MCQOPTIONS
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MCQOPTIONS
This section includes 20 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. |
Shown as in the figure below, A=60 degree and B=30 degree. Calculate the total length obtained by adding the x-axis component of both the vectors. |
| A. | 3.23m |
| B. | 4.35m |
| C. | 2.50m |
| D. | 1.5mView Answer |
| Answer» B. 4.35m | |
| 2. |
For solving of the unknown tension in the belts, which of the following equation is used? |
| A. | T2 = T1eµB |
| B. | T1 = T2eµB |
| C. | T2 = T1eB |
| D. | T2 = T1eµ |
| Answer» B. T1 = T2eµB | |
| 3. |
All the vector quantities in the solving of the unknown in the belt force system 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 |
| Answer» B. Parallelogram law of multiplication | |
| 4. |
The solving for the unknown forces in the belts requires vector math. So if a vector is multiplied by a scalar in the belt system of forces then_________ |
| A. | Then its magnitude is increased by the square root of that scalar’s magnitude |
| B. | Then its magnitude is increased by the square of that scalar’s magnitude |
| C. | Then its magnitude is increased by an amount of that scalar’s magnitude |
| D. | You cannot multiply the vector with a scalar |
| Answer» D. You cannot multiply the vector with a scalar | |
| 5. |
Determine the magnitude of the projection of the vector force F = 100N acting over a particular point on the belt, onto the v axis, from the figure given below. |
| A. | 96.6N |
| B. | 60N |
| C. | 100N |
| D. | 70.7NView Answer |
| Answer» B. 60N | |
| 6. |
There are mainly two types of forces which are being stated in the free body diagram of the belts, they are generally the resultant forces which are being acted over the body over which the belt is rolling. Which are they? |
| A. | Normal and Frictional |
| B. | Normal and Vertical |
| C. | Vertical and Frictional |
| D. | Normal and Fractional |
| Answer» B. Normal and Vertical | |
| 7. |
We show the net forces acting on the belts with the help of __________ forces. |
| A. | Rotational |
| B. | Linear |
| C. | Helical |
| D. | Resultants |
| Answer» E. | |
| 8. |
We first make equilibrium equations of the belts by considering all the three dimensional forces acting on the section chosen and then the free body diagram is made and solved. |
| 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» E. | |
| 9. |
Which one is not the condition for the equilibrium in free body diagram for the belts as considered for calculation of the normal forces, consider all forces to be straight and linear? |
| A. | ∑Fx=0 |
| B. | ∑Fy=0 |
| C. | ∑Fz=0 |
| D. | ∑F≠0 |
| Answer» E. | |
| 10. |
Dry friction in the belt is also called ___________ |
| A. | Column Friction |
| B. | Coulomb Friction |
| C. | Dry column friction |
| D. | Surface friction |
| Answer» C. Dry column friction | |
| 11. |
What is B in the equation T2 = T1eµB ? |
| A. | Angle of the belt to surface contact in radians |
| B. | Angle of the belt to surface contact in degrees |
| C. | Angle of the belt in radians |
| D. | Angle of the belt in degrees |
| Answer» B. Angle of the belt to surface contact in degrees | |
| 12. |
Vector shown in the figure below has a length of 3m and the angles shown A and B are 60 and 30 degrees each. Calculate the X-axis and Y-axis components. |
| A. | 2.59m and 1.50m respectively |
| B. | 1.50m and 2.59m respectively |
| C. | 3cos60 and 3sin30 respectively |
| D. | 3sin60 and 3sin30 respectivelyView Answer |
| Answer» B. 1.50m and 2.59m respectively | |
| 13. |
We show the net forces acting on the belts with the help of __________ forces.$ |
| A. | Rotational |
| B. | Linear |
| C. | Helical |
| D. | Resultants |
| Answer» B. Linear | |
| 14. |
6N |
| A. | 60N |
| B. | 100N |
| C. | 70.7N |
| Answer» B. 100N | |
| 15. |
For making the equilibrium equations for the belt the normal forces that are being acted over them are in which direction in the free body diagrams? |
| A. | Vertically Upward |
| B. | Vertically Downward |
| C. | Horizontally Right |
| D. | Horizontally Left |
| Answer» E. | |
| 16. |
Dry friction in the belt is also called ___________ |
| A. | Column Friction |
| B. | Coulomb Friction |
| C. | Dry column friction |
| D. | Surface friction |
| Answer» E. | |
| 17. |
What is B in the equation T2 = T1eµB ?$ |
| A. | Angle of the belt to surface contact in radians |
| B. | Angle of the belt to surface contact in degrees |
| C. | Angle of the belt in radians |
| D. | Angle of the belt in degrees |
| Answer» B. Angle of the belt to surface contact in degrees | |
| 18. |
59m and 1.50m respectively |
| A. | 1.50m and 2.59m respectively |
| B. | 3cos60 and 3sin30 respectively |
| C. | 3sin60 and 3sin30 respectively |
| Answer» C. 3sin60 and 3sin30 respectively | |
| 19. |
The frictional force in the belts always acts ____________ to the surface of the application of the friction. |
| A. | Tangential |
| B. | Perpendicular |
| C. | Parallel |
| D. | Normal |
| Answer» B. Perpendicular | |
| 20. |
____________ is the phenomena that resist the movement of the two surfaces in contact, in some of the cases it could be the belts and the rolling cylinders. |
| A. | Friction |
| B. | Motion |
| C. | Circular movement |
| D. | Rotation |
| Answer» B. Motion | |