engineering mechanics: statics in si units, 12e [compatibility mode].pdf3/15/2015 4 copyright ©...

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Equilibrium of a Particle33

Engineering Mechanics:

Statics in SI Units, 12e

Copyright © 2010 Pearson Education South Asia Pte Ltd


Chapter Objectives

• To introduce the concept of the free-body diagram for a particle

• To show how to solve particle equilibrium problems using the equations of equilibrium

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Chapter Outline

1. Condition for the Equilibrium of a Particle

2. The Free-Body Diagram

3. Coplanar Systems

4. Three-Dimensional Force Systems


3.1 Condition for the Equilibrium of a Particle

• Particle at equilibrium if

- At rest

- Moving at a constant velocity

• Newton’s first law of motion

∑F = 0

where ∑F is the vector sum of all the forces acting on the particle

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3.1 Condition for the Equilibrium of a Particle

• Newton’s second law of motion

∑F = ma

• When the force fulfill Newton's first law of motion,

ma = 0

a = 0

therefore, the particle is moving in constant velocity or at rest


3.2 The Free-Body Diagram (Continued)

• Best representation of all the unknown forces (∑F) which acts on a body

• A sketch showing the particle “free” from the surroundings with all the forces acting on it

• Consider two common connections in this subject –

– Spring

– Cables and Pulleys

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• Spring– Linear elastic spring: change in length is directly

proportional to the force acting on it

– spring constant or stiffness; k: defines the elasticity of the spring

– Magnitude of force when spring is elongated or compressed

� F = k s



• Cables and Pulley– Cables (or cords) are assumed negligible weight and

cannot stretch

– Tension always acts in the direction of the cable

– Tension force must have a constant magnitude for

equilibrium

– For any angle θ, the cableis subjected to a constant tension T

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3.2 The Free-Body Diagram (Concluded)

Procedure for Drawing a FBD

1. Draw outlined shape

2. Show all the forces

- Active forces: particle in motion

- Reactive forces: constraints that prevent motion

3. Identify each forces

- Known forces with proper magnitude and direction

- Letters used to represent magnitude and directions


Example 3.1

The sphere has a mass of 6kg and is supported as. Draw a free-body diagram of the sphere, the cord CE and the knot at C.

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WE = 6kg ××××9.81 m/(s^2)= 58.86 N≅≅≅≅ 58.9 N

FCE

FCE= 58.9 N

FCBA FCD

FCBAFCBA

FCBA

FCD


WE = 6kg ××××9.81 m/(s^2)= 58.86 N≅≅≅≅ 58.9 N

FCE= 58.9 N

FCBA FCD

FCBA Sin 60°°°°

FCBA Cos 60°°°°FCD

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FCBA Sin 60°°°° = FCE= 58.9 N

(√√√√3)/2

Hence, FCBA = 67.96 N

FCD =FCBA Cos 60°°°°

1/2

Hence, FCD = 67.96 * 0.5 = 33.98 N


FCE= 58.9 N

FCD

FCBA

Graphical SolutionWE

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Solution

FBD at Sphere

Two forces acting, weight and the force on cord CE.

Weight of 6kg (9.81m/s2) = 58.9N

Cord CE

Two forces acting: sphere and knot

Newton’s 3rd Law: FCE is equal but opposite

FCE and FEC pull the cord in tension

For equilibrium, FCE = FEC


Solution

FBD at Knot

3 forces acting: cord CBA, cord CE and spring CD

Important to know that the weight of the sphere does not act directly on the knot but subjected to by the cord CE

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3.3 Coplanar Force Systems

• A particle is subjected to coplanar forces in the x-y plane

• Resolve into i and j components for equilibrium

∑Fx = 0

∑Fy = 0

• Scalar equations of equilibrium require that the algebraic sum of the x and y components to equal to zero


3.3 Coplanar Systems

• Procedure for Analysis1. Free-Body Diagram

- Establish the x, y axes

- Label all the unknown and known forces

2. Equations of Equilibrium

- Apply F = ks to find spring force

- When negative result force is the reserve

- Apply the equations of equilibrium

∑Fx = 0 ∑Fy = 0

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Example 3.4

Determine the required length of the cord AC so that the 8kg lamp is suspended. The undeformed length of the spring AB is l’AB = 0.4m, and the spring has a stiffness of kAB = 300N/m.


Solution

FBD at Point A

Three forces acting, force by cable AC, force in spring AB and weight of the lamp.

If force in spring AB is known, stretch of the spring can be found using F = ks.

+→ ∑Fx = 0; TAB – TAC cos30º = 0

+↑ ∑Fy = 0; TAC sin30º – 78.5N = 0

Solving,

TAC = 157.0 N

TAB = 135.9 N

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SolutionTAB = kAB sAB ; 135.9 N = 300N/m*(sAB)

Hence, sAB = 0.453N

For stretched length,

ιAB = ι′AB+ sAB

ιAB = 0.4m + 0.453m

= 0.853m

For horizontal distance BC,

2m = ιAC cos30°+ 0.853m

ιAC = 1.32m


1

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Problem 3-5

30°°°°

45°°°°

30°°°°

5 kNF

∑∑∑∑Fx = 0, 5 kN sin 45°°°°

Hence, T = 13.32 kN ≅≅≅≅ 13.3 kN

T

8 kN

8 kN

F

5 kN cos 45°°°°

T sin 30°°°°

T cos 30°°°°

-T cos 30°°°°+8 kN + 5 kN sin 45°°°° = 0,

+ ve

∑∑∑∑Fy = 0,

+ ve

x

y

- 5 kN cos 45°°°° + F -T sin 30°°°° = 0,

Hence, F = 10.2 kN


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FABFAC

W

θθθθββββ

AB = √√√√[(3^2)+(4^2)]= 5 m A

BC

AC = √√√√[(3^2)+(3^2)] = 3√√√√2 m

4 m3 m

3 m

Sin θθθθ = 4m/AB = 4m/5m = 0.8

Cos θθθθ = 3m/AB = 3m/5m = 0.6

Sin ββββ = 3m/AC = 3m/(3√√√√2 m) = 1/√√√√2

Cos ββββ = 3m/AC = 3m/(3√√√√2 m) = 1/√√√√2

W = 2 * 9.81 N = 19.62 N

(2)

(3)

(4)

(5)

(1)

Copyright © 2010 Pearson Education South Asia Pte LtdCopyright © 2010 Pearson Education South Asia Pte Ltd

FABFAC

W

θθθθββββ

BC

Substituting by the values of Sin θθθθ & Sin ββββ from

Eqs. (2) and (4) in Eq. (6) to get

FAB Sin θ

FAB Cos θ

FAC Cos β

FAC Sin β

x

y

∑∑∑∑Fx = 0,

+ ve

FAB Sin θθθθ - FAC Sin ββββ = 0,

∑∑∑∑Fy = 0,

+ ve

FAB Cos θθθθ + FAC Cos ββββ - W = 0,

(6)

(7)

0.8 *FAB - FAC (1/√√√√2) = 0, Hence, FAC = 0.8 * √√√√2 FAB ,

(8)

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Substituting by the values of W, Cos θθθθ , Cos ββββ &

(FAC in terms of FAB ) from Eqs. (1), (3), (5) & (8)

in Eq. (7) to get


FAB * (0.6) + (0.8 * √√√√2 FAB )* 1/√√√√2 - 19.62 N = 0,

Hence, (1.4 * FAB ) = 19.62 N,

i.e., FAB = (19.62 /1.4) = 14.014 N

Hence, FAC = 0.8 * √√√√2 FAB = 0.8 * √√√√2 * 14.014 N = 12.2815 N,

i.e.,FAB ≅≅≅≅ 14.01 N

&FAC ≅≅≅≅ 12.28 N


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A30°°°°

30°°°°45°°°°B

C

D

FCDFBD

W

30°°°°45°°°°

FCDSin 30°°°°

FBDSin 45°

FBDCos 45° FCDCos 30°

Point D; Equation of Equilibrium

∑∑∑∑Fx = 0,

+ ve

FCD Cos 30°°°° - FBD Cos 45°°°° = 0, (9)


FCD (√√√√3 /2) - FBD (1/√√√√2) = 0,

Hence, FBD = (√√√√3/2) FCD = 1.225 FCD(10)

∑∑∑∑Fy = 0,

+ ve

(1.225 FCD ) (1/√√√√2)+0.5 *FCD - W = 0,

(11)

Substituting by FBD in terms of FCD from Eq. (10)

in Eq. (11) to get

FBD Sin 45°°°°+ FCD Sin 30°°°°- W = 0,

[(1.225 /√√√√2 )+ 0.5 ]*FCD - W = 0,

Hence, FCD = 0.731954 *W , (12)

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Substituting by FCD from Eq. (12) in Eq. (10) to get

Hence, FBD = 1.225 * 0.731954*W

i.e., FBD = 0.896644*W (13)

m = 50 kg, Hence,

W = mg = 50*9.81= 490.5 N = 0.49 kN ,

Hence, FCD = 0.731954*W = 0.731954* 0.49 kN ,

i.e., FCD = 0.359 kN , (12′)

i.e., FBD = 0.896644* 0.49 kN = 0.439 kN (13′)


A

30°°°°30°°°°45°°°°B

C

D

FBD

FBD Sin 45°

FBD Cos 45°

FAB Cos 30°

Point B; Equation of Equilibrium

∑∑∑∑Fy = 0,

+ ve

FAB Sin 30°°°° - FBD Sin 45°°°° = 0, (14)

FAB Sin 30°

FBC

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∑∑∑∑Fx = 0,

+ ve

FBD Cos 45°°°° + FBC - FAB Cos 30°°°° = 0,

(16)

Substituting by FBD in terms of W from Eq. (13) in Eq.

(15) to get:

FAB = (Sin 45°°°° / Sin 30°°°°) FBD = 1.414214 FBD

i.e.,

(15)

Hence, FBC = FAB Cos 30°°°° - FBD Cos 45°°°°

FAB = 1.414214 * (0.896644*W )= 1.268046 W

From Eq. (16) From Eq. (13)


Hence, FBC = (1.268046 W ) Cos 30°°°° - (0.896644*W) Cos 45°°°°

i.e., FBC = (1.268046 W ) Cos 30°°°° - (0.896644*W) Cos 45°°°°

i.e., FBC = 0.464137 W

FAB = 1.268046 W = 1.268046 * 0.49 kN = 0.621 kN

i.e.,

(17)

(16′)i.e.,

i.e., FBC = 0.464137 W= = 0.464137*(0.49 kN) = 0.2274 kN

(17′)

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Examples and Problems

• Review the following examples: 3.2 and 3.3

• Resolve the following problems for Homework: 3.27, 3.28 and 3.40



3.4 Three-Dimensional Force Systems

• For particle equilibrium

∑F = 0

• Resolving into i, j, k components

∑Fx i + ∑Fy j + ∑Fz k = 0

• Three scalar equations representing algebraic sums of the x, y, z forces

∑Fx i = 0

∑Fy j = 0

∑Fz k = 0

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3.4 Three-Dimensional Force Systems

• Procedure for Analysis Free-body Diagram

- Establish the x, y, z axes

- Label all known and unknown force magnitudes and directions on the digram

Equations of Equilibrium

- Apply ∑Fx = 0, ∑Fy = 0 and ∑Fz = 0- Substitute vectors into ∑F = 0 and set i, j, k components = 0

- Negative results indicate that the sense of the force is opposite to that shown in the FBD.


Example 3.7: Determine the force developed in

each cable used to support the 40kN crate.

FB FC FD

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Solution

FBD at Point A

To expose all three unknown forces in

the cables.


Expressing each forces in Cartesian vectors,

FB = FB (rB / rB)

rB = √(3m) 2 +(8m) 2 +(4m) 2 = 9.434 m

rB = ABx i + ABy j + ABz k

= (-3m) i + (-4m) j + (8m) k

Hence,FB = FB [{(-3m) i + (-4m) j + (8m) k} / 9.434 m ]

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i.e. FB = -0.318FB i – 0.424FB j + 0.848FB k

Similarly,

FC = FC (rC / rC)

rC = √(3m) 2 +(8m) 2 +(4m) 2 = 9.434 m

rC = ACx i + ACy j + ACz k

= (-3m) i + (4m) j + (8m) k

Hence,

FC = FC [{(-3m) i + (4m) j + (8m) k} / 9.434 m ]

i.e. FC = -0.318FC i + 0.424FC j + 0.848FC k

FD = FD i

W = -40 k


For equilibrium, ∑F = 0; FB + FC + FD + W = 0(-0.318FB i – 0.424FB j + 0.848FB k) + (- 0.318FC i

+ 0.424FC j + 0.848FC k) + FD i – 40 k = 0

∑Fx = 0; -0.318FB - 0.318FC + FD = 0 ∑Fy = 0; – 0.424FB + 0.424FC = 0 ∑Fz = 0; 0.848FB + 0.848FC - 40 = 0

From Eq. (19), FC =FB (21)

Substituting by FC in terms of FB from Eq. (21) in Eq. (20)to get:

(18)

(19)

(20)

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2* 0.848FB - 40 kN = 0 Hence, FB = FC = 20/0.848 kN = 23.6kN (22)

Substituting by FB & FC from Eq. (22) in Eq. (18) to get:-0.318*23.6kN - 0.318* 23.6kN + FD = 0 i.e.FD = 2* 0.318* 23.6kN =15.0kN (23)


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Solution

FBD at Point A


the cables.


Expressing each forces in Cartesian

vectors,

FB = FB (rB / rB)

rB = √(2 ft) 2 +(1 ft) 2 +(2 ft) 2 = 3.0 ft

rB = ABx i + ABy j + ABz k

= (- 2 ft) i + (1 ft) j + (2 ft) k

Hence,FB = FB [{(- 2 ft) i + 1 ft) j + (2 ft) k} / 3.0 ft ]

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i.e.,FB = -0.666 FB i + 0.333 FB j + 0.666 FB kSimilarly,FC = FC (rC / rC) rC = √(2 ft) 2 +(2 ft) 2 +(1 ft) 2 = 3.0 ftrC = ACx i + ACy j + ACz k

= (- 2 ft) i + (-2 ft) j + (1 ft) k

Hence,FC = FC [{(- 2 ft) i + (-2 ft) j + (1 ft) k} / 3.0 ft ]i.e. FC = -0.666 FC i - 0.666 FC j + 0.333 FC kFD = FD i

W = -300 k


For equilibrium, ∑F = 0; [FB] + (FC) + FD + W = 0[-0.666 FB i + 0.333 FB j + 0.666 FB k] + (-0.666 FC i -0.666 FC j + 0.333 FC k) + FD i – 300 k = 0

∑Fx = 0; -0.666FB -0.666 FC + FD = 0 (24)

∑Fy = 0; 0.333 FB - 0.666 FC = 0 (25)

∑Fz = 0; 0.666 FB + 0.333 FC - 300 = 0 (26)

From Eq. (25), FC = 0.5 FB (27)

Substituting by FC in terms of FB from Eq. (27) in Eq. (26)to get:0.666 FB + (0.333 * 0.5 FB ) - 300 = 0

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i.e., 0.666 FB + (0.333 * 0.5 FB ) - 300 = 0,Hence, FB = 360.36 Ib, (28)

Hence, From Eq. (27), FC = 180.18 Ib (29)

Substituting by FB &FC from Eqs. (28) & (29) in Eq. (24)to get:-0.666 * 360.36 Ib -0.666 * 180.18 Ib + FD = 0,Hence, FD = 360.36 Ib, (30)


Example 3.5:

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Solution

FBD at Point A


the cables.



vectors,

FC = FC (rC / rC)

FC = FC [Cos α i + Cos β j + Cos γ k ]C

= FC [(- 4/5) i + (0) j + (3/5) k] = FC [- 0.8 i + 0.6 k]

= - 0.8 FC i + 0.6 FC k


FD = FD (rD / rD)

FD = FD [Cos α i + Cos β j + Cos γ k ]D

= FD [Cos 60° i - Cos 30° j + 0 k] = FD [(1/2) i - (√3/2) j]

= 0.5 FD i - 0.866 FD j

W = -90 k

FB = FB j

For equilibrium, ∑F = 0; FB + (FC) + [FD ]+ W = 0FB j + (- 0.8 FC i + 0.6 FC k) + [0.5 FD i - 0.866 FD j ] – 90 k = 0

∑Fx = 0; - 0.8 FC +0.5 FD = 0 (31)

∑Fy = 0; FB - 0.866 FD = 0 (32)

∑Fz = 0; 0.6 FC - 90 = 0 (33)

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Fron Eq. (33), FC = 90/0.6=150 Ib (34)

Fron Eq. (31), FD = 1.6 FC = 240 Ib (35)

Fron Eq. (32), FB = 0.866 FD = 207.84 Ib (36)

But, FB = kAB xAB

Hence, 207.84 Ib = (500 Ib/ft) xAB ,

i.e. xAB = 207.84/500 = 0.416 ft.


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Solution

FBD at Point A

To expose all three unknown forces in the cords.



vectors,

FC = FC (rC / rC)

FC = FC [Cos α i + Cos β j + Cos γ k ]C

= FC [Cos 120° i + Cos 135° j + Cos 60° k ]

= FC [- 0.5 I -0.708 j + 0.5 k]

= - 0.5 FC i - 0.708 FC j + 0.5 FC k


Similarly,FD = FD (rD / rD) rD = √(1 m) 2 +(2 m) 2 +(2 m) 2 = 3.0 mrD = ADx i + ADy j + ADz k

= (- 1 m) i + (2 m) j + (2 m) k

Hence, FD = FD [{(- 1 ft) i + (2 ft) j + (2 ft) k} / 3.0 ft ]i.e. FD = -0.333 FD i + 0.666 FD j + 0.666 FD kFB = FB i

W = -mg = -100*9.81 k = -0.981 kN kFor equilibrium, ∑F = 0; FB + (FC) + [FD ]+ W = 0[FB i] + (- 0.5 FC i - 0.708 FC j + 0.5 FC k) + [-0.333 FD i + 0.666 FD j + 0.666 FD k ] -0.981 kN k = 0 ∑Fx = 0; FB - 0.5 FC- 0.333 FD = 0 (37)

∑Fy = 0; - 0.708 FC + 0.666 FD = 0 (38)

∑Fz = 0; 0.5 FC + 0.666 FD -0.981 kN = 0 (39)

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From Eq. (38),FD = (0.708/ 0.666) FC = 1.063 FC (38′)Substituting by FD in terms of FC from Eq. (38′) in Eq. (39) to get,0.5 FC + 0.666 FD =0.981 kNi.e., 0.5 FC + 0.666 * 1.063 FC =0.981 kNi.e., FC = 0.812 kN (40)Substituting by FC from Eq. (40) in Eq. (38′) to get,FD = 1.063 * 0.812 kN= 0.863 kN (41)Substituting by FC & FD from Eqs. (40) & (41) in Eq. (37) to getFB = 0.5 FC + 0.333 FD = 0.5* 0.812 kN + 0.333* 0.863 kN i.e., FB = 0.6934 kN (42)

Examples and Problems

• Review the following examples: 3.5 and 3.8

• Resolve the following problems for Homework: 3.45, 3.52 and 3.63

• Test No (1) Tuesday 4/1/1436 h at 11.00 AM.

• Chapters, 1,2 and 3.


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QUIZ

1. When a particle is in equilibrium, the sum of forces acting on it equals ___ . (Choose the most appropriate answer)

A) A constant B) A positive number C) Zero D) A negative number E) An integer

2. For a frictionless pulley and cable, tensions in the cables are related as

A) T1 > T2

B) T1 = T2

C) T1 < T2

D) T1 = T2 sin θ

T1

T2


QUIZ

3. Assuming you know the geometry of the ropes, you cannot determine the forces in the cables in which system above?

4. Why?

A) The weight is too heavy.

B) The cables are too thin.

C) There are more unknowns than equations.

D) There are too few cables for a 100 kg weight.

100 N100 N

100 N

( A ) ( B ) ( C )

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QUIZ

5. Select the correct FBD of particle A.

A 40°

100 kg

30°

30°A)

A

100 kg

B)40°

A

F1 F2

C) 30

°

A

F

100 kg

A

30

°

40°F1

F2

100 kg

D)


QUIZ

6. Using this FBD of Point C, the sum of forces in the x-direction (Σ FX) is ___ . Use a sign convention of + → .

A) F2 sin 50° – 20 = 0

B) F2 cos 50° – 20 = 0

C) F2 sin 50° – F1 = 0

D) F2 cos 50° + 20 = 0

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QUIZ

7. Particle P is in equilibrium with five (5) forces acting on it in 3-D space. How many scalar equations of equilibrium can be written for point P?

A)2 B) 3 C) 4 D) 5 E) 6

8. In 3-D, when a particle is in equilibrium, which of the following equations apply?A) (Σ Fx) i + (Σ Fy) j + (Σ Fz) k = 0

B) Σ F = 0

C) Σ Fx = Σ Fy = Σ Fz = 0

D) All of the above.

E) None of the above.


QUIZ

9. In 3-D, when you know the direction of a force but not its magnitude, how many unknowns corresponding to that force remain?

A) One B) Two C) Three D) Four

10. If a particle has 3-D forces acting on it and is in static equilibrium, the components of the resultant force ___ .

A) have to sum to zero, e.g., -5 i + 3 j + 2 k

B) have to equal zero, e.g., 0 i + 0 j + 0 k

C) have to be positive, e.g., 5 i + 5 j + 5 k

D) have to be negative, e.g., -5 i - 5 j - 5 k

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QUIZ

11. Four forces act at point A and point A is in equilibrium. Select the correct force vector P.

A) {-20 i + 10 j – 10 k}lb

B) {-10 i – 20 j – 10 k} lb

C) {+ 20 i – 10 j – 10 k}lb

D) None of the above.

12. In 3-D, when you don’t know the direction or the magnitude of a force, how many unknowns do you have corresponding to that force?

A) One B) Two C) Three D) Four

z

F3 = 10 N

P

x

A

F2 = 10 N

y

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