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# Multiple choice question for engineering

## Set 1

1. Express the vector in the Cartesian Form, if the angle made by it with y and z axis is 60˚ and 45˚ respectively. Also it make an angle of α with x-axis. The magnitude of the force is 200N.
a) 100i + 100j + 141.4k N
b) 100i – 100j + 141.4k N
c) 100i + 100j – 141.4k N
d) 100i – 100j – 141.4k N

Answer: a [Reason:] When you will resolve the vector in its x, y and z-axis components, you will get an equation containing cosα. After getting the correctly, you need to directly put that value in the previous equation of components.

2. In the figure given below, with F1 = 60j + 80k and F2= 50i -100j + 100k, what is the value of β? (β is the angle made by the resultant vector with the y-axis )

a) 76˚
b) 102˚
c) 19˚
d) 130˚

Answer: b [Reason:] Just add the two vectors in their Cartesian form only. After that you will get the resultant vector in the Cartesian form. After that just divide the y-axis component’s magnitude by the resultant of the vector. You will get cosβ.

3. In the figure given below, calculate the value of cosγ. (γ is the angle made by the resultant vector and the z-axis)

a) 0.866
b) -0.354
c) 0.540
d) 0.354

Answer: a [Reason:] Well, the question is just a tricky one. Just subtract 60 from 90, you will get the γ, which is 30. And then apply the cosine formula for getting the components of respective axis.

4. What is the magnitude of the vector, 12i – 8j – 24k?
a) 18
b) 28
c) 38
d) 48

Answer: b [Reason:] 122 + 82 + 242 = 282 . After you get this, the 28 is the magnitude. This is the using of distance formula between two Cartesian points. Considering the point (12, -8, -24) and then calculating the distance from the origin.

5. The coordinate of the Force vector AB is A (2, 0, 2) and B (-2, 3.46, 3). What are its directions?
a) -0.742i + 0.643j + 0.186k
b) 0.742i – 0.643j – 0.186k
c) -0.742i – 0.643j + 0.186k
d) -0.742i + 0.643j – 0.186k

Answer: a [Reason:] First subtract the coordinates of the A form B. And place the x, y and z-axis coordinate separated by i, j and k respectively. Like Ai + Bj + Ck , if A, B and C are the x, y and z-axis coordinates. You will get a Cartesian vector, just find its magnitude and divide it with the vector.

6. The coordinate of the Force vector AB is A (2, 0, 2) and B (-2, 3.46, 3). It has a magnitude of 750N. Which is the best Cartesian representation of the vector AB?
a) -557i + 482j + 139k N
b) -557i – 482j + 139k N
c) -557i + 482j – 139k N
d) 557i – 482j – 139k N

Answer: a [Reason:] First try to make the unit vector of AB. Then you will get the Cartesian form of unit vector, after that just multiply 750 scalar quantity with it. Which means you are first taking out the desired direction and then multiplying the scalar to get the desired vector in desired direction.

7. A force vector is along 4i – 4k direction and has a magnitude 100N and another force vector is along 4i +2j -4k and has a magnitude of 120N. What is the resultant of both forces?
a) 80i + 40j – 80k N
b) 80i – 40j – 80k N
c) 151i + 40j – 80k N
d) 151i+ 40j – 151k N

Answer: d [Reason:] Just make the unit vectors of the given directions of the vectors and then multiply them with their respective magnitudes, after than you will get two vectors in the Cartesian form. Just add them.

8. The value cos-1(-3/7) + cos-1(2/7) + cos-1(6/7) is ____________
a) 215.4˚
b) 273.4˚
c) 188.4˚
d) 219.4˚

Answer: d [Reason:] It’s a calculative question, but still can be solved if you construct the triangle of the respective cosine inverses given, then convert them in the tangent form then use log table, if necessary. Or use of calculator is also good if permissible. Or you can make use of triangle laws to obtain the same.

9. Two vectors emerging from a point are always in a single plane.
a) True
b) False

Answer: a [Reason:] Yes, whenever two vectors are there, they are always in a single plane. If the emerging vectors are more than three then saying about their plane is somewhat difficult. But for two vectors it seems easy to understand about their plane.

10. Three vectors emerging from a point are always in a single plane.
a) True
b) False

Answer: b [Reason:] Yes, whenever two vectors are there, they are always in a single plane. But you cannot say about the three vectors, they may or may be. Because if the emerging vectors are more than three then saying about their plane is somewhat difficult. But for two vectors it seems easy to understand about their plane.

11. What is the magnitude of unit vector?
a) √3
b) √2
c) 1
d) 0

Answer: c [Reason:] The magnitude of the unit vector is always = 1. This means unit vector is the vector in the direction of the original vector and the magnitude of the same is unity. And unit vector is always in the direction of its vector.

12. What is the difference between position vector and unit vector?
a) Position vector has magnitude = 1 and direction, while unit vector has magnitude = 0 and no direction
b) Position vector has magnitude = 0 and direction, while unit vector has magnitude = 0 and no direction
c) Position vector has some magnitude and direction, while unit vector has magnitude = 0 and no direction
d) Position vector has some magnitude and direction, while unit vector has magnitude = 1 and a specified direction

Answer: d [Reason:] As position vector defines the position of the point in the Cartesian plane wit respect of the origin, it will definitely have a direction and also a magnitude. Unit vector is a vector whose magnitude is = 1 and is having a specific direction.

13. What if we multiply a scalar to the unit vector?
a) The direction will change accordingly
b) The magnitude will change accordingly
c) The magnitude will not change accordingly
d) The direction will change by a factor of square root of the scalar

Answer: b [Reason:] The magnitude of the unit vector will change by the amount of the scalar multiplied. Not the direction. Because the direction is being fixed after the application of the unit vector’s calculation. That is the vector divided by its magnitude.

14. Which is true for the vector provided only position coordinates given?
a) (Final position coordinates + initial positions coordinates) gives the vector form of the vector
b) (Final position coordinates – initial positions coordinates) gives the vector form of the vector
c) (Initial positions coordinates – Final position coordinates) gives the vector form of the vector
d) (Initial positions coordinates + Final position coordinates) gives the vector form of the vector

Answer: b [Reason:] We find the vector form from subtracting the initial position coordinates from final position coordinates. Not from subtracting the final position coordinates from initial position coordinates.

15. A vector can always have_____________
a) Only one component along any of the axis
b) Only two components along any of the axis
c) Only three components along any of the axis
d) A unit vector along the direction perpendicular to its direction

Answer: c [Reason:] All vectors have three components, whether they are zero or not. And unit vector is always in the direction of its vector. And the unit vector has the magnitude equal to unity. This means unit vector is the vector in the direction of the original vector and the magnitude of the same is unity.

## Set 2

1. ____________ 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: 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.

2. 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: 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.

3. Vector shown in the figure below have 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 respectively

Answer: a [Reason:] The sine and the cosine components of the given vectors considering the angle B as the only angle of consideration comes 1.5m and 2.59m.

4. 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: a [Reason:] For solving of the unknown tension in the belts, T2 = T1eµB equation is used. In this the R.H.S tension is the maximum tension of the two tensions. While the other one is the smaller one. And the µ is coefficient of friction between the belt and the surface. And the B is the angle of belt to the surface in radians.

5. Dry friction in the belt is also called ___________
a) Column Friction
b) Coulomb Friction
c) Dry column friction
d) Surface friction

Answer: b [Reason:] The dry friction is acted upon the surfaces. Whatever may be the surfaces. And they are tangential to each other. As we know the friction is the phenomena that defines that there is a resistance which is present there between the two surfaces. The dry friction is also termed as the Coulomb friction as it was given by C.A. Coulomb.

6. 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: 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.

7. 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: 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.

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: 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.

9. We show the net forces acting on the belts 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.

10. There are main 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: 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.

11. 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) 70.7N

Answer: a [Reason:] The component of the force in the v axis, it is equal to 100cos(15˚). This is the application of the triangle over the figure. Try to resolve the components of the given force. It will be easy. This is one of the simplest example of solving for the forces acting over the belts.

12. 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 amount of that scalar’s magnitude
d) You cannot multiply the vector with a scalar

Answer: c [Reason:] If a vector is multiplied by a scalar then its magnitude is increased by amount of that scalar’s magnitude. When multiplied by a negative scalar it going to change the directional sense of the vector. Vector math is unchanged throughout.

13. All the vector quantities in the solving of the unknown in the belt force system obey _____________
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: a [Reason:] All the vectors quantities obey parallelogram law of addition. Two vectors A and B (can be called as component vectors) are added to form a resultant vector. R = A+B.

14. 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: a [Reason:] For solving of the unknown tension in the belts, T2 = T1eµB equation is used. In this the R.H.S tension is the maximum tension of the two tensions. While the other one is the smaller one. And the µ is coefficient of friction between the belt and the surface.

15. 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) 1.5m

Answer: a [Reason:] After getting the cosine components of the given vectors we obtain the total length of the x-axis components to be 3cos60 + 2cos30 = 3.23.

## Set 3

1. The journal bearings are generally used in __________
a) Belts
b) Columns
c) Beams
d) Supports

Answer: d [Reason:] The journal bearings are being used in the supports. They are used as to support the loadings. This means that the more the bearings the more is the stable the structure. Thus the use of journal bearings.

2. A phenomena is there in these collar bearings and pivots, which is very helpful in the supporting of the machines, and that is _____________
a) Column Friction
b) Coulomb Friction
c) Dry column friction
d) Surface friction

Answer: b [Reason:] The dry friction is acted upon the surfaces. And they are tangential to each other. As we know the friction is the phenomena that defines that there is a resistance which is present there between the two surfaces. This Coulomb friction is also known as dry friction.

3. Determine the moment about the point Q by the force shown as 400N.

a) -98.6kN
b) 98.6kN
c) -98.6iN
d) -98.6jN

Answer: a [Reason:] As we know that the moment is the cross product of the force and the distance between the point of contact of the force and the point about which moment needs to be calculated. Thus forming the distance vector and then crossing it with the force will give us the answer. Remember force also needs to be in the vector form for doing the cross product.

4. At what conditions does the Coulomb friction occurs between the surfaces in contact of journal bearing?
a) When there is no lubricating fluid
b) When there is no friction fluid
c) When there is no adhesive fluid
d) When there is no cohesive fluid

Answer: a [Reason:] When there is no lubricating fluid present between the surfaces in contact, the dry friction occurs. Thus as the name suggests, dry friction. When there is no liquid present over there the surfaces are going to be said dry only. Thus the term dry friction.

5. We first make equilibrium equations of the sections involving the journal bearings 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 of journal bearings 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.

6. To know the rotations involved in the journal bearings we use right handed coordinate system, in that system which of the following is followed (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.

7. The moment is the cross product of which two vectors in the calculations of the journal bearings?

Answer: b [Reason:] The cross product needs to take in the proper sequence. If not taken then the answer is just the opposite of the true answer. That’s why, the answer is not the Force and Radius vectors, but the Radius and Force vectors. Because the moment has its direction, as many of the cross products have, and thus precaution needs to be taken.

8. Determine the moment about the point P.

a) 460Nm
b) 500Nm
c) 705Nm
d) 0Nm

Answer: a [Reason:] As we know that the moment is the cross product of the force and the distance between the point of contact of the force and the point about which moment needs to be calculated. Thus forming the distance vector and then crossing it with the force will give us the answer. Remember force also needs to be in the vector form for doing the cross product.

9. The basic type of motion of a body is not the translation motion only.
a) True
b) False

Answer: a [Reason:] The basic types of the motion of the body are translation and rotational. In which translation means the motion is a straight motion while the in the rotational motion the body is moving of the body along the axis. The pivots do circular motion.

10. Sometime the principle of transmissibility is somehow applied to journal bearings. It states that:
a) It states that the force acting on the body is a sliding vector
b) It states that the force acting on the body is a rolling vector
c) It states that the force acting on the body is a wedging vector
d) It states that the force acting on the body is a unit vector

Answer: a [Reason:] Principle of transmissibility states that the force acting on the body is a sliding vector. That is it can be applied at any point of the body. It will give the same effect as if applied at any point other that the specific point. Thus the answer.

11. Coplanar forces are not easily simplified in the simplification of the force and couple system in the calculations of forces in the journal bearings.
a) True
b) False

Answer: b [Reason:] The coplanar forces can be easily simplified in the simplification of the force and couple system. This is not only for the coplanar system of the forces, but for any of the dimension vector. Which means the answer is false.

12. There are some of the applications of the two force and three force members in the journal bearings. 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.

13. Which of the following is true for the determination of the rotational effect of the bearings?
a) If we are considering the clockwise direction to be positive then the rotation along the counter clockwise direction is negative
b) If we are considering the clockwise direction to be negative then the rotation along the counter clockwise direction is also negative as no such assumptions can be taken
c) If we are considering the clockwise direction to be positive then the rotation along the counter clockwise direction is also positive as no such assumptions can be taken
d) If we are considering the clockwise direction to be negative then the rotation along the counter clockwise direction is neither positive nor negative as we only pre assumed the clockwise direction

Answer: a [Reason:] We can take such assumptions. We are free to do so. We just need to specify if clockwise is positive then the counter clockwise is to be negative. Also such assumption are very useful in 2D calculation of the moments caused by various forces on the body.

14. The doors are having hinges to support them. This hinge is too a type of bearing or is a
Device which is using effect like bearing to support. So, if you are opening the door, then the force you apply is greater if you open it from the part closer to the hinges.
a) True
b) False

Answer: a [Reason:] As the moment is the product of the force and the perpendicular distance of the point of action and the axis, thus more the distance more will be the moment. And hence to overcome that you have to apply more energy. Thus the answer.

15. Which of the following is true for the bearing’s body?
a) Total moment of various forces acting on the body is the vector sum of all moments in 3D
b) Total moment of various forces acting on the body is the algebraic sum of all moments in 3D
c) Total moment of various forces acting on the body is always zero in any dimension
d) Total moment of various forces acting on the body is the vector sum of all moments which is perpendicular to each other forces whatever be the dimensions

Answer: a [Reason:] When we consider about the dimensions we need to be careful. 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.

## Set 4

1. In general cases the screws are used as __________
a) Fasteners
b) Main components
c) Simpler
d) Upstreamers

Answer: a [Reason:] The screws are generally used as the fasteners. The main thing about the screws is that the friction helps the holding of the structures. Also screws help make the structure join fast and efficiently. Thus the name fasteners.

2. Which type of the screws are used in the machines?

Answer: a [Reason:] The square threaded screws are used in the machines. As the screws are generally used as the fasteners. The main thing about the screws is that the friction helps the holding of the structures. Also screws help make the structure join fast and efficiently.

3. Why are square threaded screws are used in the machines?
a) Large forces act along the axis of the screws
b) Small forces act along the axis of the screws
c) Large forces act perpendicular the axis of the screws
d) Small forces act perpendicular the axis of the screws

Answer: a Answer: The square threaded screws are used in the machines. As these screws are generally used as the fasteners. The main thing about these screws is that the friction helps the holding of the structures. Also screws help make the structure join fast and efficiently. And in these screws forces act along the axis of screws.

4. The angle of the threading in the screw is determined by which of the following trigonometric function?
a) Tangent Inverse
b) Sine
c) Cosine
d) Secant

Answer: a [Reason:] The angle of the threading in the screw is determined by inverse tangent trigonometric function. This is the generalised form of the function which is being used so as to get the value of the angle. In wedges the same function is used so as to find the angle of wedge.

5. In the determination of the angle of the screw, the ratio kept in the inverse tangent function is the ratio of ________
a) Vertical and Horizontal distance of the screw
b) Horizontal and Vertical distance of the screw
c) Vertical and Linear distance of the screw
d) Linear and Horizontal distance of the screw

Answer: a [Reason:] The angle of the threading in the screw is determined by inverse tangent trigonometric function. This is the generalised form of the function which is being used so as to get the value of the angle. In wedges the same function is used so as to find the angle of wedge. And in both the case the ratio is the vertical is to horizontal.

6. The horizontal distance used in the inverse trigonometry function is called as _____________
c) Major
d) Cut-off

Answer: a [Reason:] The angle of the threading in the screw is determined by inverse tangent trigonometric function. This is the generalised form of the function which is being used so as to get the value of the angle. In wedges the same function is used so as to find the angle of wedge. And thus the horizontal distance is called as lead.

7. In screws there is application of two and three force members. The basic difference between the two and the three force members used 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. Determine the moment of the force along point P (There is screw present at P, about which the rotation is there).

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.

9. When we apply the equilibrium equations in the free body diagram of the screw, for two force members to be in equilibrium then _________
a) They both must not be collinear
b) They must be of different magnitude
c) They must be in the same direction
d) They must be in the opposite direction cancelling each other

Answer: d [Reason:] Simply, for the equilibrium the net force must be zero. That is the forces must be collinear if equilibrium is to be established. And they must be in the opposite directions. They also must be of same magnitude, cancelling each other.

10. Which of the following statement for the three force system applied in the screws is true?
a) The moment need not to be seen for the equilibrium, they are always in equilibrium
b) The forces are not always in the equilibrium
c) The forces are always in equilibrium
d) The moments are always in equilibrium

Answer: b [Reason:] The forces in the three force system are not always in the equilibrium. The equilibrium is established when the forces cancels out each other. Also when the net moment is zero. Then only the equilibrium is established in the three force system.

11. Which of the following is correct for a screw being rotated?
a) The couple moment depends on the axis of rotation
b) The couple moment depends directly on the radius vector of forces
c) The couple moment depends only on the distance vector between the forces
d) The couple moment’s direction is given by the left hand rule

Answer: c [Reason:] The couple moment depends only on the distance vector between the forces. The radius vector of the forces are not making the couple depended over it. But the distance vector is driven by subtracting the radius vectors of the forces. This means that, moment of the couple doesn’t directly depend on the radius vector of forces.

12. If the forces acting on the couple acting on the screw are in the same direction, that is they are not in the opposite direction as always they are, then?
a) The direction of the forces doesn’t determine the moment
b) The couple moment will be maximum
c) The couple is not possible
d) No change occurs

Answer: c [Reason:] The forces are always in the opposite direction for the couple moment. But if they don’t act like the same that is they change their directions and align themselves in the same direction then the couple is not possible.

13. For equilibrium the net force acting on the screw is zero.
a) True
b) False

Answer: a [Reason:] The equilibrium is only attained if the net force on the body tends to be equal to zero. Thus the forces cancel 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.

14. In the screw collinear forces act, what is a collinear system of forces?
a) The force system having all the forces parallel to each other
b) The force system having all the forces perpendicular to each other
c) The force system having all the forces emerging from a single point
d) Forces cannot form a collinear system of forces, it is not possible

Answer: c [Reason:] The force system having all the forces emerging from a point is called the collinear system of force. This is a type of system of the force, which is easy in the simplification. This is because as the forces are the vector quantity, the vector math is applied and the simplification is done.

15. Determine the smallest force applied at R which creates the same moment about P as by 75N. (There is presence of screw at P, about which the rotation is being there)

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.

## Set 5

1. The determination of the internal loading 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.

a) Method of sections
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 method of section 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.

3. In method of sections we probably use the joints and take out the forces in the beams.
a) True
b) False

Answer: b [Reason:] In method of sections the determination of the forces is done over the trusses or the beams. It is not done on the joints. But yes the joints are used as the forces acting on the joints are used in the calculations. This is again used for the designing of the beams.

4. The ____________ forces are used for the making of the free body diagram of the beams so as to apply the method of sections.
a) Internal rotational
b) Couple rotational
c) Translational
d) External

Answer: d [Reason:] The free body diagram is using the external forces which are acting on the beam as the main purpose is to design the beam. Design so as to withstand the loads which are going to be added to the beams. Thus the loads which are being added externally are being used in the free body diagrams.

5. Rotational moments are not counted in the free body diagrams and the evaluation of the internal forces by the method of sections.
a) True
b) False

Answer: b [Reason:] In method of sections the determination of the forces is done over the trusses or the beams. It is not done on the joints. But yes the joints are used as the forces acting on the joints are used in the calculations. This is again used for the designing of the beams.

6. The force acting perpendicular to the beam is called _______________ force.
a) Perpendicular
b) Orthogonal
c) Normal
d) Distortion

Answer: c [Reason:] The forces which are acting perpendicular to the body of the beams are called as the normal forces. These are the forces which are supporting the beam. These forces are being calculated by the help of equilibrium equations. Thus the forces which are perpendicular to the surface are the normal forces.

7. The force acting parallel/tangential to the beam is called _______________ force.
a) Perpendicular
b) Orthogonal
c) Shear
d) Distortion

Answer: c [Reason:] The forces which are acting tangentially to the body of the beams are called as the shear forces. These are the forces which are giving moment to the beams. These forces are being calculated by the help of equilibrium equations. Thus the forces which are parallel to the surface are the shear forces.

8. The couple moment produced in the beams are called as ____________________
a) Moment of distance
b) Moment of line
c) Moment of bending
d) Bending moment

Answer: d [Reason:] The forces are making the body tend to rotate. This is in general making the body rotate or make generate the moment in the beams. This moment is the same thing as the couple moment. But in beams it is referred as bending moment. And is again used for the designing of beams.

9. The force components prevents the beam from which of the following process?
a) Relative translation
b) Relative rotation
c) Relative transformation
d) Relative collapse

Answer: a [Reason:] The force components are the forces which are the ones which prevents the beam from relative translation. This means that the shearing generated will be held strongly by the help of the forces generated. The main motto is to make the beam stable and don’t allow the relative translation occur.

10. The couple moment prevents the beam from which of the following process?
a) Relative translation
b) Relative rotation
c) Relative transformation
d) Relative collapse

Answer: b [Reason:] The couple moments are the moments which are the ones which prevents the beam from relative rotation. This means that the rotation generated will be held strongly by the help of the opposite moments generated. The main motto is to make the beam stable and don’t allow the relative rotation to occur.

11. According to the _________ law of Newton the loadings at the sections must act opposite to the forces applied.
a) First
b) Second
c) Third
d) Fourth

Answer: c [Reason:] As the law state that there is always a reaction to every action applied. Thus if the forces are applied at the section of the beams, i.e. the external forces than there must be reactive forces. And these reactive forces act in opposite directions of the forces which are being applied externally.

12. For the determination of the forces at the supports which part is to be considered in the calculations?
a) The one having one unknown and two known forces
b) The one having two unknown and two known forces
c) The one having three unknown and two known forces
d) The one having two unknown and one known forces

Answer: a [Reason:] The determination is usually done in this manner only. The selection is done as by selecting the one having one unknown and two known forces. This will ensure that the unknown value will be found. And also we have less number of unknowns which are needed to be founded.

13. What is not the condition for the equilibrium in the beams if considered a section on which the method of sections is applied?
a) ∑Fx=0
b) ∑Fy=0
c) ∑Fz=0
d) ∑F≠0

Answer: d [Reason:] For the equilibrium in beams by considering any of the 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. Thus the conditions for the equilibrium of the beams.

14. Determine the moment at the point R of the beam shown.

a) 225Nm
b) 25Nm
c) 22Nm
d) 23Nm