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# Multiple choice question for engineering

## Set 1

1. Determine the smallest force applied at R which creates the same moment about P as by 75N.

a) 37.5N
b) 112.5N
c) 60N
d) 0N

Answer: a [Reason:] As we know that the moment is the cross product of the distance and the force we will try to apply the same here. We see that the perpendicular distance is 3m. Thus we get the distance. And hence multiply it with the force, the moment = 112.5Nm. Because the force component perpendicular to the distance need to be taken.

2. What does the moment of the force measure?
a) The tendency of rotation of the body along any axis
b) The moment of inertia of the body about any axis
c) The couple moment produce by the single force acting on the body
d) The total work done on the body by the force

Answer: a [Reason:] The moment of the force measures the tendency of the rotation of the body along any axis, whether it be the centroid axis of the body, or any of the outside axis. The couple moment is produced by two forces, not by a single force. The total work done is the dot product of force and distance not the cross.

3. If a car is moving forward, what is the direction of the moment of the moment caused by the rotation of the tires?
a) It is heading inwards, i.e. the direction is towards inside of the car
b) It is heading outwards, i.e. the direction is towards outside of the car
c) It is heading forward, i.e. the direction is towards the forward direction of the motion of the car
d) It is heading backward, i.e. the direction is towards back side of the motion of the car

Answer: a [Reason:] When you curl your wrist in the direction in which the tires are moving then you will find that the thumb is pointing outwards. That is outwards the body of the car. This phenomena is also observed in rainy seasons. When cars travel on the roads, the water is thrown outside from the tires, due to moment.

4. The tendency of rotation of the body along any axis is also called ___________
a) Moment of inertia
b) Moment of couple
c) Torque
d) Force

Answer: c [Reason:] The tendency of rotation of the body along any axis also called the torque. It is the moment of the force acting perpendicular to the direction of the axis of rotation. If the axis and the force are meeting at any point then there is no moment applied by the force.

5. The moment of the force is the product of the force and the perpendicular distance of the axis and the point of action of the force.
a) True
b) False

Answer: a [Reason:] The moment is the product of the force applied to the body and the perpendicular distance of the point of action of the force to the axis about which the body is being rotated. That is the moment is the cross product of the force and the distance between the axis and the point of action.

6. Determine the moment about the point T.

a) 0Nm
b) 350Nm
c) 100Nm
d) 200Nm

Answer: a [Reason:] As we know that the moment is the cross product of the distance and the force we will try to apply the same here. We see that the perpendicular distance is 0m. Thus we get the distance. And hence multiply it with the force, 0x50.

7. The moment axis is in the direction perpendicular to the plane of the force and the distance.
a) True
b) False

Answer: a [Reason:] The moment axis is always perpendicular to the planes of the force and the distance of the axis and the point of action of the force on the body. This means that the moment is the cross product of the force and the distance between the axis and the point of action of the force.

8. Find the moment along T.

a) 1200Nm
b) 600Nm
c) 0Nm
d) 1400Nm

Answer: b [Reason:] As we know that the moment is the cross product of the distance and the force we will try to apply the same here. But here we have number of forces components acting on the T point. So by adding all the moments caused by all the forces will give us the value as 600Nm.

9. If you are getting to know about the direction of the moment caused by the force applied on the body by using your wrist and curling it in the direction of the rotation then which of the following is not right?
a) The thumb represents the direction of the force
b) The thumb represents the direction of the moment
c) The fingers represents the direction of the force
d) The direction in which you curl your wrist is towards the direction of the distance from point of contact of force to the axis of rotation.

Answer: b [Reason:] The curled hand represents various thing. The direction of the moment axis is given by the thumb. The direction of the force is given by the fingers. As we place the fingers on the force and curl towards the rotational direction of the body about the axis.

10. The moment axis, force and the perpendicular distance in the moment of the force calculation is lying in____________
a) Two planes perpendicular to each other
b) A single plane in the direction of the force
c) A single plane in the direction of the perpendicular distance
d) A single line in the direction of the force

Answer: a [Reason:] The moment axis, force and the perpendicular distance is lying in the three dimensional Cartesian. It doesn’t lye on the single plane. It also doesn’t lye in a single line. Nor in the direction of the force. Thus they all lye in the planes which are perpendicular to each other.

11. If the rotation is clockwise in this page, suppose, then in which direction will the thumb project if you curl your hand in the same direction of the rotation?
a) It will point to the direction perpendicular to the plane of paper and towards you
b) It will point to the direction perpendicular to the plane of paper and away from you
c) It will point to the direction parallel to the plane of paper and towards right
d) It will point to the direction parallel to the plane of paper and towards left

Answer: b [Reason:] As the curling will give the direction perpendicular to the paper. But it does depend upon the rotation sense. In this example, the sense is clockwise. Thus the thumb goes into the paper. That is it goes away from the viewer. Thus the answer.

12. Which of the following is true?
a) Total moment of various forces acting on the body is the vector sum of all moments
b) Total moment of various forces acting on the body is the algebraic sum of all moments
c) Total moment of various forces acting on the body is always zero
d) Total moment of various forces acting on the body is the vector sum of all moments which is perpendicular to each other forces

Answer: a [Reason:] The moment is the vector quantity. Thus the value of the total moment caused by various forces acting on the body is the vector sum of all the vectors. Also the moments are not perpendicular to each other, unless it is specified. Thus assumptions cant be taken for the direction of the moment.

13. Determine the moment about the point R.

a) 0Nm
b) 350Nm
c) 100Nm
d) 200Nm

Answer: c [Reason:] As we know that the moment is the cross product of the distance and the force we will try to apply the same here. We see that the perpendicular distance is 7m. Thus we get the distance. And hence multiply it with the force, 7×50.

14. Find the moment along Q.

a) 1200Nm
b) 600Nm
c) 0Nm
d) 1400Nm

Answer: a [Reason:] As we know that the moment is the cross product of the distance and the force we will try to apply the same here. But here we have number of forces components acting on the T point. So by adding all the moments caused by all the forces will give us the value as 1200Nm.

15. Determine the moment of the force along point P.

a) 110Nm
b) 112.5Nm
c) 60Nm
d) 0Nm

Answer: b [Reason:] As we know that the moment is the cross product of the distance and the force we will try to apply the same here. We see that the perpendicular distance is 1.5 (3cos60). Thus we get the distance. And hence multiply it with the force, 75×1.5. Because the force component perpendicular to the distance need to be taken.

## Set 2

1. Determine the moment’s magnitude produce by the force as shown in the diagram, which tends to rotate the rod ORQP along QP.

a) 80.49 Nm
b) 72.12 Nm
c) -36.67 Nm
d) 36.67 Nm

Answer: a [Reason:] The use of the formula A.(rxF) gives the answer. In which A is 0.89i + 0.447j m. And the force is 300N, which is being applied at the end of the rod. Thus, after finding the equation of the axis and then replacing it in the equation shown above we get the answer. Actually, the main task is to know the axis equation in the vector form. Then get the magnitude of the moment.

2. The calculation of the moment about the axis and the moment about any point by a force applied on the body are different from each other.
a) True
b) False

Answer: a [Reason:] The calculation of the moment about the axis and the moment about any point by a force applied on the body are different from each other. It is because both of the calculations require different formulas. And both of the formulas have different inputs, which is obviously different.

3. In the calculation of the moment of the force about the axis, the cross product table, i.e. the 3X3 matrix which is made for doing the cross product having 3 rows, contains three elements. Which are they from top to bottom?
a) Axis coordinates, point coordinates and the force coordinates
b) Point coordinates, axis coordinates, and the force coordinates
c) Axis coordinates, force coordinates and the point coordinates
d) Force coordinates, point coordinates and the axis coordinates

Answer: a [Reason:] The 3X3 matrix which is being made is having axis coordinates, point coordinates and the force coordinates. They are from top to bottom placed. The order cannot be changed. Or if changed then one needs to apply the negative sign appropriately. Negative because the directions gets reversed.

4. Which of the following is correct? (For A representing the vector representation of the axis of rotation, r the radius vector and F the force vector)
a) A.(rxF)
b) Ax(rxF)
c) A.(r.F)
d) Fx(r.F)

Answer: a [Reason:] The correct form of the equation is given by A.(rxF). Where A represents the vector representation of the axis of rotation, r the radius vector and F the force vector. This is usually done for determining the moment of the force about the axis. That is if body is being rotated by the force about an axis.

5. In the equation A.(rxF), the r vector is what?
a) The magnitude of the axis, i.e. the length of the axis
b) The length of the force vector
c) The length of the radius

Answer: d [Reason:] The r in the equation A.(rxF) is the radius vector. That is it is the vector which is having the start point at the axis and the end point at the point of action of the force on the body. This vector is being crossed by the force vector, which is then followed by the dot product with the axis vector.

6. The axis vector in the calculation of the moment along the axis of rotation is the axis which is collinear with the force vector.
a) True
b) False

Answer: b [Reason:] The axis of the rotation cannot be collinear with the force vector. If it does so then the rotation of body is not possible. That is the moment of the force is zero. Which means no rotation being given by the force along the axis of rotation of the body.

7. Which of the following is correct?
a) We cannot represent the moment caused by the force along any axis in vector form
b) We can represent the moment caused by the force along any axis in scalar form
c) We cannot represent the moment caused by the force along any point in vector form
d) We can represent the moment caused by the force along any axis in vector form

Answer: d [Reason:] As moment is a vector, we can represent it in the vector form easily. Whether it may be the moment of the force acting on the body about the axis, or about the point. We can also convert the same in the Cartesian form too. The only thing we need to do is the cross product of the radius vector and the force vector.

8. What if the perpendicular distance from the axis is infinity?
a) The rotation is not possible
b) The rotation is possible but the moment generated is very less
c) The force applied will be very much high for even a small rotation
d) No rotation unless the contact is being broken

Answer: c [Reason:] The long distance means a huge force which one needs to apply. Because distance increased will also increase the resistance inertia which will obviously increase the force required for the rotation. Though we know that the larger the distance the small is the force applied for rotation. But inertia must be taken into the considerations some times.

9. In the equation A.(rxF) the r is heading from ______________ and ending at _____________
a) Axis of rotation, Force vector
b) Axis of rotation, Force vector’s point of action on the body
c) Force vector, Axis of rotation
d) Force vector’s point of action on the body, Axis of rotation

Answer: b [Reason:] It is the radius vector. The radius vector is always from the axis of rotation to the point of action of the force on the body. Which means that the radius vector is not on any point on the force vector. Rather it ending at the point on the force vector, where it is being in contact of the body.

10. What if the moment of the force calculated about the axis is negative?
a) It means that the force is applied in the opposite direction as imagined
b) It means that the direction of the motion is in the opposite sense as imagined
c) It means that the radius vector is in the opposite sense as imagined
d) Such calculation means that the calculations are wrongly done

Answer: b [Reason:] It means that the direction of the motion is in the opposite sense as imagined. We can’t say about the direction of the force or the direction of the radius vector. But yes we can say about the direction of the rotation as it the thing which is going to be calculated. Rest all the parts are fixed. They can’t be altered.

11. Determine the moment MQP in the vector form produce by the force as shown in the diagram, which tends to rotate the rod ORQP along QP.

a) 72i + 36j Nm
b) 72i – 36j Nm
c) -72i – 36j Nm
d) -72i + 36j Nm

Answer: a [Reason:] The use of the formula A.(rxF) gives the answer. In which A is 0.89i + 0.447j m. And the force is 300N, which is being applied at the end of the rod. Thus, after finding the equation of the axis and then replacing it in the equation shown above we get the answer. Actually, the main task is to know the axis equation in the vector form.

12. Determine the moment of the force F along the segment QP of the pipe assembly shown in the figure.

a) 110Nm
b) 100Nm
c) 500Nm
d) 510Nm

Answer: b [Reason:] The use of the formula A.(rxF) gives the answer. In which A is 0.6i + 0.8j m and the r is 0.5i + 0.5k. And the force is 300N, which is being applied at the end of the rod. Thus, after finding the equation of the axis and then replacing it in the equation shown above we get the answer. Actually, the main task is to know the axis equation in the vector form. Then get the magnitude of the moment.

13. Determine the magnitude of the moment of the force about the axis PQ.

a) -72Nm
b) 82Nm
c) 90Nm
d) 50Nm

Answer: a [Reason:] The use of the formula A.(rxF) gives the answer. In which A is 0.6i + 0.8j m and the r is -0.2k, and the force which is being applied at the end of the rod. Thus, after finding the equation of the axis and then replacing it in the equation shown above we get the answer. Actually, the main task is to know the axis equation in the vector form. Then get the magnitude of the moment.

14. Determine the magnitude of the moment of the force about the y-axis.

a) -72Nm
b) 82Nm
c) 210Nm
d) 50Nm

Answer: c [Reason:] The use of the formula A.(rxF) gives the answer. In which A is 1j m and the r is -3i + 4j + 2k and the force which is being applied at the end of the rod. Thus, after finding the equation of the axis and then replacing it in the equation shown above we get the answer. Actually, the main task is to know the axis equation in the vector form. Then get the magnitude of the moment.

15. The ___________ forces do not cause the rotation if the rotation is considered in about the axis of the body or the centroid axis of the body.
a) Non-concurrent
b) Concurrent
c) Parallel
d) Non-Parallel

Answer: b [Reason:] The concurrent forces are the which are somewhere touching the axis of rotation. If any of the force is touching that axis, that force is not considered, or is insufficient to cause a rotation. If a force is concurrent then the perpendicular distance of the force from the line of axis is zero, thus no rotation. As we know rotation is caused by moment.

## Set 3

1. Whenever the distributed loading acts perpendicular to an area its intensity varies __________
a) Linearly
b) Non-Linearly
c) Parabolically
d) Cubically

Answer: a [Reason:] The load intensity is varying linearly in the structures. Thus the intensity is not varying parabolically nor is it cubically. It cannot be a vector also. Thus the intensity is linearly varied.

2. Determine the moment of inertia of the area about y-axis.

a) 0.273m2
b) 11m2
c) 0.141m2
d) 0.811m2

Answer: a [Reason:] Parallel axis for any area is used to add the two mutually perpendicular moment of inertias for areas. It gives a moment of inertia perpendicular to the surface of the body. That is the moment of inertia perpendicular to the surface in considerance.

3. The calculation of the moment of the body due to the loadings involve a quantity called ____________
a) Moment
b) Inertia
c) Moment of Inertia
d) Rotation

Answer: c [Reason:] The calculation of the moment of the body due to the loadings involve a quantity called moment of inertia. This is having much significance in the various fields in the engineering sector. The main types are the ‘I’ section structures which are being much used.

4. Moment of Inertia is the integration of the square of the distance of the centroid and the del area along the whole area of the structure.
a) True
b) False

Answer: a [Reason:] The moment of inertia of the section is the integration of the square of the distance of the centroid and the del area along the whole area of the structure. This is having much significance in the various fields in the engineering sector. The main types are the ‘I’ section structures which are being much used.

5. There is perpendicular axis theorem for the area.
a) True
b) False

Answer: b [Reason:] There is no perpendicular axis theorem for the area. In spite there is the theorem as parallel axis for any area. Thus we have the theorem which is used to add the two mutually perpendicular moment of inertias.

6. What is parallel axis theorem and to whom it is applied?
a) Theorem used to add the two mutually perpendicular moment of inertias for areas
b) Theorem used to add the two mutually perpendicular moment of inertias for volumes
c) Theorem used to add the two mutually perpendicular moment of inertias for linear distances
d) Theorem used to add the two mutually perpendicular moment of inertias for vectors

Answer: a [Reason:] Parallel axis for any area is used to add the two mutually perpendicular moment of inertias for areas. It gives a moment of inertia perpendicular to the surface of the body. That is the moment of inertia perpendicular to the surface in considerance.

7. The parallel axis theorem gives the moment of inertia ______________ to the surface of considerance.
a) Linear
b) Non-Linear
c) Perpendicular
d) Parallel

Answer: c [Reason:] Parallel axis for any area is used to add the two mutually perpendicular moment of inertias for areas. It gives a moment of inertia perpendicular to the surface of the body. That is the moment of inertia perpendicular to the surface in considerance.

8. The parallel axis theorem can add any angle varied moment of inertias to give the perpendicular moment of inertia.
a) True
b) False

Answer: b [Reason:] Parallel axis for any area is used to add the two mutually perpendicular moment of inertias for areas. It gives a moment of inertia perpendicular to the surface of the body. That is the moment of inertia perpendicular to the surface in considerance.

9. The parallel axis theorem uses the ____________ of the distance.
a) Square root
b) Square
c) Cube root
d) Cube

Answer: b [Reason:] Parallel axis for any area is used to add the two mutually perpendicular moment of inertias for areas. It gives a moment of inertia perpendicular to the surface of the body. And uses the square of the distance from the axis of rotation.

10. The distance in the parallel axis theorem is multiplied by ___________
a) Area
b) Volume
c) Linear distance
d) Area/Volume

Answer: a [Reason:] Parallel axis for any area is used to add the two mutually perpendicular moment of inertias for areas. It gives a moment of inertia perpendicular to the surface of the body. And uses the square of the distance from the axis of rotation multiplied by the area.

11. One of the use of the centre of mass or centroid is as in the moment of inertia is that the net force acts at the ___________ of the loading body.
a) Centroid
b) The centre axis
c) The corner
d) The base

Answer: a [Reason:] In the moment of inertia calculations we see that the net force acts at the centroid of the loading body. That is if the loading system is in the form of the triangle then the at the distance 2 by 3 of the base the net force of the loading will act. And the load will be half the area of the loading.

12. If the non-Uniform loading is of the type of parabola then for calculating the moment of inertia for areas?
a) The net load will not be formed as all the forces will be cancelled
b) The net force will act the centre of the parabola
c) The net force will act on the base of the loading horizontally
d) The net force will act at the centroid of the parabola

Answer: d [Reason:] The net force will act at the centroid of the parabola. Whether it be a parabola or the cubic curve the centroid is the only point at which the net force act. Force can’t be acted horizontally if the loading is vertical. Hence whatever be the shape of the loading, the centroid is the point of action of net force. Thus the use of centroid.

13. If any external force also is applied on the structure and we are determining the moment of inertia then what should we consider?
a) The net force will act at the centroid of the structure only
b) The net load will not be formed as all the forces will be cancelled
c) The net force will act on the base of the loading horizontally
d) The net force will not to be considered, there would be a net force of the distribution, rest will be the external forces

Answer: d [Reason:] The external forces are treated differently. They are not added by the force of the distributed loading. That is the force not only acts at the centroid always. It can be shifted also. Depending on the external forces. Thus the use of centroid or centre of mass.

14. The body is sometimes acted by two or three force members and we need to find the moment of inertia for the same. The difference between the two and the three force members is:
a) The former is collinear and the latter is parallel
b) The former is parallel and the latter is perpendicular
c) The former is perpendicular and the latter is collinear
d) The former is acting on two points in the body while the latter is on three points

Answer: d [Reason:] The definition of the two force member only defines that the forces are being acted on the two points on the body. So does is the definition of the three forces members. The points of action of the three forces are three.

15. Determine the moment of inertia of the area about x-axis.

a) 0.111m2
b) 11m2
c) 0.141m2
d) 0.811m2

Answer: a [Reason:] Parallel axis for any area is used to add the two mutually perpendicular moment of inertias for areas. It gives a moment of inertia perpendicular to the surface of the body. That is the moment of inertia perpendicular to the surface in considerance.

## Set 4

1. If the non-Uniform loading is of the type of parabola then for calculating the moment of inertia for areas inclined at an axis?
a) The net load will not be formed as all the forces will be cancelled
b) The net force will act the centre of the parabola
c) The net force will act on the base of the loading horizontally
d) The net force will act at the centroid of the parabola

Answer: d [Reason:] The net force will act at the centroid of the parabola. Whether it be a parabola or the cubic curve the centroid is the only point at which the net force act. Force can’t be acted horizontally if the loading is vertical. Hence whatever be the shape of the loading, the centroid is the point of action of net force. Thus the use of centroid.

2. Determine the orientation of the principle axis for the cross section of area of member shown whose width is 100mm.

a) 57.1˚
b) 5.1˚
c) 7.1˚
d) 52.1˚

Answer: a [Reason:] The net force will act at the centroid of the parabola. Whether it be a parabola or the cubic curve the centroid is the only point at which the net force act. Force can’t be acted horizontally if the loading is vertical. Hence whatever be the shape of the loading, the centroid is the point of action of net force. Thus the use of centroid.

3. The product of Inertia for an area is required so as to__________
a) Determine Maximum moments of inertia for an Area
b) Determine Maximum moments of inertia for a Line
c) Determine Maximum moments of inertia for a Volume
d) Determine Maximum moments of inertia for a Rectangle

Answer: a [Reason:] The product is required as to know what is the maximum and the minimum moment of inertia of an area. This data is required as to design the structure of the body. This means that the designing of the body is majorly done by the help of the determination of the product of the moment of Inertia.

4. Moment of Inertia about an inclined axis is the integration of the square of the distance of the centroid and the del area along the whole area of the structure and after this calculations we multiply the moment of areas.
a) True
b) False

Answer: a [Reason:] The moment of inertia of the section is the integration of the square of the distance of the centroid and the del area along the whole area of the structure. This is having much significance in the various fields in the engineering sector. The main types are the ‘I’ section structures which are being much used.

5. The product of moment of inertia is the sum of _____________ and _________________. It can be used in the calculations of moments of inertia for an area about inclined axis.
a) Area and volume
b) Volume and linear distance
c) Moment of inertia at centroid and the product of the area and del dx and del dy
d) Moment of inertia at base and the product of the area and del dx and del dy

Answer: d [Reason:] The product of moment of inertia is required as to design the structure of the body. This means that the designing of the body is majorly done by the help of the determination of the product of the moment of Inertia.

6. There is perpendicular axis theorem for the area, and it is can be used to determine the moment of inertia of an area about inclined axis.
a) True
b) False

Answer: b [Reason:] There is no perpendicular axis theorem for the area. In spite there is the theorem as parallel axis for any area. Thus we have the theorem which is used to add the two mutually perpendicular moment of inertias.

7. If any external force also is applied on the structure and we are determining the moment of inertia for areas about inclined axis then what should we consider?
a) The net force will act at the centroid of the structure only
b) The net load will not be formed as all the forces will be cancelled
c) The net force will act on the base of the loading horizontally
d) The net force will not to be considered, there would be a net force of the distribution, rest will be the external forces

Answer: d [Reason:] The external forces are treated differently. They are not added by the force of the distributed loading. That is the force not only acts at the centroid always. It can be shifted also. Depending on the external forces. Thus the use of centroid or centre of mass.

8. Determine the magnitude of the principle moment of inertia for the cross section of area of member shown whose width is 100mm.

`
a) 7.54 x 109 mm2
b) 54 x 109 mm2
c) 3.4 x 109 mm2
d) 1.54 x 109 mm2

Answer: a [Reason:] The net force will act at the centroid of the parabola. Whether it be a parabola or the cubic curve the centroid is the only point at which the net force act. Force can’t be acted horizontally if the loading is vertical. Hence whatever be the shape of the loading, the centroid is the point of action of net force. Thus the use of centroid.

9. The calculation of moment of inertia about inclined axis due to the loadings involve a quantity called ____________
a) Moment
b) Inertia
c) Moment of Inertia
d) Rotation

Answer: c [Reason:] The calculation of the moment of the body due to the loadings involve a quantity called moment of inertia. This is having much significance in the various fields in the engineering sector. The main types are the ‘I’ section structures which are being much used.

10. Whenever the distributed loading acts perpendicular to an area its intensity varies __________ for the determination of moment of inertia about the inclined axis.
a) Linearly
b) Non-Linearly
c) Parabolically
d) Cubically

Answer: a [Reason:] The load intensity is varying linearly in the structures. Thus the intensity is not varying parabolically nor is it cubically. It cannot be a vector also. Thus the intensity is linearly varied.

11. The distance in the parallel axis theorem for the use in the determination of the moment of inertia about an inclined axis is multiplied by:
a) Area
b) Volume
c) Linear distance
d) Area/Volume

Answer: a [Reason:] Parallel axis for any area is used to add the two mutually perpendicular moment of inertias for areas. It gives a moment of inertia perpendicular to the surface of the body. And uses the square of the distance from the axis of rotation multiplied by the area.

12. One of the use of the centre of mass or centroid is as in the determination of the moment of inertia about an inclined plane is that the net force acts at the ___________ of the loading body.
a) Centroid
b) The centre axis
c) The corner
d) The base

Answer: a [Reason:] In the moment of inertia calculations we see that the net force acts at the centroid of the loading body. That is if the loading system is in the form of the triangle then the at the distance 2 by 3 of the base the net force of the loading will act. And the load will be half the area of the loading.

13. What is parallel axis theorem and to whom it is applied so that it can give the product of inertia of an area of an area inclined about an axis?
a) Theorem used to add the two mutually perpendicular moment of inertias for areas
b) Theorem used to add the two mutually perpendicular moment of inertias for volumes
c) Theorem used to add the two mutually perpendicular moment of inertias for linear distances
d) Theorem used to add the two mutually perpendicular moment of inertias for vectors

Answer: a [Reason:] Parallel axis for any area is used to add the two mutually perpendicular moment of inertias for areas. It gives a moment of inertia perpendicular to the surface of the body. That is the moment of inertia perpendicular to the surface in considerance.

14. The body is sometimes acted by two or three force members and we need to find the moment of inertia about the inclined axis for the same. The difference between the two and the three force members is:
a) The former is collinear and the latter is parallel
b) The former is parallel and the latter is perpendicular
c) The former is perpendicular and the latter is collinear
d) The former is acting on two points in the body while the latter is on three points

Answer: d [Reason:] The definition of the two force member only defines that the forces are being acted on the two points on the body. So does is the definition of the three forces members. The points of action of the three forces are three.

15. The parallel axis theorem can add any angle varied moment of inertias to give the perpendicular moment of inertia and it can be used in the determination of the moment of inertia about inclined axis.
a) True
b) False

Answer: b [Reason:] Parallel axis for any area is used to add the two mutually perpendicular moment of inertias for areas. It gives a moment of inertia perpendicular to the surface of the body. That is the moment of inertia perpendicular to the surface in considerance.

## Set 5

1. For making the equilibrium equations the normal forces acts in which direction in the free body diagrams?
a) Vertically Upward
b) Vertically Downward
c) Horizontally Right
d) Horizontally Left

Answer: a [Reason:] As the loads are being acting in the downward direction. Thus to make the forces balance, the normal forces act in the vertically upward direction. As we know that when there is no lubricating fluid present between the surfaces in contact, the dry friction occurs. This friction magnitude is taken out from these normal forces.

2. Which one is not the condition for the equilibrium in free body diagram 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: d [Reason:] For the equilibrium in the three dimensional system of axis we have all the conditions true as, ∑Fx=0, ∑Fy=0 and ∑Fz=0. Also we have the summation of the forces equal to zero. Which is not a non-zero value.

3. We first make equilibrium equations of the body 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: d [Reason:] We first make the free body diagram and then we make the equilibrium equations to satisfy the given conditions. This helps us to solve the question easily. As this reduces the part of imagination and increases accuracy too.

4. We show the net forces by the help of __________ forces.
a) Rotational
b) Linear
c) Helical
d) Resultants

Answer: d [Reason:] The net forces acting on the body is shown by the help of the resultant forces. This is because the resultant forces have the sum of all the forces which are acting on the direction which is same. Thus the resultant forces are used to show the net forces acting in the body.

5. There are main two types of forces which are being stated in the free body diagram, they are generally the resultant forces which are being acted over the body. Which are they?
a) Normal and Frictional
b) Normal and Vertical
c) Vertical and Frictional
d) Normal and Fractional

Answer: a [Reason:] The net forces acting on the body is shown by the help of the resultant forces. There are two types, first the frictional and the second is the normal. This is because the resultant forces have the sum of all the forces which are acting on the direction which is same.

6. In the problems regarding the friction and normal force calculations there is no scope of the two force and three force systems.
a) True
b) False

Answer: b [Reason:] The two and three force member system can be acted upon the body even if there is friction calculations involved. The main thing is that the frictional forces are the phenomena in which there is a resistance between the surfaces. And the resistance magnitude is given by those calculations. So if two or three system does comes, it has no effect.

7. The difference between the two and the three force members is:
a) The former is collinear and the latter is parallel
b) The former is parallel and the latter is perpendicular
c) The former is perpendicular and the latter is collinear
d) The former is acting on two points in the body while the latter is on three points

Answer: d [Reason:] The definition of the two force member only defines that the forces are being acted on the two points on the body. So does is the definition of the three forces members. The points of action of the three forces are three.

8. There is the application of the Newton’s third law of motion in the free body diagrams of the friction calculations.
a) True
b) False

Answer: a [Reason:] There is some calculations involved in which we have to make use of the Newton’s third law of motion. The requirement of the third law is important in the equilibrium of the body. The body particles are in the equilibrium and are thus facing the forces and to be in the equilibrium they also react and apply the opposite force and thus the third law of newton.

9. What does the Newton’s third law states?
a) The rate of change of momentum is equal to the force applied
b) For every reaction there is an opposite reaction
c) The body is tend to be rotated if the force is applied tangentially
d) The body is rest until a force is applied

Answer: b [Reason:] The requirement of the third law is important in the equilibrium of the body. The body particles are in the equilibrium and are thus facing the forces and to be in the equilibrium they also react and apply the opposite force and thus the third law of Newton. Thus there is some involvement of the Newton’s third law of motion.

10. 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: c [Reason:] The net moment of the body is zero that doesn’t means that the distance between the force and the rotational axis is zero. This thing is used when there is any rotational effect possible during the friction. For e.g. if we have a wedge and the block is placed over it then.

11. As we are using vector math to solve for the unknown forces in the free body diagrams, what is k.i?
a) 0
b) 1
c) -1
d) ∞

Answer: a [Reason:] As the dot product of only the same Cartesian component is unity, i.e. i.i = 1 and j.j =1, rest all remaining dot product will give 0(i.j = 0 and j.k = 0). Cross product of the same plane vectors always give zero. This is generally done because the normal forces are perpendicular to the surfaces of action of the friction.

12. The three force system can also be in the equilibrium if:
a) All the forces are parallel to each other heading towards the same direction
b) The force components cancel each other
c) The forces are very small in magnitude
d) The forces are very huge in magnitude

Answer: b [Reason:] Even if the forces components cancels each other than the body is said to be in equilibrium. Also it should satisfy whether the moments caused by the forces are equal to zero or not. This means that we need to take components of the normal and frictional forces so as to make the desired calculations.

13. The normal forces and the forces of friction are collinear.
a) True
b) False

Answer: b [Reason:] They are not collinear. They are just acting perpendicular or parallel to the surface of action. Thus they are not even in the same plane. So as to make the calculations proper we need to see the same and then apply the equations.

14. Determine the horizontal force acting in the given figure.

a) 80cos30
b) 60cos30
c) 40cos30
d) 8cos30