MCQOPTIONS
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MCQOPTIONS
This section includes 35 Mcqs, each offering curated multiple-choice questions to sharpen your Engineering Mechanics knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
The main beam along the wing of an airplane is swept back at an angle of 25°. From load calculations it is determined that the beam is subjected to couple moments aMx = 25,000 lb ft and My = 17,000 lb ft. Determine the equivalent couple moments created about the x' and y' axis. |
| A. | Mx' = 26.0 kip-ft, My' = 29.8 kip-ft |
| B. | Mx' = 29.8 kip-ft, My' = -4.84 kip-ft |
| C. | Mx' = 26.0 kip-ft, My' = 15.47 kip-ft |
| D. | Mx' = 29.8 kip-ft, My' = 4.84 kip-ft |
| Answer» D. Mx' = 29.8 kip-ft, My' = 4.84 kip-ft | |
| 2. |
Determine the direction (0° 180°) of the 30-lb force F so that the moment of F about point A has the maximum magnitude. |
| A. | 0 |
| B. | 0 |
| C. | 0 |
| D. | 0 |
| Answer» B. 0 | |
| 3. |
A force and couple act on the pipe assembly. Replace this system by an equivalent single resultant force. Specify the location of the resultant force along the y axis, measured from A. The pipe lies in the x-y plane. |
| A. | R = 25k lb, y = -0.9 ft |
| B. | R = 25k lb, y = 0.9 ft |
| C. | R = 25k lb, y = 3.9 ft |
| D. | R = 25k lb, y = -3.9 ft |
| Answer» B. R = 25k lb, y = 0.9 ft | |
| 4. |
This structural connection is subjected to the 8,000-lb force. Replace this force by an equivalent force and couple acting at the center of the bolt group, O. |
| A. | F = (6400i+4800j) lb, M = (-3.20i+7.20j) kip-ft |
| B. | F = (-6400i-4800j) lb, M = -400k kip-ft |
| C. | F = (-6400i-4800j) lb, M = (-3.20i-7.20j) kip-ft |
| D. | F = (6400i+4800j) lb, M = 400k kip-ft |
| Answer» C. F = (-6400i-4800j) lb, M = (-3.20i-7.20j) kip-ft | |
| 5. |
Replace the loading by an equivalent force and couple moment acting at point O. |
| A. | R = 90 kN, M = 473 kN-m CW |
| B. | R = 90 kN, M = 338 kN-m CW |
| C. | R = 45 kN, M = 203 kN-m CW |
| D. | R = 135 kN, M = 270 kN-m CW |
| Answer» C. R = 45 kN, M = 203 kN-m CW | |
| 6. |
The resultant force of a wind loading acts perpendicular to the face of the sign as shown. Replace this force by an equivalent force and couple moment acting at point O. |
| A. | F = -120i lb, M = (-3000j+1800k) lb-ft |
| B. | F = -120i lb, M = 3500i lb-ft |
| C. | F = -120i lb, M = -3500j lb-ft |
| D. | F = -120i lb, M = 3500k lb-ft |
| Answer» B. F = -120i lb, M = 3500i lb-ft | |
| 7. |
The boys A, B and C stand near the edges of a raft as shown. Determine the location (x, y) of boy D so that all four boys create a single resultant force acting through the raft's center O. Provided the raft itself is symmetric, this would keep the raft afloat in a horizontal plane. the mass of each boy is indicated in the diagram. |
| A. | x = 4.5 m, y = 1.5 m |
| B. | x = 3.0 m, y = 3.0 m |
| C. | x = 1.5 m, y = 4.5 m |
| D. | x = 3.0 m, y = 4.0 m |
| Answer» D. x = 3.0 m, y = 4.0 m | |
| 8. |
Determine the direction (0° 180°) of the 30-lb force F so that the moment of F about point A has the maximum magnitude. |
| A. | = 36.9° |
| B. | = 53.1° |
| C. | = 127.0° |
| D. | = 90.0° |
| Answer» B. = 53.1¬∞ | |
| 9. |
A twist of 4 N-m is applied to the handle of the screwdriver. Resolve this couple moment into a pair of couple forces F exerted on the handle. |
| A. | F = 133 N |
| B. | F = 120 N |
| C. | F = 60 N |
| D. | F = 266 N |
| Answer» B. F = 120 N | |
| 10. |
The three forces acting on the water tank represent the effect of the wind. Replace this system by a single resultant force and specify its vertical location from point O. |
| A. | R = 600 lb, d = 65.0 ft below O |
| B. | R = 600 lb, d = 130.4 ft below O |
| C. | R = 600 lb, d = 65.0 ft above O |
| D. | R = 600 lb, d = 130.4 ft above O |
| Answer» E. | |
| 11. |
The pole supports a 22-lb traffic light. Using Cartesian vectors, determine the moment of the weight of the traffic light about the base of the pole at A. |
| A. | MA = 216k lb-ft |
| B. | MA = (-132i-229j-216k) lb-ft |
| C. | MA = (-229i+121j) lb-ft |
| D. | MA = (-132i+229j) lb-ft |
| Answer» D. MA = (-132i+229j) lb-ft | |
| 12. |
Determine the magnitude and direction of the moment of the force at A about point P. |
| A. | Mp = 143.5 ft-lb |
| B. | Mp = 1191 ft-lb |
| C. | Mp = 1104 ft-lb |
| D. | Mp = 1200 ft-lb |
| Answer» D. Mp = 1200 ft-lb | |
| 13. |
The distributed loadings of soil pressure on the sides and bottom of a spread footing are shown. Simplify this system to a single resultant force and couple moment acting at A. What is the resultant force F and the couple moment M? |
| A. | F = (72i + 125j)N, M = 83.5 N-m CCW |
| B. | F = (144i + 125j)N, M = 135.7 N-m CCW |
| C. | F = (144i + 125j)N, M = 83.5 N-m CCW |
| D. | F = (72i + 125j)N, M = 102.1 N-m CCW |
| Answer» E. | |
| 14. |
Replace the force at A by an equivalent force and couple moment at P. Express the results on Cartesian vector form. |
| A. | F = (-50i+20j+30k) N, M = (50i-130j+170k) N-m |
| B. | F = (-50i+20j+30k) N, M = (-50i+60j+60k) N-m |
| C. | F = (50i-20j-30k) N, M = (-50i+130j-170k) N-m |
| D. | F = (50i-20j-30k) N, M = (50i-60j-60k) N-m |
| Answer» D. F = (50i-20j-30k) N, M = (50i-60j-60k) N-m | |
| 15. |
Replace the two forces acting on the tree branches by an equivalent force and couple moment acting at point O. |
| A. | R = (60i+198j-36k) N, M = (747i+18j-924k) N-m |
| B. | R = (60i+18j-324k) N, M = (747i+18j-924k) N-m |
| C. | R = (60i+198j-36k) N, M = (360i+1260j-414k) N-m |
| D. | R = (60i+18j-324k) N, M = (360i+1260j-414k) N-m |
| Answer» C. R = (60i+198j-36k) N, M = (360i+1260j-414k) N-m | |
| 16. |
Determine the couple moment. Use a vector analysis and express the result as a Cartesian vector. |
| A. | M = (800i-4800j-800k) lb-ft |
| B. | M = (4000i-2000j+4000k) lb-ft |
| C. | M = (-4000i+7000j-4000k) lb-ft |
| D. | M = (-800i-4800j-800k) lb-ft |
| Answer» C. M = (-4000i+7000j-4000k) lb-ft | |
| 17. |
The man at B exerts a force of 140N on the rope attached to the end of beam AC as shown. Determine the moment of this force about the base of the beam at A. |
| A. | MA = (720i-360k) N-m |
| B. | MA = (-720i+360k) N-m |
| C. | MA = (-720i-360k) N-m |
| D. | MA = (720i+360k) N-m |
| Answer» D. MA = (720i+360k) N-m | |
| 18. |
Three parallel forces act on the rim of the tube. If it is required that the resultant force FR of the system have a line of action that coincides with the central z axis, determine the magnitude of FC and its location on the rim. What is the magnitude of the resultant force FR? |
| A. | Fc = 361 lb, = 56.3°, R = 861 lb |
| B. | Fc = 500 lb, = 54.0°, R = 1000 lb |
| C. | Fc = 500 lb, = 36.0°, R = 1000 lb |
| D. | Fc = 361 lb, = 36.9°, R = 861 lb |
| Answer» B. Fc = 500 lb, = 54.0¬∞, R = 1000 lb | |
| 19. |
Replace the force and couple system by an equivalent single force and couple acting at point P. |
| A. | F = (-8.66i-15.00j) N, M = 50.0 N-m |
| B. | F = (8.66i-15.00j) N, M = 42.0 N-m |
| C. | F = (8.66i-15.00j) N, M = 50.0 N-m |
| D. | F = (-8.66i-15.00j) N, M = 42.0 N-m |
| Answer» C. F = (8.66i-15.00j) N, M = 50.0 N-m | |
| 20. |
A force of 50 N is applied to the handle of the door as shown. Determine the projection of the moment of this force about the hinged axis z. Neglect the size of the doorknob. Suggestion: Use a scaler analysis. |
| A. | Mz = 15.9 N-m |
| B. | Mz = -15.9 N-m |
| C. | Mz = -27.6 N-m |
| D. | Mz = 27.6 N-m |
| Answer» C. Mz = -27.6 N-m | |
| 21. |
Determine the moment of force F2 about point A on the beam. |
| A. | M2 = 800 ft-lb |
| B. | M2 = 50 ft-lb |
| C. | M2 = 600 ft-lb |
| D. | M2 = 37.5 ft-lb |
| Answer» D. M2 = 37.5 ft-lb | |
| 22. |
The wind has blown sand over a platform such that the intensity of load can be approximated by the function . Simplify this distributed loading to a single concentrated force and specify the magnitude and location of the force measured from A. |
| A. | R = 1250 N, x = 6.67 m |
| B. | R = 2500 N, x = 6.67 m |
| C. | R = 1250 N, x = 8 m |
| D. | R = 2500 N, x = 8 m |
| Answer» D. R = 2500 N, x = 8 m | |
| 23. |
Replace the force F, having a magnitude of F = 40 lb and acting at B, by an equivalent force and couple moment at A. |
| A. | F = (32i-24k) lb, M = (-120i+96j-160k) lb-ft |
| B. | F = (32i-24k) lb, M = 233k lb-ft |
| C. | F = (32i-24k) lb, M = -72k lb-ft |
| D. | F = (32i-24k) lb, M = (-120i+160j+40k) lb-ft |
| Answer» B. F = (32i-24k) lb, M = 233k lb-ft | |
| 24. |
Determine the moment of the force at A about point P. Use a vector analysis and express the result in Cartesian vector form. |
| A. | MP = (160i+240j+40k) N-m |
| B. | MP = (380i+160j+400k) N-m |
| C. | MP = (280i+200j+400k) N-m |
| D. | MP = (40i+80k) N-m |
| Answer» E. | |
| 25. |
Determine the moment of force F1 about point A on the beam. |
| A. | M1 = 1600 ft-lb |
| B. | M1 = 100 ft-lb |
| C. | M1 = 100 ft-lb |
| D. | M1 = 1600 ft-lb |
| Answer» E. | |
| 26. |
The bricks on top of the beam and the supports at the bottom create the distributed loading shown in the second figure. Determine the required intensity w and dimension d of the right support so that the resultant force and couple moment about point A of the system are both zero. |
| A. | w = 175.0 N/m, d = 1.5 m |
| B. | w = 138.2 N/m, d = 1.9 m |
| C. | w = 125.0 N/m, d = 2.1 m |
| D. | w = 154.4 N/m, d = 1.7 m |
| Answer» B. w = 138.2 N/m, d = 1.9 m | |
| 27. |
Determine the magnitude and direction of the moment of the force of the movement if the force at A about point P. |
| A. | Mp = 400 Nm |
| B. | Mp = 400 Nm |
| C. | Mp = 200 Nm |
| D. | Mp = 200 Nm |
| Answer» B. Mp = 400 Nm | |
| 28. |
Determine the magnitude and direction of the couple moment. |
| A. | M = 3900 lb-ft CW |
| B. | M = 3900 lb-ft CCW |
| C. | M = 3120 lb-ft CW |
| D. | M = 3120 lb-ft CCW |
| Answer» E. | |
| 29. |
Replace the force at A by an equivalent force and couple moment at P. |
| A. | F = 15 kN @ = 36.9° , Mp = 67.1 kN-m |
| B. | F = 15 kN @ = 36.9° , Mp = 30 kN-m |
| C. | F = 15 kN @ = 143.1° , Mp = 67.1 kN-m |
| D. | F = 15 kN @ = 143.1° , Mp = 30 kN-m |
| Answer» B. F = 15 kN @ = 36.9¬∞ , Mp = 30 kN-m | |
| 30. |
Determine the magnitude and direction of the couple shown. |
| A. | M = 22.6 kN-m CCW |
| B. | M = 22.6 kN-m CW |
| C. | M = 21.9 kN-m CCW |
| D. | M = 21.9 kN-m CW |
| Answer» D. M = 21.9 kN-m CW | |
| 31. |
Determine the magnitude of the projection of the moment cause by the force about the aa axis. |
| A. | Maa = 16.97 kN-m |
| B. | Maa = 12.00 kN-m |
| C. | Maa = 6 kN-m |
| D. | Maa = 8.48 kN-m |
| Answer» D. Maa = 8.48 kN-m | |
| 32. |
Replace the loading system acting on the post by an equivalent force and couple system at point O. |
| A. | F = (-8i-66j) lb, M = 220 lb-ft |
| B. | F = (-8i-66j) lb, M = 160 lb-ft |
| C. | F = (4i78j) lb, M = 220 lb-ft |
| D. | F = (4i78j) lb, M = 100 lb-ft |
| Answer» E. | |
| 33. |
Express the moment of the couple acting on the pipe on Cartesian vector form. What is the magnitude of the couple moment? |
| A. | Mc = (37.5i-25j) N-m, Mc = 45.1 N-m |
| B. | Mc = (25i-37.5j) N-m, Mc = 45.1 N-m |
| C. | Mc =45.1kN-m, Mc = 45.1 N-m |
| D. | Mc =-45.1kN-m, Mc = 45.1 N-m |
| Answer» B. Mc = (25i-37.5j) N-m, Mc = 45.1 N-m | |
| 34. |
Determine the moment of force F3 about point A on the beam. |
| A. | M3 = 6930 ft-lb |
| B. | M3 = 6000 ft-lb |
| C. | M3 = 3000 ft-lb |
| D. | M3 = 4000 ft-lb |
| Answer» D. M3 = 4000 ft-lb | |
| 35. |
The main beam along the wing of an airplane is swept back at an angle of 25°. From load calculations it is determined that the beam is subjected to couple moments aMx = 25,000 lb • ft and My = 17,000 lb • ft. Determine the equivalent couple moments created about the x' and y' axis. |
| A. | Mx' = 26.0 kip-ft, My' = 29.8 kip-ft |
| B. | Mx' = 29.8 kip-ft, My' = -4.84 kip-ft |
| C. | Mx' = 26.0 kip-ft, My' = 15.47 kip-ft |
| D. | Mx' = 29.8 kip-ft, My' = 4.84 kip-ft |
| Answer» D. Mx' = 29.8 kip-ft, My' = 4.84 kip-ft | |