Class 11 KPK G11 Physics Chapter 2 Notes Vectors and Equilibrium conceptual questions, and Comprehensive Questions,
Conceptual Questions KPK G11 Physics Chapter 2
Q.1) Is it possible to add three vectors of equal magnitude but different directions to get a null vector? Illustrate with the diagram.
Answer:
Yes, if three vectors are making equilateral triangle their resultant will be a null vector because in this case the sum of two vectors will be equal in magnitude but opposite in direction to the third vector.
Consider three vectors A, B and C of equal magnitudes as shown in fig;
If we add vectors A and B, their resultant can be shown as;
From fig it can be seen that the resultant of A and B i.e. R is equal in magnitude but opposite in direction to C i.e. R = – C.Thus,
(A + B) + C = R + C ………….. eq (i)
As R = – C so eq (i) becomes;
A + B + C = – C + C = 0
where R = – C
Q.2) The magnitudes of three vectors are 2 m, 3 m, and 5 m, respectively. The directions are at your disposal. Can these vectors be added to yield zero? Illustrate with a diagram.
Answer:
Yes, these vectors can be added to yield zero.
Say A = 2 m, B = 3 m and C =5 m.
In order to yield zero the resultant of two vectors should be equal in magnitude but opposite in direction to the third vector. Thus, If the magnitudes of vectors A and B are added up, the magnitude of their resultant ‘R‘ is given by;
| R | = | A + B |
| R | = √ [( Rx)2 + (Ry)2] …………. (i)
where Rx = Ax + Bx …………. (ii)
and Ry = Ay +By …………. (iii)
Read more: Physics notes for class 11 KPK Board Chapter 1 | FSc Pre Engineering
From fig, vector A is along x-axis and vector B is making an angle θ with x- axis.
eq (ii) becomes;
Rx = A cosθ + B cosθ
= A cos0° + B cosθ
Rx = A + B cosθ
eq (iii) becomes;
Ry = A sinθ + B sinθ
= A sin0° + B sinθ
Ry = B sinθ
Thus eq (i) becomes;
⇒ | R | = √ [( A+B cosθ)2 + (B sinθ)2]
⇒ | R |= √ [( A)2 + (B)2 + 2AB cosθ]
for θ ≈ 10°
| R | = 5 m
which is equal in magnitude to that third vector C. In order to yield zero, the direction of R should be opposite to that of C i.e. R should be equal to – C as shown in fig;
Hence A + B + C = R + C
or A + B + C = – C + C
A + B + C = 0
Q.3) What units are associated with the unit vectors î, ĵ, and k̂ ?
Answer:
Unit vectors î, ĵ, and k̂ represent the x, y, z directions of vectors i.e. direction of vectors in Cartesian plane and are defined as;
Unit vector = Vector / Magnitude of that vector
For example, If a force F is applied on an object in x direction, its unit vector î is given by;
î = F / |F|
Since F and |F| have same units so î is unit less.
Q.4) Can a scalar product of two vectors be negative? Provide a proof and give an example.
Answer:
Yes, a scalar product of two vectors can be negative.
Scalar product of two vectors is defined as the product of magnitude of one vector and the component of the other vector in the direction of first vector.
Hence the scalar product of vectors A and B is;
A.B = ABcosθ
since cosθ is negative between 90° and 270° so the scalar product can be negative for these values of θ.
For Example, Consider two vectors A and B that are anti parallel to each other, the angle between them is 180°.
Hence,
A.B = ABcos(180° )
A.B = -AB
Q.5) A and B are two non-zero vectors. How can their scalar product be zero? And how can their vector product be zero?
Answer:
Scalar product of two vectors is defined as the product of magnitude of one vector and the component of the other vector in the direction of first vector.
Hence the scalar product of two non-zero vectors A and B is;
A.B = AB cosθ
If A and B are mutually perpendicular vectors, the angle between them is 90°. Thus,
A.B = AB cos90°
since cos90° = 0
⇒ A.B = 0
Vector product of two vectors is defined as the product of magnitude of one vector and the component of the other vector perpendicular of first vector.
Hence the vector product of two non-zero vectors A and B is;
A×B = AB sinθ
If two vectors are parallel or antiparallel to each other, the angle between them is θ = 0° or 180°. As sin0° = sin 180° = 0
⇒ A×B = 0
Q.6) Suppose you are given a known nonzero vector A. The scalar product of A with an unknown vector B is zero. Likewise, the vector product of A with B is zero. What can you conclude about B?
Answer:
Scalar product of vectors A and B is given by;
A.B = AB cosθ
Vector product of vectors A and B is given by;
A×B = AB sinθ
According to given conditions, both, scalar and vector products of these vectors are zero i.e. A.B = 0
and A×B = 0
Since A ≠ 0
⇒ B = 0 to give, both, scalar and vector products to be zero
i.e. B is a null vector
Q.7) Why a particle experiencing only one force cannot be in equilibrium?
Answer:
An object is said to be in complete equilibrium when sum of all forces and torques acting on it do not change its transnational as well as rotational motion. i.e the net force and net torque acting on it should be zero.
Fnet = 0 and τnet = 0
Thus if only one force is acting on the particle, the condition of net force and net torque to be zero cannot be fulfilled. Hence the particle cannot be in equilibrium .
Q.8) To open the door that has the handle on the right and the hinges on the left a torque must be applied. Is the torque clockwise or counterclockwise when viewed from the above? Does your answer depend on whether the door opens towards or away from you?
Answer:
Yes, it depends on whether the door opens towards or away from us.
If the door opens towards us, we need to pull the door handle i.e the direction of force is towards us. Thus if we align our fingers of right hand along ‘r’ (the distance between the hinges and the line of action of force) and curl them in the direction of ‘F’, our thumb points in the downward direction i.e. a clockwise torque is produced.
If the door opens away from us, we need to push the door handle i.e. the direction of force is away from us. Thus if we align our fingers of right hand along ‘r’ and curl them in the direction of ‘F’, our thumb points in the upward direction i.e. a counterclockwise torque is produced.
Q.9) Explain the warning ‘Never use a large wrench to tighten a small bolt’.
Answer:
For a large wrench the distance between the line of action of force and the axis of rotation will be large. So, even a small amount of force can produce a large amount of torque which may result in the breakage of bolt.
Q.10) A central force is one that is always directed toward the same point. Can a central force give rise to a torque about that point?
Answer:
No, a central force cannot give rise to torque because the line of action of force passes through the axis of rotation i.e. the angle between r and F is 0° . Hence the torque is zero
τ = r × F
τ = rFsinθ
= rFsin0°
= 0
Comprehensive Questions
Q.1) How are vectors added and subtracted geometrically?
Answer:
Addition of Vectors:
Vectors may be added geometrically by following these steps;
1. Draw the vectors to a common scale.
2. Now place them head to tail such that put the tail of the second vector on the head of the first vector according to selected scale in the given direction.
3. Then join the tail of the first vector with the head of the last which will give another vector which is the sum of these vectors called resultant vector.
Example:
Addition of two Vectors: Consider two vectors A and B, drawn to same scale making certain angles θA and θB with the x – axis respectively as shown in figure;
To add these vectors we redraw them to a common scale and place them head to tail as mentioned above. Such that the tail of vector B is on head of vector A. Joining the tail of the A with the head of the B will give another vector which is the sum of these vectors called resultant vector R as shown in figure below. The resultant will have the same effect as the combined effect of both vectors.
These can be added by head to tail rule as before; i.e. placing them head to tail such that tail of each vector is on the head of previous vector. Then draw the resultant vector by joining the tail of first vector to the head of last one. such that;
R = A + B + C + D
Subtraction of vectors:
Subtraction of one vector from another vector means addition of the negative of the vector with the first.
Example:
If vector B is to be subtracted from vector A, first find the negative of vector B (which is –B). Then follow the rules of vector addition to get the resultant R as shown in Figure;
Q.2) If a vector is multiplied to a positive scalar, how is the result related to the original vector? What if the scalar is zero? Negative?
Answer:
If a vector A is multiplied to a positive scalar ‘n’, the length of the vector becomes n times the length of vector A. While the direction of vector remains unchanged. i.e.
If n > 0 then +n•A = nA.
Example:
Consider vector ‘A‘ as shown in fig 1
When this vector ‘A‘ is multiplied by a scalar n= 2, the length of the vector increases two times and the direction remains the same as shown in fig 2;
If the scalar n is zero, the vector multiplied to it becomes a null vector. i.e.
If n= 0 then nA = 0.
If a vector A is multiplied to a negative scalar ‘-n’, the length of the vector is n times the length of vector A. While the direction of vector reverses. i.e.
If n< 0 then -n•A = -nA.
Example:
When the vector A shown in fig 1 is multiplied by a scalar n = – 2, the length of the vector increases two times and the direction reverses.
Q.3) What are rectangular components of a vector? How rectangular components are used to represent a vector?
Answer:
Rectangular Components of a Vector:
If a vector is split or resolved into two or more components that are perpendicular to each other, such components are known as rectangular components of that vector.
Consider vector A as shown in fig. The perpendicular vectors Ax and Ay are the rectangular components of A.
A = Ax + Ay
or A = Ax î + Ay ĵ ……………… (i)
Rectangular Components to Represent a Vector:
From fig 1 consider a triangle OPQ, without considering the sides as vectors, as shown in fig 2. This forms a right angle triangle OPQ for which we have.
cosθ = Base / Hyp = Ax / A
⇒ Ax = A cosθ …………….. (ii)
and
sinθ = Perp / Hyp = Ay / A
⇒ Ay = A sinθ …………….. (iii)
equation (ii) and equation (iii) are used to represent the components in terms of its vector. Putting equation (ii) and equation (iii) in equation (i) we get
A = A cosθ î + A sinθ ĵ …………….. (iv)
From a right angle triangle Δ OPQ, using Pythagoras theorem
(hyp)2 = (base)2 + (perp)2
√ hyp =√ [ (base)2 + (perp)2]
or hyp = √ [ (base)2 + (perp)2]
therefore
|A| = √ (Ax2 + Ay2) …………….. ( v)
The magnitude of vector can now be determined from eq (v) if the values of the magnitudes of components are known.
Also to determine the direction in right angle triangleΔOPQ, we have
tanθ = perp / base = Ay / Ax
and θ = tan-1 ( Ay / Ax ) ………………. (vi)
In three-dimensional space vector A can be written as
|A| = √ (Ax2 + Ay2 + Az2) ………………… (vii)
Hence the rectangular components are used to represent a vector.
Q.4) Explain addition of vectors by rectangular components.
Answer:
Addition of Vectors by Rectangular Components:
The basic rule for addition of vectors is, by definition, the head-to-tail rule. But sometimes it gets difficult to sketch the vectors according to head-to-tail rule. Another method of addition of vectors is called analytical method also known as rectangular component method.
“Rectangular components of a vector are effective values of vector in all three directions.”
And similarly,
Ry = Ay + By
So for any number of coplanar vectors, we can write
Rx = Ax + Bx + Cx + Dx+…
Similarly,
Ry = Ay + By + Cy + Dy + …
So magnitude of ‘R’ is given by,
|R| = √ (Rx2 + Ry2)
Putting values of Rx and Ry, we get
Direction of R:
Direction of R is given by,
Q.8) What is mechanical equilibrium? Explain different types of equilibrium.
Answer:
Mechanical equilibrium:
The state of a body in which under the action of several forces and torques acting together there is no change in translational motion as well as rotational motion is called equilibrium.
In other words the acceleration of such object is zero.
Types of Equilibrium:
Equilibrium is of two types;
- Static Equilibrium
- Dynamic Equilibrium
Dynamic equilibrium is further divided into;
- Dynamic Translational Equilibrium
- Dynamic Rotational Equilibrium
Static equilibrium:
When a body is at rest under the action of several forces acting together the body is said to be in static equilibrium.
For example, a book resting on the table is in static equilibrium; the weight “mg” of the book is balanced by a normal reaction force from the table surface.
Dynamic equilibrium:
When a body is moving at uniform velocity under the action of several forces acting together the body is said to be in dynamic equilibrium.
Dynamic Translational Equilibrium:
When a body is moving with uniform linear velocity the body is said to be in dynamic translational equilibrium.
For example a paratrooper falling down with constant velocity is in dynamic translational equilibrium.
Dynamic Rotational Equilibrium:
When a body is moving with uniform angular velocity the body is said to be in dynamic rotational equilibrium.
For example, a compact disk (CD) rotating in CD Player with constant angular velocity is in dynamic rotational equilibrium.
Q.9) What type of equilibrium is guaranteed by each condition of equilibrium?
Answer:
According to first condition of equilirium the vector sum of all the forces acting on the body should be ZERO.
i.e. Fnet = 0 anet = 0
SO, when the first condition is satisfied this means that there is no net force acting on the body, so it will represent translational equilibrium only.
According to second condition of equilibrium the vector sum of all the torques acting on the body should be ZERO.
i.e. τnet = 0 αnet = 0
So, when the second condition is satisfied this means that there is no net torque acting on the body, so it will represent rotational equilibrium only.
For complete equilibrium both the first and second conditions of equilibrium must be satisfied.
