1© Manhattan Press (H.K.) Ltd. Scalar and vector quantities Manipulation of vectors Manipulation of...

© Manhattan Press (H.K.) Ltd. 1

• Scalar and vector quantitiesScalar and vector quantities

• • Manipulation of vectorsManipulation of vectors

1.3 Vectors and 1.3 Vectors and resolution of forcesresolution of forces

• • Forces in equilibriumForces in equilibrium


Scalar and vector quantities

1.3 Vectors and resolution of forces (SB p. 47)

A scalar quantity is one which can be described fully by just stating its magnitude.

1. Scalar quantities

e.g.mass, time, length, temperature, density, speed, energy, volume




A vector quantity is one which can only be fully described if its magnitude and direction are stated.

2. Vector quantities

e.g.displacement, velocity, acceleration, force, momentum, magnetic flux, electric intensity





Represented by:


Manipulation of vectors


1. Addition of vectors (P and Q)

parallelogram of vectors

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Common Error

Resultant

P + Q




1. Addition of vectors (P and Q)

triangle of vectors




2. Subtraction of vectors (P and Q)




3. Resolution of vector

vector R = resultant of 2 vectors (A, B or C, D or E, F)

A, B or C, D or E, F are components of R





R to 2 perpendicular directions

(resolution of vector)





Rx = R cosθ; Ry = R sinθ

tanθ = Ry/Rx

R = 22yx RR





e.g.

(a) a horizontal component

X = F cosθ

(b) a vertical component

Y = F sinθ


Forces in equilibrium


Note:

For simplicity, the vector notation ( → ) is omitted in describing the forces in the following chapters.

1. Addition of forces

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More to Know 7More to Know 7

P, Q, R, S, T in equilibrium?




1. Addition of forces

The polygon of forces states that if the forces acting on a point can be represented in magnitude and direction by the sides of a polygon, then the forces are in equilibrium.

In equilibrium (can form polygon)




2. Resultant of a number of forces

Not in equilibrium (cannot form polygon)

R = resultant of forces

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Example 3Example 3

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Example 4Example 4




3. Forces in equilibrium

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Example 5Example 5

In equilibrium,

algebraic sum of components = 0

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Common Error

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Example 6Example 6

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Example 7Example 7Go to

Example 8Example 8

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Example 9Example 9


End



The magnitude of the resultant force is not necessarily greater than those of its components of forces.

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Concurrent & coplanar forces

The forces shown in Fig. 1.22(a) are concurrent and coplanar forces. When the force vectors all intersect at one point, the forces are known as concurrent. When the problem involves forces in two dimensions only, the forces are said to be coplanar as they all lie on the same plane.

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Q:Q: The forces acting on a point O are shown in the figure below. Find the magnitude and direction of the resultant.

Solution



Solution:Solution:

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Algebraic sum of components of forces along Ox,

X = A cos15° + B cos30° + C cos120° + D cos225°

= 4 cos15° + 3 cos30° + 3 cos120° + 2 cos225° = 3.55 N

Algebraic sum of components of forces along Oy,

Y = 4 sin15° + 3 sin30° + 3 sin120° + 2 sin225° = 3.72 N

The magnitude of the resultant

R = = 5.14 N

tanθ=

∴ θ= 46.3°

The resultant is 5.14 N at an angle 46.3° to the direction Ox.

2222 723553 ..YX

553723..

XY



Q:Q: Sam, Brian and John attempt to push an

object in the direction Ox. Sam exertsa force of 200 N at a direction 30° to Ox, and Brian exerts a force of 400 N at60° to Ox as shown in the figure.What are the magnitude and direction of the smallest force that John should exert such that the resultant of all three forces acts along Ox?

Solution



Solution:Solution:

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In the figure, PQ represents the force FB = 400 N, QR represents the force FA = 200 N and PS along the Ox direction represents the resultant of the force FA + FB + FC, where FC is the smallest force that John exerts.

Therefore, the direction of FC is in the negative Oy direction and the magnitude is:

400 cos30° – 200 cos60°

= 246.4 N



Although the forces are in equilibrium, the object is not necessary to be in “static equilibrium”. We will learn the conditions for static equilibrium in Section 1.5. Return to

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Q:Q: The figure shows a point O in equilibrium

under the action of 5 coplanar forces. Calculate the values for θand x.

Solution



Solution:Solution:Since the forces are in equilibrium, the algebraic sum of the componentsof the forces in any direction is zero.Algebraic sum of components of forces along OX = 03x sinθ + 2x cos90° + 4x cos180° + 13 cos270° + 2x sinθ = 05x sinθ − 4x = 0

sinθ = 4/5θ = 53° 8’

Algebraic sum of components of forces along OY = 03x cosθ + 2x − 2x cosθ − 13 = 02x + x cosθ = 132 x + x (3/5) = 13

(13/5) x = 13 ∴ x = 5


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Q:Q: The figure shows a body S of weight W

hanging vertically by a thread tied at L to the string KLM. Find the tension in the section KL if the system is in equilibrium.

Solution



Solution:Solution:Since the forces W, T and F are in equilibrium, the resultant of the three forces must be zero. Therefore, the algebraic sum of the components of the forces in any direction must be zero.Algebraic sum of components of forces along LX = 0.F cos30° + (− T) = 0

F cos30° − T = 0T = F cos30° ............... (1)

Algebraic sum of components of forces along LY = 0.F sin30° + (− W) = 0

F sin30° − W = 0W = F sin30° ................ (2)

(1)/(2)


60tan30tan

30tan1

30sin30cos

WWT

WT

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Q:Q: The figure shows a picture of weight 10.0 N

hanging freely by a cord EFG. Find the tension T in the cord.

Solution



Solution:Solution:

Since the picture is in equilibrium, algebraic sum of vertical components of the forces = 0.

T sin60° + T sin60° − 10.0 = 0

2T sin60° = 10.0

∴ T =

= 5.77 N


60sin 2010.

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Q:Q: A mass of 5 kg hangs from two light strings of

lengths 3 m and 4 m from two points at the same level and 5 m apart. Find the tension in each of the strings. (Assume g = 9.81 m s−2.)

Solution



Solution:Solution:

Since the forces are in equilibrium, algebraic sum of the forces along CX = 0.

T2 sinθ−T1 cosθ= 0

T2 = ..........................(1)

Also, algebraic sum of the forces along CY = 0.

T2 cosθ+ T1 sinθ−W = 0

Substituting T2 from (1),


N 429N 23943

tan1

sincos

N 23954)8195(sin

sinsincos

sinsincos

112

1

22

1

11

2

..TTT

..WT

WT

WTT

θθθθ

θθθ

θθθ

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1sincos Tθ

θ


Q:Q: An object of weight W rests in

equilibrium on top of a smooth plane inclined at an angle θ to the horizontal as shown in Fig. (a).

(a) Find F1 and R1 in terms of W

and θ.(b) If the forces acted on the object are as shown in Fig. (b)

and the object is in

equilibrium, find F2 and R2

in terms of W and θ.

Solution


Fig. (a)

Fig. (b)


Solution:Solution:

(a) Since the object is in equilibrium, algebraic sum of components of forces along the inclined plane = 0.

F1 + (− W sinθ) = 0

∴ F1 = W sinθ

Algebraic sum of components of forces perpendicular to the inclined plane = 0.

R1 + (− W cosθ) = 0

∴ R1 = W cosθ


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Solution (cont’d):Solution (cont’d):

(b) Since the object is in equilibrium, algebraic sum of components of forces along the inclined plane = 0.

F2 cosθ + (−W sinθ) = 0

∴ F2 = = W tanθ

Algebraic sum of components of forces perpendicular to the inclined plane = 0.

R2 + (−F2 sinθ) + (−W cosθ) = 0

∴ R2 = F2 sinθ + W cosθ

Substituting F2, = (W tanθ) sinθ + W cosθ


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θθ

cossinW

θθθθ

coscoscossin 22 WWW

1© Manhattan Press (H.K.) Ltd. Scalar and vector quantities Manipulation of vectors Manipulation of...

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