Units and Dimensions

There are 3 commonly used units utilized in various systems. These 3 units are:

M.K.S. system of units (Metre-Kilogram-Second)

As the initials suggest, this system expresses mass in kilogram, length in meter and time in seconds. With the help of this unit expression, we can see the expression of unit of force depicted as kgf and isexpressed as kilogram force.

C.G.S. system of units (Centimetre-Gram Second)

In this system too, according to the suggested initials, depiction of mass is in gram, length in centimeter and time in seconds. With the help of its unit expression, we can see its expression of unit force as dyne. As per its definition, dyne is the force that acts on one gram mass and produces an acceleration of 1cm / sec ².

S.I. system of units (Systeme International)

Systeme International or S. I. unit is also referred as International System of units.The unit depiction of this system is done with mass and time. In this case too, time is depicted with seconds, and meter mass is represented with kilogram. Unit force of this system is illustrated with Newton and in equational term is showed with the help of N.

According to S.I system, any force acting on a certain mass of 1 kilogram, whose outcome is an acceleration of 1 m / s ² is known as Newton.

In fact on proceeding further with the relation of S.I. system of units and C.G.S. system of units, it is found that,

1 N = mass x 1 m / s ²

      = 1 kg x 1 m / s ²

This on equating further will give us,

= 1000 gm. x 100 cm / s²

= 1000 x 100 x gm. x cm / s²

= 10⁵ dyne

Therefore,

1 Newton = 10⁵ dyne

Now, in such instances where magnitude of force is huge, utilization of unit of force is described in mega newton or kilo newton.

On converting Kilo newton to newton, its value comes to be,

1 kilo newton = 10 ³ newton

Representation of kilo newton is kN.

After finding out the value of kilo newton, value of mega newton is,

1 mega newton = 10 ⁵ newton

As it is stated above, that magnitude of forces of large quantities have depiction in greater value, all such quantities are represented in,

• Tera
• Kilo
• Giga
• Mega

In case of smaller quantities, they are represented as,

• Pico
• Nano
• Micro
• Milli

Representation and valuation of smaller quantities

• Pico = p = 10 ^-12
• Nano = n = 10 ^-9
• Micro = µ = 10 ^-6
• Milli = m = 10 ^-3
• Depiction of the above values with Newton

Depiction of the above values with Newton

• Pico = pN = 10 ^-12N
• Nano = nN = 10 ^-9N
• Micro = µN = 10 ^-6N
• Milli = mN = 10 ^-3N

Representation and valuation of these huge quantities

• Tera = T = 10 ¹²
• Kilo = K = 10 ³
• Giga = G = 10 ⁹
• Mega = M = 10 ⁶

Depiction of the above values with Newton

• Tera = TN = 10 ¹² N
• Kilo = KN = 10 ³ N
• Giga = GN = 10 ⁹ N
• Mega = MN = 10 ⁶ N

Relation between Newton (N) and kilogram force (kgf) is represented by,

1 kgf = 9.81 N

According to this formula, this value is similar to a body weight which is attracted towards the earth with a certain force. Formulation of it is done through,

m = mass of a body in kilogram

w = weight of that body

g = gravity

W= (m x g) N

If we assume mass to be 1 Kg, its weight will be,

W = 1 kg x 9.81 m /s² = 9.81 N

Basic Units with their dimensions

Physical quantity                          Symbol or Dimension       Unit or Notation

Mass                            kg                                kilogram

Length                                             m                                     Metre

Electric Current   A                              Ampere

Time                                                  s                                                Second

Luminous Intensity                             cd                                             Candela

Temperature        K                                         Kelvin

Derived Units with their dimensions

Momentum of forceNm                                             Newton metre

Power                                                  W=J/s                                      Watt

Torque                                                 Nm                                          Newton metre

Frequency                                           Hz= s^-1                                      Hertz

Angular velocity                                 rad/s                                        radian/second       Acceleration                                        m/s²                                         metre/second²

Force                                                   N =kg m/s²                              Newton

Pressure                                               Pa = N/m²                               Pascal

Angular acceleration                           rad/s²                                     radian / second²

Work, Energy                                      J = Nm =kg m²/s²                     Joule

Supplementary Units with their dimensions

Solid angle                                       sr                                                Steridian

Plane angle                                          rad                                          Radian

Few sub multiples and multiples of S. I units prefixes

Prefix                                 Symbol                 multiplying factor

Pico                                        p                     10^-12 = 0.000,000,000,001

Nano                                       n                      10^-9 = 0.000,000,001

Micro       µ                      10^-6 = 0.000,001

Milli                                        m                     l0^-3 = 0.001

Kilo                                        k                      10³ = 1,000

Mega       M                     10⁶ = 1,000,000

Giga                                      G                       10⁹ = 1,000,000,000

Tera                                         T                       10¹² = 1,000,000,000,000

Problems and solutions

Problem 1

There are 2 forces which are equal and are acting at same point. The angle between those 2 forces is60°. You are provided with its resultant force which is 20 x √3 N. Calculatemagnitude of each of those forces.

Solution

Provided resultant is,

R = 20 x √3

Angle between forces,

a = 60°

Let us assume that each of the magnitudes of the 2 forces is P.

So, with the help of these derivations,we can easily calculate the magnitude.

Its equation will be,

R = (2P cos a / 2)x (20 x √3)

    = 2P x cos (60° / 2)

    = 2P cos 30°

    = 2P x (√3 / 2)

    = P x √3

Therefore,

P = (20 x √3) / √3

= 20 N

Here, 20N is the magnitude of each force.

Problem 2

At a certain point, there are 2 forces that are acting on it with magnitudes 8 N and 10 N. Suppose, the given angle between the 2 forces is 60°, find the magnitude of resultant force.

Solution

We are aware of the values of the magnitudes of both forces. Let us name those forces as P and Q. Let us assume P to be 10 N and Q is 8 N. given angle is 60°.

R = √p2 + Q2 + 2PQ cos a

= √((10)2 +  (8)2 +  2  x  10  x 8  x cos⁡〖60°〗 )

= √(100 +  64 +  160  x 0.5)

= √(100 + 64 +  80)

= √(164 +  80)

= √244

= 15.62

Problem 3

According to the figure created with the help of law of triangle of forces, there are 2 forces which are acting on a certain fixed point (figure 1).There is another angular figure with certain values apart from the previous one (figure 2).Values of both forces in the second one are 100 N and 50 N respectively. Angle between those 2 forces is30°. And the other angle between one of the forces and plane line is 15°. Find out both of their direction and magnitudes.

Solution

Let us assume that fixed point to be O and the 2 forces to be ‘p’ and ‘q’.

Let value of p = 50 N and q = 100 N

As per figure 2, angle between p and q is depicted with value 30°.

Equation to find out magnitude of resultant R is,

R = √(P2 + Q2 + 2PQ cos a)

It is with the help of figure 2;we can see this resultant.

Angle value found between the direction and resultant of force p is performed with the help of this equation,

tan θ = q sin a / p + q sin a

θ = tan ^-1 (q sin a / p + q sin a)

   = tan ^-1 (100 x sin 30° / 50 x 100 cos 30°)

= tan ^-1 0.366

= 20.10°

Angular value with x axis made with resultant R is,

θ + 15° = 20.10° + 15°

= 35.10°

Problem 4

Angular value of 2 forces is 60°, and the resultant of those 2 forces is 14 N. these forces when acting on a 90° angled triangle will give resultant as  N. Calculate magnitude of both the forces.

Solution

To find out magnitude of both the forces, we need to divide the problem into 2 parts.

In the first part, (#1)

Angle  = 60°

Resultant of 1 of the forces = 14 N

i.e., R1 = 14 N

In case of second part, (#2)

Let us assume magnitude of both the forces be P and Q.

So using the equation to find out resultant force,

R = √(P2 + Q2 + 2 PQ cos α)

R1 = √(P2 + Q2 + 2 PQ cos 60)°

14 = √(P2 + Q2 + 2 PQ x 0.5)

14 2= P2 + Q2 + PQ

196 = P2 + Q2 + PQ (equation 1.1)

Now, utilizing similar equation in #2, we will get,

R = √(P2 + Q2)

In this case, R2 is √136 N

So,

√136  = √(P2 + Q2)

Therefore,

136 = P ² + Q ²(equation 1.2)

Now, subtracting equation (1.2) from equation (1.1),

P2 + Q2 + PQ – (P2 + Q2) = 196 – 136

PQ =60 (equation 1.3)

In the next step, when we multiply this new equation (1.3) with 2, we get,

2 PQ = 120 equation (1.4)

Now, we will add equation (1.4) with equation (1.2). on doing so, we will get,

P² + Q² + 2PQ = 136 + 120

P² + Q² + 2PQ= 256

(P + Q)² = (16)²

P + Q =16

P = (16- Q)

On substituting P’s value with equation (1.3), we will get,

60 = (16 – Q) x Q

= 16Q – Q²

     = Q² – 16Q + 60

= 0

This entire equation is known as quadratic equation.

After getting all these equations, we will calculate the values of both the forces from these.

So, with the help of above equation,

= ((16 – 4))/2 and((16 + 4))/2

= 6 and 10

These are the values of Q

In case of P, we will substitute Q’s value in this equation,

P = (16 – 6) or (16 – 10)

So here the answers are 10 and 6.

We can clearly see that values of both equations for P and Q are same. So, values of 2 forces are 6 N and 10 N.

Problem 5

There are 2 concurrent forces whose combined value or resultant is 180 N and of forces (P + Q) are 270 N. Providedangle between resultant force R and one of the concurrent forces is 90°. Determine the following things,

1. a) Magnitude of P and Q
2. b) Angle between P and Q

Solution

Resultant or R, in this case, is 180 N.

(P + Q) = 270 N

Angle between P and R is 90°

Let us name that angle as θ. So θ = 90°

So now to find out magnitude of P and Q, we will use the equation with tan θ.

tan θ = Q sin α / P + Q cos α

Here we have assumed the angle between P and Q is α.

tan = 90° Q sin α / P + Q cos α

But we know that tan 90° is infinity. This instance can only happen when,

P + Q cos α = 0

Therefore,

P = – Q cos α

Although it is one of the methods, the above result can be found with the help of another method too.

Alternative method

Let us take the help of figure 2 in problem 3.

Suppose in that triangle ∠OAC, θ = 90°

So,

∠OAC = 180 – α

∠ACO = α- θ

= α – 90°

Now, we will be using sine rule and from that we get,

sin 90° / Q = sin (180 –α) / R = sin (α – 90) / P

Taking the help of initial term and last one, we get the equation,

sin 90° / Q = sin (α – 90) / P

It can also be written as,

1 / Q = – cos α/ P [sin (α – 90) = sin [-(90 –α)] (90 –α)] = – sin (90 –α) = – cos α]

Its resultant value will come as,

P = – Q cos α (equation 1)

Now, using the equation for resultant and squaring it, we will get,

R2 = P2 + Q2 + 2PQ cos α

= P2 + Q2 + 2P x (-P)

We will take the help of (equation 1) but with a little change. In this case, the equation will stand as,

Q cos α = – P

Now, utilizing this above equation,

R2 = P2 + Q2 -2P2

= P2 – Q2

= (P + Q) (P – Q)

Given value of R is 180 and (P + Q) is 270

On putting the values in their designated places,

1802 = 270(P – Q)

32400 = 270 (P – Q)

(P – Q) = 32400/270

We have at present, values of both (P + Q) and (P – Q). Now putting the values on the above equation we will get,

2Q = (P + Q) + (P – Q)

2Q = 270 + 120

2Q = 390

Q = 195

So, 195 N is the magnitude of Q.

For the magnitude of P,

P = (P + Q) – Q

= 270 – Q

= 270 – 195

= 75

As we have already assumed the angle between P and Q to be α, we will be substituting P and Q’s values in equation 1.

So,

P = – Q cos α

75 = – 195 cos α

cosα = -75 / 195

= -0.3846

α = cos ^-1 (- 0.3846)

= 112.618°

Problem 6

There are 2 concurrent forces having its resultant force as 1500 N. Angle between both the forces is 90°. Angle between one of the forces and the resultant is 36°. Determine the magnitude of both forces.

Solution

Let us assume the names of concurrent forces be P and Q.

Angle between P and Q is 90° and is represented as α = 90°

Angle between one force and resultant force is θ = 36°

We will use tan rule in this case,

tan θ = Q sin α/ P + Q cos 90°

tan 36° = Q sin 90° / P + Q cos90°

= Q x 1 / P + Q x 1

= Q /P

0.726 = Q / P

Q = 0.726 P (equation 1)

Using the equation for resultant and squaring it,

R2 = P2 + Q2 + 2PQ cos α

(1500)2 = P2 + (0.726 P)2 + 2P x (0.726) x cos 90°

(1500)2 = P2 + 0.527 P2 + 0

= 1.527 P2

Putting the value of P in equation 1,

Q = 0.726 P

Q = 0.726 x 1213.86

= 881.26 N

To find the magnitude of P, we will use sine rule and consider ∠OAC (figure 2 problem 3).

From this rule we will get,

sin 90° / R = sin 36°Q = sin 54° / P

Now,

sin 90° / R = sin 36°Q

Q = R sin 36° / sin 90°

Here, given value of R = 1500 N

Q = 1500 x 0.5877 / 1

= 881.67 N.

We also have,

sin 90° / R = sin 54° / P

Now,

P = R sin 54° / sin 90°

= 1500 x 0.809 / 1

= 1213.52 N

Problem 7

There are 2 chains of length 3 m and 4 m. both these chains are supporting a weight of 900 N. find the tension in individual chains.

Solution

Let C be the weight 900 N

Given lengths are,

AC = 4 m

BC = 3 m

As its formation depicts a triangular formation, so the length of the third side will be AB = 5 m

In ∠ABC,

AC2 + BC2 = (4)2 + (3)2

= 16 + 9 = 25

AB2 =52 = 25

AB2 =AC2 + BC2

From the formulation it is clear that it is a right angle triangle,

So, ∠ACB is 90°

sina = BC / AB

= 3/ 5

= 0.6

a = 36° 52’

(a + b) =90°

Now,

b = 90° – a

= 90° – (36° 52’)

= 53° 8’

We will be calculating tensions of both chains BC and AC.

We will assume tension of AC as Tn1 and BC as Tn2.

In case of ∠BDC, which is a right angled triangle, its working will be,

θ = 90° – P

= 90°- 53° 8′

= 36° 52′

In ∠ACE,

= 180° – P

= 180° – 53° 8′

= 126° 52′

In ∠BCE,

= 180°-θ

= 180°-36° 52′

= 143° 8′

In ∠ACE, angle is 90°

Next step will be application of Lami’s theorem. On its application on C, we will be getting,

(Tn1 / sin of∠BCE) = (Tn2 / sin of ∠ACE) = (90° / sin 90°)

Tn2 / sin 143° 8’ = Tn2 / sin 126° 52’ = 90° / 1

Tn2 = 90° x sin 126° 52’

= 720 N

Problem 8

From a plane surface, there are 2 hanging chains which together are bearing a weight of 1000 N. One chain is bigger in length than the other. The point of connect shows it to be a right angle with angle of 30° between on side and center imaginary line. The other angle is 60°. Calculate tension of both chains individually.

Solution

Points from where both chains hang from, are A and B.

∠ACB = 90°

So, ∠CBA = 60°

∠CAB = 30°

Suppose the point at the plane from which the weight is hanging is D.

Now,

∠BCE =180° – 30° = 150°

∠BCD = 90° – 60° = 30°

∠ACE = 180° – 60° = 120°

We will assume tension of chain 1 as Tn1 and tension of chain 2 as Tn2

On application of Lami’s theorem at the point of weight (C),

(Tn1 / sin 150°) = (Tn2 / sin 120°) = (1000 / sin 90°)

(Tn1 / sin 150°) = (Tn2 / sin 120°) = 1000 / 1

Tn2 = 1000 sin 150° = 1000 x 0.5 = 500 N

Tn2 = 1000 sin 120° = 1000 x 0.866

= 866 N

Links of Previous Main Topic:-

• Fluid statics introduction
• Manometers measurement pressure
• Fluid kinematics
• Bernoullis equation
• Basics and statics of particles introduction
• Units and dimensions

Links of Next Mechanical Engineering Topics:-

• Laws of mechanics in basics and statics of particles
• Lamis theorem in basics and statics of particles
• Parallelogram and triangular law of forces
• Vectors
• Resolution and composition of a force
• Coplanar forces
• Resultant of coplanar forces
• Equilibrium of a particle
• Equilibrium of a rigid body
• Forces in space
• Equilibrium of a particle in space
• Equivalent system of forces
• Principle of transmissibility
• Single equivalent force
• Highlights of basics and statics of particles
• Equilibrium of rigid bodies introduction