the above

Week- 5

Application of Static Equilibrium Equations

1. In the figure below, if a=5 feet, L=14 feet, and the respective weights of the diver and the board are 200 lb and 90 lb, find the reactions onto the uniform board at the pin A and the roller B.

Used with permission from “Engineering Mechanics: Statics,” McGill/King, 4th Ed., 2003

  • a. Ax = 0; Ay = 396 lb down; and B = 686 lb up
  • b. Ax = 0; Ay = 174 lb down; and B = 464 lb up
  • c. Ax = 0; Ay = 396 lb up; and B = 686 lb up
  • d. None of the above

2. Determine the reactions at the supports A and B for the jib crane shown below. Also identify if the cable BC is in compression (C) or tension (T). Neglect the weight of the members of the jib crane.

  • a. Ax = 104 lb to the right; Ay = 60.8 lb up; and TBC = 174 lb (T)
  • b. Ax = 188 lb to the right; Ay = 58.8 lb up; and TBC = 235 lb (T)
  • c. Ax = 139 lb to the right; Ay = 95.6 lb up; and TBC = 174 lb (T)
  • d. None of the above

3. Find the force reactions, Rx, Ry, Rz, and the couple reactions, Cx, Cy, Cz at the base of the advertising sign shown below. The sign weighs 200 lb and the pole weighs 500 lb.

Used with permission from “Engineering Mechanics: Statics,” McGill/King, 4th Ed., 2003

  • a. Rx = -250\hat{i}250 i^lb; Ry = -700\hat{j}700 j^​ lb; Rz = 0Cx = -400\hat{i}400 i^lb.ft; Cy = 625\hat{j}625 j^​ lb.ft; Cz=2250\hat{k}2250 k^lb.ft
  • b. Rx = 250\hat{i}250 i^lb; Ry = 700\hat{j}700 j^​ lb; Rz = 0Cx = -400\hat{i}400 i^lb.ft; Cy = 625\hat{j}625 j^lb.ft;Cz=2250\hat{k}2250 k^lb.ft
  • c. Rx = 250\hat{i}250 i^lb; Ry = 700\hat{j}700 j^​ lb; Rz = 0Cx = 400\hat{i}400 i^lb.ft; Cy = -625\hat{j}625 j^​ lb.ft; Cz=-2250\hat{k}2250 k^lb.ft
  • d. None of the above

Conclusion

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