Inscribed angle theorem

In geometry, a central angle is an angle whose vertex is at the center of a circle. A central angle is formed by two radii (plural of radius) of a circle. The central angle is equal to the measure of the intercepted arc. An intercepted arc is a portion of the circumference of a circle encased by two line segments meeting at the center of the circle

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Inscribed angle theorem

An inscribed angle in a circle is formed by two chords that have a common end point on the circle. This common end point is the vertex of the angle. In the figure below, circle with center O has the inscribed angle ∡ABC. The other end points than the vertex, A and C define the intercepted arc AC of the circle.

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Theorem

The measure of an inscribed angle is half the measure of the intercepted arc.
Proof
Given Consider a circle C with center O , we consider an inscribed angle at B by the arc AC
To Prove
∡B= \(\frac{1}{2} \measuredangle AOC\)
Construction
Join the vertices A and C with center O. Also draw a line through B and O .

SN	Statement	Reasons
1	∆BCO is Isosceles	CO=BO
2	y=2x	Triangle exteriar angle theorem
3	b=2a	Triangle exteriar angle theorem
4	y+b=2(a+x) a+x= \( \frac{1}{2}(y+b)\) ∡B= \(\frac{1}{2} \measuredangle AOC\)	Adding 2 and 3

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Symbolic Notation

Due to the theorem given above, it is seen that, the measure of arc AC has equal influence to the measure of its central angle ∡AOC. So it is also written as
\( \overset{⏜}{AC} \cong \measuredangle AOC \) or \( \overset{⏜}{AC} \equiv \measuredangle AOC \)
Similarly, the measure of chord AC has equal influence to the measure of its central angle ∡AOC. So it is also written as
\( \overline{AC} \cong \measuredangle AOC \) or \( \overline{AC} \equiv \measuredangle AOC \)
Similarly, the measure of chord AC has equal influence to the measure of its arc AC. So it is also written as
\( \overline{AC} \cong \overset{⏜}{AC} \) or \( \overline{AC} \equiv \overset{⏜}{AC} \)

Read More

Dynamics and Statics

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Introduction

Mechanics भनेको विज्ञानको एक शाखा हो जसले forces and their effects on structure (or machine) को बारेमा अध्य्यन गर्दछ। It is study of force, deformation, motion, and the relations. Mechanics लाई (a) statics vs dynamics; (b) particle vs rigid object vs many objects (‘multi-object’); and (c) 1 vs 2 vs 3 spatial dimensions (1D, 2D and 3D) आदि तरिकाले बिभिन्न 2, 3, 3 गरि 18 संयोजनहरूमा अध्य्यन गर्न सकिन्छ।

सामान्यतया, Mechanics लाई दुई मुख्य भागहरूमा विभाजन गरि अध्य्यन गरिन्छ।

1. Dynamics
   Dynamics is the branch of mechanics जसले गतिमा रहेको भौतिक बस्तुहरुको विश्लेषण गर्दछ. This means that dynamics implies change. It study about forces on a body which is at motion: causes of motion such as force, normal force, and friction force; including Newton's law of motion
2. Statics
   Statics is the branch of mechanics जसले स्थिर अवस्थाको वा स्थिर गतिमा चल्ने भौतिक बस्तुहरुको विश्लेषण गर्दछ . This means that statics implies changelessness. It deals with the study of forces on a body which is at rest.

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Force

Solution 👉 Click Here

To understand force, we need definition of force.

1. An intuitive definition of force is, "a push or a pull" on an object with a specific magnitude and direction. Force can be represented by vectors
2. Working definition of force is, an interaction between two material objects
3. Secintific definition of force is, an agent which produces or tends to produce or tends to destroy motion, so force may have either of the two functions: produces (or) tends to produce motion.

Depending upon the qualitative and quantitative natures of interactions between material objects, forces are broadely divided into the following catagories

1. Gravitional
2. Electromagnitic
3. Nuclear
4. Work

A systematic study of forces begains with two fundamental laws.

1. Newton's law of Universal Gravitation
2. Newton's law of Motion

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Newton's law of Universal Gravitation

Every particle of matter in the universe attracts every other particle with a force that is directionally proportional ro the product of the masses and inversly proportional to the square of the distance between them.
In symbol, it is written as
\(F=G \frac{m_1m_2}{d^2}\)

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Force acting at a Point

A realization of force may come from its effect. Commonly a force makes an object move or stop or combination of two or more of these effects. In general, the effect of force depends on

1. How large or how hard is the force?
2. In which direction the force is exerted?
3. Where the force is exerted?

In other words, a complete description of a force needs

1. its magnitude
2. its direction
3. its point of application

A force is geometrically represented by a line segment drawn from a point O representing its point of application at a point (or particle), in a direction from O to A pointing its direction and length OA proportional to its magnitude.
Since a force has both magnitude and direction, it is a vector quantity, and can be represented by
\(\overrightarrow{OA} \)
Moreover
If it originates from O, so as a position vector.
then the force is denoted by a single capital letter such as P, Q etc. with an arrow head avove it. viz. \(\vec{P},\vec{Q} \) and its magnitude by P,Q.

If there is system of concurrent forces they may lie

1. along a line, i.e., collinear
2. on the same plane, i.e., coplannar
3. on the different planes, i.e., non-coplannar

In nature, two or more forces may act simultaneously at a point. If the effect produced by a single force is the same as that produced by number of forces acting at a point,

1. the single force is called the resultant force of separate forces
2. separate forces is called components of single forces

Also, the process of finding the resultant force is called the composition of forces, and the converse process is called the resolution of forces

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Law of Forces

Solution 👉 Click Here

Law of forces

The basic laws of forces were first formulated by Newton. Newton’s three laws are as follows.

1. First law (or) law of inertia:
   First Law states that when a body is in motion (or) is at rest unless until it is forced to change that state by forces imposed on it. The property with which a particle resists a change in its state of rest (or) movement is called its inertia. Newton’s first law is therefore also known as the law of inertia.
2. Second law of motion:
   

   Newton’s second law of motion is closely related to Newton’s first law of motion. It mathematically states the cause and effect relationship between force and changes in motion. Newton’s second law of motion is more quantitative and is used extensively to calculate what happens in situations involving a force.

   
   It states that the rate of change in momentum of a particle is proportional to the force applied to it, and takes place in the direction of that force. Second law is defined by the following formula.
   \( F = K.m.a\)
   Where,
   F= force in N, m= mass in kg, a =Acceleration in \(m/s^2\), k=Proportionality constant
   

   The acceleration of a system is directly proportional to and in the same direction as the net external force acting on the system, and inversely proportional to its mass. In equation form, Newton’s second law of motion is
   \( F_{net} = ma\)

3. Third law of motion:
   It states that for every action there is an equal and opposite reaction. If a body A exerts a force on particle (or) body B, it will exert an equal and opposite force on body A.

We can readily see Newton’s third law. Consider a swimmer pushing off from the side of a pool, as illustrated in Figure below. She pushes against the pool wall with her feet and accelerates in the direction opposite to that of her push. The wall has exerted an equal and opposite force back on the swimmer.

Note that the swimmer pushes in the direction opposite to that in which she wishes to move. The reaction to her push is thus in the desired direction

Example 1

What Acceleration Can a Person Produce When pulling a load?
Suppose that the net external force (pull minus friction) exerted is 51 N parallel to the ground. The mass of the load is 24kg. What is its acceleration?

This problem involves only motion in the horizontal direction; we are also given the net force, indicated by the single vector, but we can suppress the vector nature and concentrate on applying Newton’s second law. Since Fnet and m are given, the acceleration can be calculated directly from Newton’s second law as
\( F=ma\).
The magnitude of the acceleration a is
\( a=F/m\) .
Entering known values gives
\( a=\frac{51N}{24kg}\)
Substituting the unit of kilograms times meters per square second for newtons yields
\( a=\frac{51kg⋅m/s^2}{24kg}=2.1m/s^2\)

Significance

The direction of the acceleration is the same direction as that of the net force, which is parallel to the ground. This is a result of the vector relationship expressed in Newton’s second law, that is, the vector representing net force is the scalar multiple of the acceleration vector.

Example 2

When Does Newton’s First Law Apply to a Car?

1. A car is parked outside house. Does Newton’s first law apply in this situation? Why or why not?
   When car is parked, all forces on the car must be balanced; the vector sum is \( 0N\) . Thus, the net force is zero, and Newton’s first law applies. The acceleration of the car is zero, and in this case, the velocity is also zero.
2. The car moves at constant velocity down the street. Does Newton’s first law apply in this situation? Why or why not?
   When car is moving at constant velocity down the street, the net force must be zero according to Newton’s first law. The car’s frictional force between the road and tires opposes the drag force on the car with the same magnitude, producing a net force of zero. The body continues in its state of constant velocity until the net force becomes nonzero. Realize that a net force of zero means that an object is either at rest or moving with constant velocity, that is, it is not accelerating.

NOTE

As this example shows, there are two kinds of equilibrium. In (a), the car is at rest; we say it is in static equilibrium. In (b), the forces on the car are balanced, but the car is moving; we say that it is in dynamic equilibrium.

Example 3

What Rocket Thrust Accelerates This Sled?

Calculate the magnitude of force exerted by each rocket, called its thrust T, for the four-rocket propulsion system shown in Figure below. The sled’s initial acceleration is \( 49m/s^2\) , the mass of the system is 2100kg, and the force of friction opposing the motion is 650N.

Solution
Although forces are acting both vertically and horizontally, we assume the vertical forces cancel because there is no vertical acceleration. This leaves us with only horizontal forces. Directions are indicated with plus or minus signs, with right taken as the positive direction. See the free-body diagram in Figure above.

Since acceleration, mass, and the force of friction are given, we start with Newton’s second law. We have defined the direction of the force and acceleration as acting “to the right,” so we need to consider only the magnitudes of these quantities in the calculations. Hence we begin with
\( Fnet=ma\)
where Fnet is the net force along the horizontal direction. We can see from the figure that the engine thrusts add, whereas friction opposes the thrust. In equation form, the net external force is
\( Fnet=4T−f\)
Substituting this into Newton’s second law gives us
\( Fnet=ma=4T−f\)
Using a little algebra, we solve for the total thrust 4T
\( 4T=ma+f\)
Substituting known values yields
\( 4T=ma+f=(2100kg)(49m/s^2)+650N\)
Therefore, the total thrust is
\( 4T=103550N\)
and the individual thrusts are
\( T=\frac{103550N}{4}=25887N\)

NOTE
Newton’s second law is a relationship among acceleration, force, and mass.

Example 4

Clearing a Field: A farmer is lifting some moderately heavy rocks from a field to plant crops. He lifts a stone that weighs 180 N . What force does he apply if the stone accelerates at a rate of \( 1.5m/s^2\) ?

Solution
No forces act in the horizontal direction, so we can concentrate on vertical forces, as shown in the free-body diagram. Now, the solution is [See Figure ]
\( w=mg\)
or \( m=\frac{w}{g}=\frac{180N}{9.8m/s^2}=18kg\)
or \( F_{net}=ma\)
or \( F-w=ma\)
or \( F-180N=(18kg)(1.5m/s^2)\)
or \( F-180=27N\)
or \( F=207N\)

Example 5: Third Law of Motion

A physics professor pushes a cart of demonstration equipment to a lecture hall, as seen in Figure below. Her mass is 65.0 kg, the cart’s is 12.0 kg, and the equipment’s is 7.0 kg. Calculate the acceleration produced when the professor exerts a backward force of 150 N on the floor. All forces opposing the motion, such as friction on the cart’s wheels and air resistance, total 24.0 N.

Solution
\( F_{net} = F_{floor} − f = 150N − 24N = 126N\).
The mass is
\( m = (65.0 + 12.0 + 7.0) kg = 84 kg\)
These values of F net and m produce an acceleration of
\(a=\frac{F_{net}}{m}=\frac{126}{84}=1.5m/s^2\)

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Composition of Forces

Solution 👉 Click Here

If a number of forces P, Q, R etc are acting simultaneously on a particle, then it is possible to find out a single force F, which would replace them, i.e. F would produce the same effect as produced by all the given forces P, Q, R.

1. the combined force F is called the resultant force of P, Q, R
2. the forces P, Q, R etc are called component forces of F.

In physical phenomenon and laboratory experiments, we can see that a single force can be represented in magnitude, direction and a line of action, whose effect is the same as the sum of the total of the effects acting at a point.
We can summarize these facts as below

1. Resultant of concurrent forces acting along the same line
   
   • If two collinear forces P and Q act on a particle in the same direction the the resultant R =P+Q and acts along their compound direction
   • If two collinear forces P and Q act on a particle in the opposite direction the the resultant R =P-Q and acts in the direction of greator force
2. Resultant of two forces acting at inclination (Parallelogram law)

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Parallelogram law

If two non-parallel forces P and Q act at a point O, they have a combined effect equivalent to a single force P+Q acting at O. The single force P+Q is called the resultant force.
This is an empirical result referred to as the parallelogram law

The Parallelogram law of forces states that if two forces, acting simultaneously on a particle are represented in magnitude and direction by the two adjacent sides of a parallelogram, then their resultant force can be represented in magnitude and direction by the diagonal of the parallelogram which passes through their point of intersection

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Parallelogram law: Proof

Assume that the two forces P and Q act at a point O as shown in figure given below.

The force P and Q are represented in magnitude and direction by vector \( \vec{p}\) and \( \vec{q}\)
The angle between the two force P and Q is \( \alpha \) .

Then magnitude of the resultant force R is given by
\( R=\sqrt{p^2+q^2 +2pq \cos \alpha}\)

Then direction of the resultant force of P and Q is given by
\( \tan \beta=\frac{q \sin \alpha}{p+q \cos \alpha }\)
Here, \( \beta\) is the angle that the resultant force R makes with one of the force, say P

Proof
We construct a parallelogram OPRQ in which the resultant R is represented by \( \vec{r}\)
Also, we construct a right angle triangle PSR in which
\( PS=q \cos \alpha \)
\( RS=q \sin \alpha \)
Now in \( \triangle PSR\) we have
\( OR^2=OS^2+RS^2\)
or \( R^2=(OP+PS)^2+RS^2\)
or \( R^2= (p+q \cos \alpha )^2+(q \sin \alpha)^2 \)
or \( R=\sqrt{p^2+q^2 +2pq \cos \alpha} \)
this is the magnitude of resultant \( R\)
Next, the direction \( \beta \) of resultant \(R\) is
\( \tan \beta = \frac{q \sin \alpha}{p+q \cos \alpha }\)

Alternative Method

Let \(\vec{p}\) and \(\vec{q}\) be two forces acting at O and α be the angle between them.
Also suppose that \(\vec{r}\) is the resultant of \(\vec{p}\) and \(\vec{q}\) and β is the angle, then
The magnitude of the resulatant R is
\(\vec{r}=\vec{p}+\vec{r}\)
or \(r^2=p^2+q^2+2pq \cos \alpha\)(1)
Also
\(\vec{r}=\vec{p}+\vec{q}\)
or \(\vec{p}.\vec{r}=\vec{p}.\vec{p}+\vec{p}.\vec{q}\)
or \(pr \cos \beta=p^2+pq \cos \alpha\)
or \(r \cos \beta=p+q \cos \alpha\)(2)
Again
\(\vec{r}=\vec{p}+\vec{q}\)
or \(\vec{p} \times \vec{r}=\vec{p} \times \vec{q}\)
or \(pr \sin \beta=pq \sin \alpha\)
or \(r \sin \beta=q \sin \alpha\)(3)
Hence using (2) and (3), the direction of the resulatant R is
\(\tan \beta=\frac{q \sin \alpha}{p+q \cos \alpha}\)

Conditions

1. Resultant r is maximum when two forces collinear and are in the same direction.
   i.e., \( \alpha =0^0 \) then \( r=p+q\) is maximum.
2. Resultant r is minimum when two forces collinear but acting in the opposite direction.
   That is \( \alpha = 180^0 \) then \( r= p- q\) is minimum.
3. If \( \alpha = 90^0\) , that the forces act at right angle, then
   \( r = \sqrt{p^2+q^2}\)
4. If the two forces are equal that is,\( p = q\) then \( r= 2 p \cos( \alpha / 2) \)

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Solved Examples

Solution 👉 Click Here

1. Find the resultant of two forces 3N and 6N acting at an angle of 120⁰
   
   Let us suppose that the acting forces are P and given by
   \( \vec{p}=3N\) and \(\vec{q}=6N\)
   Also assume that, the angle between the two forces P and Q is θ, then
   \( \theta =120^o\)
   Now, the magnitude of the resultant forces of P and Q is
   R²=P²+Q²+2PQ cosθ
   or R²=3²+6²+2.3.6 cos 120⁰
   or \(R= 3 \sqrt{3}\)N
   Also, the direction \( \beta\) of the resultant forces of P and Q is
   \( \beta = \frac{Q \sin \theta}{P+Q \cos \theta}\)
   or \( \beta = \frac{6 \frac{\sqrt{3}}{2}}{3-6 \frac{1}{2}}\)
   or \( \beta = 90^0\)
   This completes the solution

2. The resultant of two forces kN and 5N is 7N. If the direction of force 5N is reversed, the resultant will be \(\sqrt{19}\)N. Find the value of k.
   Solution

   Given that two forces kN and 5N are acting at a point.
   So, suppose that \(\alpha \) is the angle between then, then
   The resultant of kN and 5N is 7N, this implies
   \(7^2=k^2+5^2+2.k.5\cos \alpha\)(1)
   Also, if the direction of force 5N is reversed, the resultant will be \(\sqrt{19}\)N, this implies that
   \(19=k^2+5^2+2.k.5\cos (\pi-\alpha)\)(2)
   Solving (1) and (2), we get
   k=3N

3. If the resultant of two forces acting on a particle be at right angles to one of them, and its magnitude be half of the the magnitude of other, then show that the ratio of greater force to smaller is \(2:\sqrt{3}\)
   Solution

   Given that two forces P and Q acting at a point.
   So, suppose that \(\alpha \) is the angle between then, then
   The resultant of P and Q is R, this implies
   \((\frac{1}{2} Q)^2=P^2+Q^2+2.P.Q\cos \alpha\)(1)
   Also
   \(\tan 90=\frac{Q \sin \alpha}{P+Q \cos \alpha} \)(2)
   Solving (1) and (2), we get
   \(Q:P=2:\sqrt{3}\)

4. Forces equal to 7P,5P and 8P acting on a particle are in equilibrium. Find the angle between later pair of forces
   Let us suppose that α is the angle between two acting forces 5N and 8N, then
   \( \vec{p}=5N\) and \(\vec{q}=8N\)
   Also given that, the magnitude of the resultant forces of P and Q is 7N, thus we have
   R²=P²+Q²+2PQ cosα
   or 7²=5²+8²+2.5.8 cos α
   or \( \alpha =120^0\)
   This completes the solution
5. The resultant of two forces P and Q is R. If Q is doubled, the new resultant is perpendicular to P, prove that Q=R
   
   Given that, the resultant of two forces P and Q is R.
   Let us suppose that α is the angle between two forces P and Q, then
   R²=P²+Q²+2PQ cosα (1)
   Now, if Q is doubled, the new resultant is perpendicular to P, then
   \( \tan \alpha = \frac{2Q \sin \alpha}{P+2Q \cos \alpha}\)
   or \( \tan 90 = \frac{2Q \sin \alpha}{P+2Q \cos \alpha}\)
   or \( P+2Q \cos \alpha=0\)
   or \( \cos \alpha=-\frac{P}{2Q}\) (2)
   Using (2) in (1), we get
   Q=R
   This completes the solution
6. Two forces 2P and P act on a particle. If the first be doubled and second be increased by 12N, the direction of the resultant force be unaltered. Find the value of P.
   
   Let α is the angle between two forces then
   Case 1
   P=2P
   Q=P
   α=α
   Let us suppose that direction of the resultant force is θ, then
   \( \tan \theta=\frac{Q \sin \alpha}{P+Q \cos \alpha}\)
   or \( \tan \theta=\frac{P \sin \alpha}{2P+Q \cos \alpha}\) (1)
   Case 2
   P=4P
   Q=P+12
   θ=θ
   \( \tan \theta=\frac{Q \sin \alpha}{P+Q \cos \alpha}\)
   or \( \tan \theta=\frac{(P+12) \sin \alpha}{4P+(P+12) \cos \alpha}\) (2)
   Equating (1) and (2), we get
   \( \frac{P \sin \alpha}{2P+Q \cos \alpha}=\frac{(P+12) \sin \alpha}{4P+(P+12) \cos \alpha} \)
   P=12N
   This completes the solution

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Three or more coplanar forces acting at a point

Solution 👉 Click Here

Consider the three forces F1, F2 and F3 shown in Figure below. The procedure to obtain resultatnt force is simply to join the vectors F1, F2 and F3 end-on-end; then the resultant R corresponds to the vector joining the start of F1 to the end of F3.
The resultant vector R corresponds to the vector addition
\(R = F1 + F2 + F3\)
In terms of Cartesian components:
\(R_x = F1_x + F2_x + F3_x\)
and
\(R_y = F1_y + F2_y + F3_y \)

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The three forces F1, F2 and F3 will be in equilibrium if their resultant R is zero. In this case, joining the vectors end-on-end, the end of F3 will coincide with the start of F1. Thus we have the triangle of forces, which states that three coplanar forces acting at a point are in equilibrium if their vectors joined end-on-end correspond to the sides of a triangle, as illustrated in Figure below

Referring to Figures above, we notice that the angle in the triangle between F1 and F2 is 180◦ minus the angle between the forces F1 and F2 acting at the point O, and similarly for the other angles.

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Let OA and OB represent two forces P and Q respectively acting at a point O.
Now, we construct a parallelogram OACB in which diagonal OC represent the resultant R of the force P and Q.
Since, OB and AC being parallel and equal, their projections OM and LN on OX are equal in magnitude.

Thus
R_x=P cos α1+ Q cos α2
R_y=P sin α1+ Q sin α2
Then,the magnitude of resultant R is given by
\(R =\sqrt{R_x^2+R_y^2}\)
Also, the direction of resultant R is given by
\(\tan \theta =\frac{R_y}{R_x}\)

Similarly,
Let \(P_1,P_2,P_3,...\) are the forces acting at a point O inclined with \( \alpha_1,\alpha_2,\alpha_3,...\) with OX respectively.
If The resultant R makes an angle θ with OX, then
\(R_x=P_1 \cos \alpha_1+ P_2 \cos \alpha_2+...\)
\(R_y=P_1 \sin \alpha_1+ P_2 \sin \alpha_2+...\)
Thus, the magnitude of resultant R is given by
\(R =\sqrt{R_x^2+R_y^2}\)
Also, the direction of resultant R is given by
\(\tan \theta =\frac{R_y}{R_x}\)

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Solved Examples

Solution 👉 Click Here

1. Three forces of magnitudes 2N,3N, and 4N act in a plane at a point. The angle between any two of them is 120⁰, find the magnitude and direction of the resultant force R
2. A garden roller is pulled with a force of 200 \(\sqrt{2}\)N acting at an angle of 45⁰ with the lawn. Find the effective force pulling the roller along the ground (Friction Neglected)
3. Forces P-Q, P, and P+Q act at a point in the direction parallel to the sides of an equilateral triangle taken in order. find the magnitude and direction of the resultant force R
4. ABCDEF is a regular hexagon. Forces of Magnitudes \(2, 4\sqrt{3},8,2\sqrt{3},4\)N acts at A in the directions AB,AC,AD,AE and AF respectively. Find the magnitude and direction of the resultant force R

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Resolution of a Force

Solution 👉 Click Here

A given force F, it can be resolved into (or replaced by) two or more forces, which together produces the same effects that of force F. These forces are called the components of the force F. This process of replacing a force into its components is known as resolution of a force into components.

A force can be resolved into two components, which are either perpendicular to each other or inclined to each other. If the two components are perpendicular to one another, then they are known as rectangular components and when the components are inclined to each other, they are called as inclined components.

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Resolution of Forces into rectangular components

Let OC represent given force \(\vec{r} \) which is to be resolved into two directions OX and OY making angles α and β with OC.
Here,β=90-α.

Then
\( a=r \cos \alpha\)
\( b=r \sin \alpha \)

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Resolution of Forces into inclined components

Let OC represent given force \( \vec{r}\) which is to be resolved into two directions OX and OY making angles α and β with OC.

Now we construct a parallelogram OACB in which OA and OB represent the required components \( \vec{p}\) and \( \vec{q}\) of the force \( \vec{r}\).
In a triangle ΔOAC, we have
\( \frac{OA}{\sin \beta}=\frac{AC}{\sin \alpha}=\frac{OC}{\sin (180-\alpha-\beta)}\)
or \( \frac{p}{\sin \beta}=\frac{q}{\sin \alpha}=\frac{r}{\sin [180-(\alpha+\beta)]}\)
or \( \frac{p}{\sin \beta}=\frac{q}{\sin \alpha}=\frac{r}{\sin (\alpha+\beta)}\)
Thus we have
\( p=\frac{r \sin \beta}{\sin(\alpha+\beta)} \)
\( q=\frac{r \sin \alpha}{\sin(\alpha+\beta)} \)

Alternative Method

Let \(\vec{r}\) represent given force which is to be resolved into two directions \( \vec{p}\) and \( \vec{q}\) making angles α and β with \(\vec{r}\).

Then
\(\vec{r}=\vec{p}+\vec{q} \)
or \(\vec{p} \times \vec{r}=\vec{p} \times \vec{q} \)
or \(pr \sin \alpha=pq \sin (\alpha+\beta) \)
or \( q=\frac{r \sin \alpha}{\sin(\alpha+\beta)} \)

Again
\(\vec{r}=\vec{p}+\vec{q} \)
or \(\vec{r} \times \vec{q}=\vec{p} \times \vec{q} \)
or \(rq \sin \beta=pq \sin (\alpha+\beta) \)
or \( p=\frac{r \sin \beta}{\sin(\alpha+\beta)} \)

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Resolution of a Force: Solved Example

Solution 👉 Click Here

1. Resolve a force of 100N into two equal forces acting an angle of \(60^0\) to each other
   Solution

   Let \(\vec{r}=100N \) represent given force which is to be resolved into two forces represented by \( \vec{p}\) and \( \vec{q}\) making angles α and β with \(\vec{r}\).

   Given that, two component forces are equal, thus α =β =30
   Thus, the component force is
   \( p=\frac{r \sin \beta}{\sin(\alpha+\beta)} \)
   or \( p=\frac{100 \sin 30}{\sin 60} \)
   or \(p=\frac{100}{\sqrt{3}}\)

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Forces in Equilibrium

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Forces acting on a particle or on a point are said to be equillibrium if their resultant is zero.In such a situation, the state of the motion of the particle is unchanged., ie., at relative rest or uniform motion. Means, if the size and direction of the forces acting on an object are exactly balanced, then there is no net force acting on the object and the object is said to be in equilibrium.

NOTE

1. If the resultant of the forces acting on a particle is zero we say that these forces are in equilibrium.
2. For forces in equilibrium, the sum of the components of the forces in any direction must be zero.
3. An object at rest is in equilibrium.
4. Because there is no net force acting on an object in equilibrium, then from Newton's first law of motion, an object at rest will stay at rest, and an object in motion will stay in motion.

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Theorems in Equilibrium

1. Lami's Theorem
   Lami’s theorem is an equation that relates the magnitudes of three coplanar, concurrent and non-collinear forces, that keeps a body in static equilibrium. Lami’s theorem states that if three forces acting at a point are in equilibrium, each force is proportional to the sine of the angle between the other two forces.
   If three forces p, q, r acting on a particle or rigid body making angles A, B and C with each other.
   Then
   each force is proportional to the sine of the angle between the other two forces”. It is written as
   \( \frac{p} {\sin A} =\frac{q}{\sin B}=\frac{r}{\sin C}\)

   Statement

   Let ABC be a triangle in which the length of the sides opposite to A, B, C are a ,b , and c respectively, then following property holds.
   \( \frac{a} {\sin A} =\frac{b}{\sin B}=\frac{c}{\sin C}\)

   Proof
   Let ABC be a triangle in which the length of the sides opposite to A, B, C are a ,b , and c respectively.
   Here,
   \( \vec{a}+\vec{b}+\vec{c}=0\)
   or \( \vec{a}\times ( \vec{a}+\vec{b}+\vec{c} )=0\)
   or \( \vec{a}\times \vec{b}=\vec{c}\times \vec{a}\) (1)
   Similarly, we get
   \( \vec{b}\times \vec{c}=\vec{c}\times \vec{a} \) (2)
   Hence, from (1) and (2) we have
   \( \vec{a}\times \vec{b}=\vec{b}\times \vec{c}=\vec{c}\times \vec{a} \)
   Using the formula, we get
   \( a b \sin C=bc \sin A=ca \sin B \)
   or \( \frac{a} {\sin A} =\frac{b}{\sin B}=\frac{c}{\sin C}\)

2. m:n Theorem
   If A and B are two points with position vector \( \vec{a}\) and \( \vec{b}\) then the position vector P which divids AB in the ratio of m:n is \( \frac{m\vec{a}+n\vec{b}}{m+n}\) The resultant of two forces acting at a point O in direction OA and OB is represented in magnitude and direction by m OA and n OB is represented by (m+n)OP where P divides the segment AB in the ratio m:n, that is mAP=nBP
   Let \( \vec{OA}=\vec{a}, \vec{OB}=\vec{b}\) and \( \vec{OP}=\vec{p}\)
   Then
   \( m\vec{AP}=n \vec{PB}\)
   or\( \vec{p}= \frac{m\vec{a}+n\vec{b}}{m+n}\)
   This completes the proof
3. Cosine Theorem
   

   In a triangle ABC , prove that \( a^2 =b^2 +c^2 -2bc\cos A\)

   Let ABC be a triangle in which the length of the sides opposite to A, B, C are a ,b, and c respectively.

   Here,
   \( \vec{BC}+\vec{CA}+\vec{AB}=0\)
   or \( \vec{a}+\vec{b}+\vec{c}=0\)
   or \( \vec{a}=-\vec{b}-\vec{c}\)
   Squaring both side, we get
   \( \vec{a} ^ 2={{( -\vec{b}-\vec{c} )}^2}\)
   or \( \vec{a} ^ 2=( \vec{b}+\vec{c} )^2\)
   or \( a^2 =\vec{b} ^ 2+2\vec{b}\vec{c}+\vec{c} ^ 2\)
   or \( a^2 =b^2 +c^2 +2\vec{b}\vec{c}\)
   Using the formula , we get
   \( a^2 =b^2 +c^2 +2( -bc\cos A )\)
   or \( a^2 =b^2 +c^2 -2bc\cos A\)
   Similarly, we can prove that

   1. In a triangle ABC, prove that \( c^2 =a^2 +b^2 -2ab\cos C\)
   2. In a triangle ABC, prove that \( b^2 =c^2 +a^2 -2ca\cos B\)

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Solved Examples

Solution 👉 Click Here

1. Determine the magnitude and direction of the resultant of the forces shown below

   The solution is
   \( R^2= (200)^2+(150)^2+2(200)(150) \cos 150^0=102.66N\)
   \( \theta=\frac{Q \sin 150}{P+Q \cos 150}=46.9^0\)

2. Determine the magnitude and direction of the resultant of the forces shown below

   The solution is
   \( R^2 = (3)^2 + (2 )^2 +2(3)(2) \cos 60^0 = kN\)
   \( \theta=\frac{3 \sin 60^0}{2+3 \cos 60^0}=\)

3. The resultant of the two forces acting on the screw eye is known to be vertical. Determine the angle \( \theta\) and the magnitude of the resultant.

   By Lamis; law, the solution is
   \( \frac{\sin (90+\theta)}{40} =\frac{\sin 120}{60}=\frac{\sin (150-\theta)}{R}\)
   This helps to compute
   \( \theta=54.74^0\)
   \( R=69N\)

4. To support the 2-kg flower pot shown, the resultant of the two wires must point upwards and be equal in magnitude to the weight of the flower pot. Determine the angles and , if the forces in the wires are known to be 25 N and 30 N

   The solution is
   Weight of flower pot \( = (2 kg)(9.81 m/s^2)= 19.62 N\)
   Resultant, R, of forces in wires balances the weight.
   Therefore, the resultant is
   \( R = 19.62 N\)
   The, lami's theorem helps to calculate
   \(\frac{25}{\sin(180-\phi)}=\frac{30}{\sin(180-\theta)}=\frac{R}{\sin(\theta+\phi)}\)
   Also
   Cosine law helps to calculate
   \( (25 )^2= (30)^2+ (19.62)^2-2(30 )(19.62 ) \cos \phi \)
   Solving gives
   \( \phi = 55.90^0 \)
   Next, From the Lami's law, we calculate
   \( \theta = 83.6^0 \)

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