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This section includes 16 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. |
Determine the minimum amount of the force P required to remove the wedge shown. Take the value of the coefficient of friction to be 0.3. Assume that the block doesn’t slip at C. |
A. | 1.15 KN |
B. | 2.15 KN |
C. | 3.15 KN |
D. | 4.15 KN |
Answer» B. 2.15 KN | |
2. |
In the simplification of the forces applied in the wedges net force acts at the ___________ of the loading body. |
A. | Centroid |
B. | The centre axis |
C. | The corner |
D. | The base |
Answer» B. The centre axis | |
3. |
What is j.j? |
A. | 0 |
B. | 1 |
C. | -1 |
D. | ∞ |
Answer» C. -1 | |
4. |
In the free body diagrams involving the wedges we have use of vector math, so for two vectors A and B, what is A.B (if they have angle α between them)?a) |A||B| cosαb) |A||B|c) √(|A||B |
A. | |A||B| cosα |
B. | |A||B| |
C. | √(|A||B|) cosα |
D. | |A||B| sinα |
Answer» B. |A||B| | |
5. |
Determine the normal force at B in the wedge shown. Take the value of the coefficient of friction to be 0.3. Assume that the block doesn’t slip at C. |
A. | 2452 N |
B. | 215 KN |
C. | 3150 N |
D. | 415 KNView Answer |
Answer» B. 215 KN | |
6. |
Which one is not the condition for the equilibrium in free body diagram for calculation of the normal forces along the body kept over the wedge, consider all forces to be straight and linear? |
A. | ∑Fx=0 |
B. | ∑Fy=0 |
C. | ∑Fz=0 |
D. | ∑F≠0 |
Answer» E. | |
7. |
In the free body diagrams involving the wedges we have use of vector math, so for two vectors A and B, what is A.B (if they have angle α between them)?$# |
A. | |A||B| cosα |
B. | |A||B| |
C. | √(|A||B|) cosα |
D. | |A||B| sinα |
Answer» D. |A||B| sin‚âà√≠¬¨¬± | |
8. |
2452_N$ |
A. | 215 KN |
B. | 3150 N |
C. | 415 KN |
Answer» C. 415 KN | |
9. |
Sometimes in the calculations involving the wedges, there are situations when there is a toppling effect on the body. In that, if the net moment of the body is zero that means the distance between the force and the rotational axis is zero. |
A. | The first part of the statement is false and other part is true |
B. | The first part of the statement is false and other part is false too |
C. | The first part of the statement is true and other part is false |
D. | The first part of the statement is true and other part is true too |
Answer» B. The first part of the statement is false and other part is false too | |
10. |
What_is_j.j? |
A. | 0 |
B. | 1 |
C. | -1 |
D. | ‚àû |
Answer» B. 1 | |
11. |
When does the two wedge system is termed as self-locking system? |
A. | If friction forces hold the block in plane |
B. | If friction forces doesn’t hold the block in plane |
C. | If friction forces hold the block in phase |
D. | If friction forces doesn’t hold the block in phase |
Answer» B. If friction forces doesn‚Äö√Ñ√∂‚àö√ë‚àö¬•t hold the block in plane | |
12. |
The coefficient of friction between the body being slided over the wedge and the wedge surface is generally determined by ____________ |
A. | Written over the Body |
B. | Experiments |
C. | Weighing the body |
D. | Measuring length of the body |
Answer» C. Weighing the body | |
13. |
The normal force exerted by the surface of the wedge is normal to the surface of the ________ |
A. | Base of the wedge |
B. | Base of the body residing over it |
C. | Base of the body just neighbour to the wedge |
D. | Earth’s surface |
Answer» C. Base of the body just neighbour to the wedge | |
14. |
The blocks of heavy weights are being able to lift from the help of the wedges. |
A. | True |
B. | False |
Answer» B. False | |
15. |
Wedges are used to transfer heavy loads. |
A. | True |
B. | False |
Answer» B. False | |
16. |
A ____________ is a simple machine that is used as to transfer applied forces into much larger forces. |
A. | Wedge |
B. | Beam |
C. | Pillar |
D. | Bridges |
Answer» B. Beam | |