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# Multiple choice question for engineering

## Set 1

1. The determination of the internal loading in the beams is usually done so as to ______________
a) Break the beam
b) Know the length
c) Know the diameter
d) Design the beam

Answer: d [Reason:] The determination of the internal forces in the beam is done so as to design the beams as in the application purpose the beams will be subjected to many loads. This will help us to make the beam properly. And also this will ensure that the beams will not break after the loading is done on them. For this another method is the distribution load method.

2. The slope of the shear diagram is equal to__________
a) Rotational moment
b) Bending moment
c) Total weight

Answer: d [Reason:] After the application of the force equation of equilibrium to the segment of the beam, we have the above result. This is done on the very small part of the beam. That is the minimal section of the beam is to be considered and then the application of the equilibrium equations are done so as to calculate the final result.

3. The shear diagram will jump downward if the _________________
a) Rotational moment
b) Bending moment
c) Total weight
d) Force will act downwards

Answer: d [Reason:] After the application of the force equation of equilibrium to the segment of the beam, we have the above result. This is done on the very small part of the beam. That is the minimal section of the beam is to be considered and then the application of the equilibrium equations are done so as to calculate the final result.

4. Change in the shear is equal to ____________
a) Rotational moment
b) Bending moment
c) Total weight

Answer: d [Reason:] After the application of the force equation of equilibrium to the segment of the beam, we have the above result. This is done on the very small part of the beam. That is the minimal section of the beam is to be considered and then the application of the equilibrium equations are done so as to calculate the final result.

b) Method of area
c) Method of line
d) Method of volume

Answer: a [Reason:] The determination of the internal forces is done by various methods among which the distribution of load method is the one. Also we have method of joints, which is done so as to determine the internal forces which are being developed in the trusses or the frames. This is done as to design them. In this method the loading acting upward is considered to be positive.

6. The equation of change in moment equals area under shear diagram is not applied when_________
a) Rotational moment
b) Bending moment
c) Total weight
d) Couple moment acts

Answer: d [Reason:] As this is done with the help of distributed load method thus when a couple moment this equation act is not valid. This is done on the very small part of the beam. That is the minimal section of the beam is to be considered and then the application of the equilibrium equations are done so as to calculate the final result.

7. Slope of the moment diagram is equal to _________
a) Rotational moment
b) Bending moment
c) Total weight
d) Shear

Answer: d [Reason:] After the application of the force equation of equilibrium to the segment of the beam, we have the above result. This is done on the very small part of the beam. That is the minimal section of the beam is to be considered and then the application of the equilibrium equations are done so as to calculate the final result.

8. The change in moment is equal to _________
a) Rotational moment
b) Bending moment
c) Total weight
d) Area under shear diagram

Answer: d [Reason:] After the application of the force equation of equilibrium to the segment of the beam, we have the above result. This is done on the very small part of the beam. That is the minimal section of the beam is to be considered and then the application of the equilibrium equations are done so as to calculate the final result.

9. In the making of the shear force diagram or the bending moment diagrams, one method is used, distributed load. In this the distributed load will be considered positive if ____________

Answer: a [Reason:] The loading acting upward is considered to be positive. This is done as to decrease the time used for the calculations. If the method of sections is being applied at the complex sections for the determination of the unknown forces, much time is required for it. Thus the main motto is to make the calculations easy and make the less use of time.

10. Determine the shear force of the beam shown.

a) 450N
b) 50N
c) 40N
d) 45N

Answer: a [Reason:] The loading acting upward is considered to be positive. This is done as to decrease the time used for the calculations. If the method of sections is being applied at the complex sections for the determination of the unknown forces, much time is required for it. Thus the main motto is to make the calculations easy and make the less use of time.

11. The shear diagram will jump upwards if there is __________
a) Rotational moment
b) Bending moment
c) Total weight
d) Change in shear is positive

Answer: d [Reason:] After the application of the force equation of equilibrium to the segment of the beam, we have the above result. This is done on the very small part of the beam. That is the minimal section of the beam is to be considered and then the application of the equilibrium equations are done so as to calculate the final result.

12. The equation of change in moment equals area under shear diagram is not applied when?
a) Rotational moment
b) Bending moment
c) Total weight
d) A concentrated force acts

Answer: d [Reason:] As this is done with the help of distributed load method thus when a concentrated load is applied this equation is not valid. This is done on the very small part of the beam. That is the minimal section of the beam is to be considered and then the application of the equilibrium equations are done so as to calculate the final result.

13. The couple in the beam is simplified easily by the help of right hand rule. But the forces simplification is not possible as there is no such system of rules so that the forces can be simplified.
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:] Both of them are vector quantities. And both of them can be easily simplified. If taken in the vector form then the task is even easier. Thus it is not necessary for the force or the couple to be vector only, even if the magnitude is taken, the simplification is done in the 2D. Forces are simplified on the basis of the algebra.

14. The couple moment and the force is divided to get the distance of the axis from the point of action of the force.
a) True
b) False

Answer: a [Reason:] The couple is the cross product of the force and the perpendicular distance between the forces. And if the division of the both is done than the resulting quantity is the distance only. But care must be taken that the product is the cross product and hence sine angle must be divided too to get the distance.

15. Determine the normal force of the beam shown.

a) 0N
b) 50N
c) 40N
d) 45N

Answer: a [Reason:] The loading acting upward is considered to be positive. This is done as to decrease the time used for the calculations. If the method of sections is being applied at the complex sections for the determination of the unknown forces, much time is required for it. Thus the main motto is to make the calculations easy and make the less use of time.

## Set 2

1. Potential energy is stored in the body if some work is done on it. Work done is best given by:
a) dU = Fdrcosθ
b) dU = Fdrsinθ
c) dU = Fdrcotθ
d) dU = Fdrdθ

Answer: a [Reason:] The work is defined as the dot product of the force and the distance. This means that the work done does depends upon the angle of the force. That is the angle which is being made by the force vector to the surface of action of the force.

2. If a block moves from a height h above the ground then the potential energy is given by_____

Answer: a [Reason:] The potential energy is converted into the work done. Thus is the block moves from the height h to the ground, the work done is given by Negative because the work done is in the opposite direction of the motion of the body.

3. Potential energy stored in a spring to a body from moving it from x1 to x 2 distances is given by_____

Answer: a [Reason:] The potential energy is converted into the work done. Thus is the block moves from the distance x2 to x1, the work done is given by Negative because the work done is in the opposite direction of the motion of the body.

4. For equilibrium the net moment acting on the body by various forces is zero, in any of the body in the space, as in the space it is acted by various forces, like gravity.
a) True
b) False

Answer: a [Reason:] The equilibrium is only attained if the net moment on the body tends to be equal to zero. Thus the moments caused by different forces cancels out. If this happens there is no motion of the body along any direction and hence the body is said to be in equilibrium. The body here is a rigid body.

5. We use sometimes the measures to know the direction of moment in the calculations of the moments caused by the gravitational forces over the body above the ground level. Which one is right about it(consider the mentioned axis to be positive)?
a) Thumb is z-axis, fingers curled from x-axis to y-axis
b) Thumb is x-axis, fingers curled from z-axis to y-axis
c) Thumb is y-axis, fingers curled from x-axis to z-axis
d) Thumb is z-axis, fingers curled from y-axis to x-axis

Answer: a [Reason:] As right handed coordinate system means that you are curling your fingers from positive x-axis towards y-axis and the thumb which is projected is pointed to the positive z-axis. Thus visualizing the same and knowing the basic members of axis will not create much problem.

6. Determine the magnitude of the force F = 300j parallel to the direction of AB?

a) 155N
b) 257.1N
c) 200N
d) 175N

Answer: b [Reason:] Force component in the direction parallel to the AB is given by unit vector 0.286i + 0.857j + 0.429k. Now (300j).(0.286i + 0.857j + 0.429k) = 257.1N. Just try to resolve the force into it’s particular components.

7. Which statement is correct about the vector F in the showing the gravitational nature of the forces?
a) F= Fcos β + Fcos α + Fcosγ
b) F= Fsin β + Fcos α + Fcosγ
c) F= Fcos β + Fsin α + Fcosγ
d) F= Fcos β + Fcos α + Fsinγ

Answer: a [Reason:] As we know the α, β and γ are the angles made by the x, y and z-axis respectively. Thus, is the magnitude of the vector is F, the F= Fcos β + Fcos α + Fcosγ. Which means the force is the resultant of all its axis’ components.

8. What is the value of θ in the figure given below ?

a) 69˚
b) 60˚
c) 55˚
d) 90˚

Answer: a [Reason:] Fab= -(3/5)j + (4/5)k ; F= (4/5)i – (3/5)j ; θ = cos-1(Fab.F) = 69˚. This is the application of the triangle over the figure. Try to resolve the components of the given force. It will be easy.

9. Air also resist the body in the motion. The maximum value of the frictional force is called _________
a) Limiting Friction
b) Non-Limiting Friction
c) Limiting Action Friction
d) Non-Limiting Action Friction

Answer: a [Reason:] The friction is the phenomena that defines that there is a resistance which is present there between the two surfaces. The two surfaces are in contact and the friction applies at that surface only, resisting the motion of the surface. Thus the maximum values is called as limiting friction.

10. The coefficient of kinetic friction when a body is moving from a height above the ground level to a smaller height is ____________ than coefficient of static friction.
a) Smaller
b) Larger
c) Significantly larger
d) Highly larger

Answer: a [Reason:] The coefficient of kinetic friction is smaller than coefficient of static friction. The main thing about the kinetic one is that it is applied by the surface when the body is in motion. The static one is applied to the body when the body is static and is about to move.

11. Free body diagrams doesn’t play any role in making the calculations on the conditions of the equilibrium of the body when there is the involvement of the potential energy.
a) True
b) False

Answer: b [Reason:] The free body diagrams does play an important role in the formation of the conditions of the equilibrium of the rigid body. As the net forces are zero, the fbd helps us to take the measure of the same. That is to see whether the summation is really zero or not.

12. The net forces of acting on the body needs to be zero. This is also applicable for the bodies having some potential energy. This means that the work done over it is saved in the form of potential energy.
a) True
b) False

Answer: a [Reason:] The support reactions of the beam is also counted in the making of the forces zero. As far as the net force is concerned the support reaction does affect the conditions for the equilibrium of the body. Hence one needs to take care of the support reactions of the beam too.

13. Virtual Work done is saved in the form of potential energy. It is best given by ___________
a) δU = Fδrcosθ
b) δU = Fδrsinθ
c) δU = Fδrcotθ
d) δU = Fδrδθ

Answer: a [Reason:] The work is defined as the dot product of the δ force and the δ distance. This means that the work done does depends upon the angle of the force. That is the angle which is being made by the force vector to the surface of action of the force.

14. Which of the following is correct for the forces which are considered in the potential energy calculations?
a) The application of the conditions of the equilibrium of the body is valid only if the forces are collinear
b) The application of the conditions of the equilibrium of the body is valid only if the forces are parallel
c) The application of the conditions of the equilibrium of the body is valid only if the forces are perpendicular
d) The application of the conditions of the equilibrium of the body is valid throughout

Answer: d [Reason:] The application of the conditions of the equilibrium of the body is valid throughout. This means that the conditions are irrespective of the types of forces. The conditions are the basic rules that defines the equilibrium of the body and thus are applicable in any type of forces of the real axis.

15. For the conditions of the equilibrium of the body, i.e. the rigid body only the external forces defines the equilibrium. And the support reactions only cancels out the rotation part of the body and thus the potential energy of the body is stored in it.
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 application of the support reaction forces does affect the conditions of the equilibrium of the body. Not only the external but the support reaction forces that are developed by the sake of external forces does develop a tending effect on the equilibrium of the body. Thus the support reaction forces also cancels the external forces.

## Set 3

1. 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 making of the equilibrium equations 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. The forces here can be normal forces or the loading forces.

2. The net force of the body is zero that means the force are not being applied to the body at all and hence the body is in equilibrium.
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 force of the body is zero that doesn’t means that the force are not being applied to the body at all and hence the body is in equilibrium. The equilibrium is only attained if the net force on the body tends to be equal to zero. Thus the forces cancels out. If this happens there is no motion of the body along any direction and hence the body is said to be in equilibrium.

3. In the simplification of the forces for the free body diagram the net force acts at the ___________ of the loading body.
a) Centroid
b) The centre axis
c) The corner
d) The base

4. For a vector F, Fcosβ is equal to zero. What does this refer?
a) X-axis component is zero
b) Y-axis component is zero
c) Z-axis component is zero
d) β = 180˚

Answer: b [Reason:] As we know the α, β and γ are the angles made by the x, y and z-axis respectively. Thus y-axis component is zero, or β = 90˚. And thus if the angle is giving component to be zero this means the vector in that particular axis is perpendicular to that axis. This concept is used when the wedge is involved in the motion.

5. As in the determination of the various unknown forces in the free body diagram involves the use of vectors. So one of the vector law is commutative law and it is valid for the cross product of two vectors. (Commutative law: PxQ = QxP; for two vectors P and Q)
a) True
b) False

Answer: b [Reason:] This statement is wrong. It is not possible, unless we apply a negative sign to the RHS of the equation. That is PxQ = -(QxP). It is because, if you curl your wrist from one vector towards another vector, the thumb projected will give the direction of the cross product. Thus if you reverse the direction, negative sign is necessary.

6. As the free body diagram of the problem statement involves the use of forces. So in them the ___________ forces do not cause the rotation.
a) Non-concurrent
b) Concurrent
c) Parallel
d) Non-Parallel

Answer: b [Reason:] The concurrent forces are the one 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.

7. For equilibrium the normal forces acts in which direction in the free body diagrams if they are constructed for the friction part calculations?
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.

8. Which of the following is the assumption on the property of the surface on which block is present in case we are dealing with the calculations involving the friction?
a) Non-rigid
b) Coloured red
c) Having density in negative
d) The motion of that block is always horizontal

Answer: a [Reason:] The block used to explain the theory of friction is a used in a deformable surface. As we know that the theory of friction says that when there is no lubricating fluid present between the surfaces in contact, the dry friction occurs. Thus to show the same experiments are done over a non-rigid/deformable surface.

9. Calculate the Normal force developed between the body and the surface.

a) 611N
b) 116N
c) 100N
d) 180N

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.

10. There are three types of friction problems. One of them is _____________
a) Impending motion at a single point of contact
b) No-Impending motion at a single point of contact
c) Impending motion at a all points of contact
d) Apparent Impending motion

Answer: c [Reason:] There are mainly three types of the friction problems. They are no-apparent impending motion, impending motion at all points of contact and impending motion at some points of contacts. Therefore the answer is impending motion at all points of contact. There is involvement of various types of calculations which is there in all of the three types.

11. The phenomena of horizontal pull and push explains something related to the friction, among the following what is it?
a) Theory of friction
b) Theory of relativity
c) Theory of action
d) Theory of forces

Answer: a [Reason:] As we know that when there is no lubricating fluid present between the surfaces in contact, the dry friction occurs. This is the phenomena that defines that there is a resistance which is present there between the two surfaces. Thus the pull and push phenomena explains the theory of friction.

12. There are main two types of forces which are being stated in the free body diagram, and they are generally the resultant of various 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.

13. The frictional force is directly proportional to the surface of the solid.
a) True
b) False

Answer: b [Reason:] The frictional force is directly proportional to the vertical forces that is being applied normal to the surface of the body. The force of friction is not dependent on the type of the surface. Thus the only thing the frictional force does depend is the normal force.

14. Calculate the frictional force developed between the body and the surface.

a) 160N
b) 16N
c) 10N
d) 180N

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.

15. The frictional force developed always acts ____________ to the surface of the application of the friction.
a) Tangential
b) Perpendicular
c) Parallel
d) Normal

Answer: a [Reason:] The friction is the phenomena that defines that there is a resistance which is present there between the two surfaces. This friction is applied tangentially to the surfaces in contact. Thus the main thing is that the forces on both of the surfaces act tangential to each other.

## Set 4

1. 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.

2. Determine the product of inertia for the triangle in the given figure.

a) (bh)2/8
b) (bh)2/16
c) (bh)2/2
d) (bh)2/4

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.

3. 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 and after these 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.

4. There is perpendicular axis theorem for the area, and it is can be used to determine the product of moment of inertia.
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.

5. What is parallel axis theorem and to whom it is applied so that it can give the product of inertia of an area?
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.

6. The product of moment of inertia is the sum of _____________ and _________________
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.

7. The distance in the parallel axis theorem for the use in the determination of the product of moment of inertia 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.

8. One of the use of the centre of mass or centroid is as in the determination of the product of 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.

9. The product of Inertia for an area is helpful so as to _____________
a) Determine Minimum moments of inertia for an Area
b) Determine Maximum moments of inertia for a Line
c) Determine Minimum 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.

10. Determine the del product of inertia for the triangle in the given figure used for the calculations.

a) dIxy=(h2/2b2)x3dx
b) dIxy=(h/2b2)x3dx
c) dIxy=(h2/2)x3dx
d) dIxy=(h2/b2)x3dx

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.

11. If the non-Uniform loading is of the type of parabola then for calculating the product of 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.

12. If any external force also is applied on the structure and we are determining the product of 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.

13. The body is sometimes acted by two or three force members and we need to find the product of 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.

14. Whenever the distributed loading acts perpendicular to an area its intensity varies __________ for the determination of the product of moment of inertia.
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.

15. The calculation of the product of 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.

## Set 5

1. Free body diagrams doesn’t play any role in making the calculations on the conditions of the equilibrium of the body.
a) True
b) False

Answer: b [Reason:] The free body diagrams does play an important role in the formation of the conditions of the equilibrium of the rigid body. As the net forces are zero, the fbd helps us to take the measure of the same. That is to see whether the summation is really zero or not.

2. The net forces of acting on the body needs to be zero. This is also applicable for the simply supported beams. This means that the support reaction are also counted in making the net force zero.
a) True
b) False

Answer: a [Reason:] The support reactions of the beam is also counted in the making of the forces zero. As far as the net force is concerned the support reaction does affect the conditions for the equilibrium of the body. Hence one needs to take care of the support reactions of the beam too.

3. Determine the horizontal components of the reaction on the beam caused by the pin at Q.

a) 268N
b) 68N
c) 28N
d) 288N

Answer: a [Reason:] The summation of the forces needs to be zero. So does the summation of the moments need to zero. But talking about the angles, they not needed to zero. But the forces which are acting at particular angles, must needed to be equal to zero. The basic need of the forces to make the body in equilibrium.

4. If the body is in equilibrium, but it having a rotational curled ray shown in the free body diagram then:
a) The diagram is wrong
b) Such rotations can’t be shown in the fbds(free body diagrams)
c) The ray shown may be correct, but the body is not said to be in equilibrium
d) The body is said to be in equilibrium only, as the other forces will cancel out that rotation

Answer: d [Reason:] The body having equilibrium will not rotate at any cost. Yes, the diagram may contain the rotational array showing the couple being acted over the structure. But the thing is that the forces, i.e. the other forces which are outside the dependency of this rotation will cancel out this rotation and thus the body is in equilibrium.

5. If five forces are acting on the single particle and having an angle of 72˚ between each and are collinear, then:
a) The net force acting on the body is zero
b) The net force acting on the body is horizontal
c) The net force acting on the body is vertical
d) The net force acting on the body is at an angle of 45

Answer: a [Reason:] The net force acting on the body is zero. This means that the forces cancel out. This means that the body is in equilibrium and doesn’t need any of the external force to make itself in the equilibrium.

6. Which of the following is correct?
a) There is only one type of support for the beams
b) There are only two types of support for the beams
c) There are only three types of support for the beams
d) There are various types of support for the beams and they are countless

Answer: d [Reason:] The support is the thing which a human decides. That can be anything. That is the support can be given in any way. The type of support also is important as different types of supports have different conditions for the equilibrium.

7. What does the Newton’s second 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: a [Reason:] The requirement of the second law is important in the equilibrium of the body. Specially the rigid bodies. The rigid body particles are if are in motion then this law helps in making the conditions for the equilibrium of the body. As when the body will leave its state of motion and come in rest, etc.

8. Determine the vertical components of the reaction on the beam caused by the pin at Q.

a) 286N
b) 68N
c) 28N
d) 288N

Answer: a [Reason:] The summation of the forces needs to be zero. So does the summation of the moments need to zero. But talking about the angles, they not needed to zero. But the forces which are acting at particular angles, must needed to be equal to zero. The basic need of the forces to make the body in equilibrium.

9. Which of the following is correct?
a) The application of the conditions of the equilibrium of the body is valid only if the forces are collinear
b) The application of the conditions of the equilibrium of the body is valid only if the forces are parallel
c) The application of the conditions of the equilibrium of the body is valid only if the forces are perpendicular
d) The application of the conditions of the equilibrium of the body is valid throughout

Answer: d [Reason:] The application of the conditions of the equilibrium of the body is valid throughout. This means that the conditions are irrespective of the types of forces. The conditions are the basic rules that defines the equilibrium of the body and thus are applicable in any type of forces of the real axis.

10. Determine the horizontal components of the reaction on the beam caused by the pin at Q. The force 60N is multiplied by 10 and then is applied.

a) 0N
b) 445N
c) 45N
d) 40N

Answer: a [Reason:] The summation of the forces needs to be zero. So does the summation of the moments need to zero. But talking about the angles, they not needed to zero. But the forces are acting at particular angles.

11. For the conditions of the equilibrium of the body, i.e. the rigid body only the external forces defines the equilibrium. And the support reactions only cancels out the rotation part of the body.
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 application of the support reaction forces does affect the conditions of the equilibrium of the body. Not only the external but the support reaction forces that are developed by the sake of external forces does develop a tending effect on the equilibrium of the body. Thus the support reaction forces also cancels the external forces.

12. Cantilever beams are always in equilibrium, whether you form the equilibrium equations or not.
a) True
b) False

Answer: b [Reason:] The thing is that the formation of the conditions only verifies that the body is in equilibrium or not. Thus if the cantilever is in equilibrium then also the condition needs to be applied. It will be wrong to justify the equilibrium only on the basis of structure.

13. Determine the horizontal components of the reaction on the beam caused by the pin at Q.

a) 268N
b) 68N
c) 28N
d) 288N

Answer: a [Reason:] The summation of the forces needs to be zero. So does the summation of the moments need to zero. But talking about the angles, they not needed to zero. But the forces which are acting at particular angles, must needed to be equal to zero. The basic need of the forces to make the body in equilibrium.

14. Determine the vertical components of the reaction on the beam caused by the pin at Q. The force 60N is multiplied by 10 and then is applied.

a) 319N
b) 445N
c) 45N
d) 40N