statics problems by dkmamonai

SOLVED PROBLEMS

Statics Mechanics Assignment

Darya Mamonai – 09CE37

Mehran University of Engineering and Technology, Jamshoro

Mechanics Assignment Submitted to: Dr. Abdul Sami Quershi

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Problem#1: Two forces of 100+R N and 50+R N are applied of a section of beam , angle between

the two applied forces is 900 find out the resultant forces of these two forces and its direction.

SOLUTION:

As According to Pythagoras Theorem:

R2=F12 + F2

2 -> R=√ (100+37)2 + (50+37)2 -> √12500 -> 162.28 N

For Angel:

θ = tan-1( F2/F1) -> tan-1 ( 87/137)-> tan-1( 0.63) -> 32.410

Problem#2: The screw eye ( ring) is supported by two ropes making an angle of 300 force acting on

the screw eye – when magnitude of forces are 50+R N and 60+R N respectively.

SOLUTION:

By using Cosine law we will find resultant of these two forces :

As, R= √ F12 + F2

2 – 2 F1 F2 Cosθ -> √ (87)2+(97)2- 2 (87)(97) Cos 120 -> √ 16978+8439 -> √25417-> 159.4 N

For Direction using Sine law :

R/ Sin60 = 60/ Sinθ -> R Sinθ = 60 X Sin60 -> 159.4 Sinθ = 51.96 -> Sinθ= 0.325 -> θ= Sin-1(0.325) ->

θ= 19.02

Problem#3: 2 people are pushing a stoped car of mass 185KG the force they apply are respectively

are 275+R N and 395+R N with directed forward while the force of friction is 560+R N, what Is the resultant force respectively.

SOLUTION:

By Using Cosine Law:


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As, R= √ F12 + F2

2 – 2 F1 F2 Cosθ -> √ (275+37)2+(395+37)2- 2 (275+37)(395+37) Cos 00 -> √

283968+269568 -> √553536-> 744 N

Now as Friction for acting opposite to that forces, so it is taken as –ve. So, R= R1+R2 -> 744 + 597 -> 110N

Problem#4: Find the resultant of two forces 60+R N and 45+R N respectively acting at an angle

whose tangent is 12/5.

SOLUTION:

First we hence to find angle between the two forces for this as we have given.

Tanθ= 12/5 -> θ= Tan-1 (12/5) -> θ = 67.380


As, R= √ F12 + F2

2 – 2 F1 F2 Cosθ -> √ (60+37)2+(45+37)2- 2 (60+37)(45+37) Cos 67.380 -> √ 16133+6118.4

-> √22251.4-> 149.1 N

Problem#5: Find the angle between two equal forces F

(A) When their resultant is equal to one of the force

(B) Equal to half of the one of the force

SOLUTION:

(A)

Let F1= F2 = F, and also R = F ( one of the force ) , θ = ?


As, R= √ F12 + F2

2 – 2 F1 F2 Cosθ -> F = √ (F)2+(F)2- 2 (F)(F) Cosθ

Squaring on Both sides

F2=2F2+2F2 Cosθ -> 2F2Cosθ = F2 – 2F2 -> Cosθ = -F2 / 2 F2 -> Cosθ = -1/2 -> θ= Cos-1 ( -1/2) OR θ=Cos-1

(1/2) -> θ = 600

(B)


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Let F1= F2 = F, and also R = F ( one of the force ) , θ = ?


As, R= √ F12 + F2

2 – 2 F1 F2 Cosθ -> F = √ (F)2+(F)2- 2 (F)(F) Cosθ

Squaring on Both sides

F2/4=2F2+2F2 Cosθ -> 2F2Cosθ = F2 / 4 – 2F2 -> 2 F2 Cosθ = F2 – 8F2/ 4 -> 2 F2 Cosθ = -7/ 4 F2

Cosθ = -7/8 -> θ= Cos-1 ( -7/8) OR θ=Cos-1 (7/8) -> θ = 28.950

Problem#6: A push of 70 N and a pull of 50N acts on a body simultaneously find the resultant of

two forces

(a) if the angle between push and pull force is equal to 1350

(b) find the direction of resultant w.r.t +ve y-axis.

SOLUTION:

(a) For resultant

By using cosine law

R=√F12+ F2

2+2 F1 F2 cos θ

R=√F12+ F2

2-2 F1 F2 cos θ ̈θ>900

R=√(70+37)2+(50+37)2-2(70+37)(50+37) cos 1350

R=√2858-18618( cos 1350 )

R=√-45760

R= 213.915N

(b) For direction

By using sine law for direction

R/sin 135 = 50/sinθ => R sinθ =50 sin 135

sinθ 35.355/R = 35.355/213.915


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θ= sin-1 (0.166)

θ=9.510

This angle is made by resultant with respect to x- axis with respect to y-axis , we have

Θ=90-Φ

Θ=90-9.51 => θ= 80.490

This is the angle made by resultant w.r.t to y- axis.

Problem#7: The following forces shown in the figure. 30 kg are acting on a point . we have

20 kg which making an angle of 300 to word North of East and other force of 30 kg making an

angle of 45 to North of west of another force of 35 kg making an angle of 450 to the south of

west , Find.

SOLUTION:

By using principle of solution

F1x= F1cosθ1 = 20 cos 300 =17.32

F2x= F2cosθ2 = 30 cos 450 =21.21 &

F3x= F3cosθ3 = 35 cos 400=26.81

Now by resolving y- components

F1y= F1sinθ1 = 20 sin 300 =10

F2y= F2sinθ2 = 30 sin 450=15√2=21.21

F3y= F3sinθ3 = 35 sin 400=22.49

Now by using submission

∑ Fx+ F2x+ F3x = F1x=(- F2x )+(-F3x) => 17.32-21.21-26.81 => -30.7 kg

Similarly

∑Fy =F1y +F2y -F3y =>10+21.21-22.49 => 8.72 kg


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Now for resultant

R= √(∑ Fx)2+( ∑Fy)2 => √(-30.7)2+(8.72)2 => √1018.52 => 31.91 kg

For direction (Location)

As

Tan θ= Fy/Fx => Tanθ = 8.72/-30.7

θ=tan-1 (0.284)

θ=-15.850

Problem#8:The Horizontal and vertical component are given. Determine each force and Location.

(A) Px= -100 lb ,Py= 200 lb, P=?

SOLUTION:

For Resultant

By using formula

P=√Px2+Py2 => P=√(-1002)+(200)2 => P=√50000 =>223.606 lb

For Location

Tan θ= Py/Px => θ=tan-1 (Py/Px) => tan-1 (200/-100) => tan-1 (-2) => θ=-63.43o

(b) Fx= -100 lb, Fy= 200 lb, F=?

Again by using formula

F=√Fx2+Fy2 => F=√(302)+(-200)2 => √40900 => 202.23 lb

For location

Tan θ=Fy/Fx => -200/30 => tan-1 (-6.66) => θ=-81.46o

(c) Tx= -50 lb, Ty= -70 lb, T=?


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By using formula

T=√Tx2+Ty2 => T=√(-502)+(-70)2 => √7400 => 86.02 lb

For Direction

Tan θ= Ty/Tx => θ=tan-1 (Ty/Tx) => θ = tan-1 (-70/-50) => θ= 54.46o

Problem#9: Determine the magnitude of resultant force and its direction measured counter clock

were from the +ve x-axis

SOLUTION:

First we will find angle formed by F1 .

Tan θ=P/B => 3/4 => θ = tan-1 (3/4) => θ = tan-1 (0.75) => θ= 36.86

Now we will find the x and y co-ordinates of these forces

F1x= F1cosθ= 850o cos (36.80o+37) => 237.14N

F1y= F1sinθ=850 sin (36.80o+37) => 816.24

Since F2 makes an angle θ with y –axis so with respect to x-axis it will be

90-30=60o =312.5 w.r.t x –axis

F2x= F2cosθ=625 cos (60+37)= 339.5 N

F2y= F2sinθ=625 sin (60=37)= 620.34 N

Now for F2, since F3 makes an angle with y-axis , therefore w.r.t x-axis the angle will be 90-45=45o with x-axis

F3x= F3cosθ =750 cos 45 => 530.33 N

F3y= F3sinθ=750 sin 45 => 530.33 N

Now ∑ Fx & Fy will become

∑ Fx= F1+(- F2)+(- F3) => F1x- F2x - F3x => 680.62-312.5-530.33+3 => -125.21 N

∑ Fy= F1y-F2y-F3y => -509.17-541.26+530.33+37 => -483.1 N


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Now for resultant

R= √(∑ Fx)2+(∑ Fy)2 => √(-162.21)2+(-520.1)2 => (√296816.09)+37 => 581.80 N

For Direction

By using tangent ratio

Tanθ=Fy/Fx => Tanθ= -520.1/-162.21 => θ = tan-1(3.206) => θ= 72.670

Now with respect to +ve x-axis the angle will become

Φ=180+θ => 180+72.67 => 252.67o

Problem#10: Determine the magnitude and Direction θ of F1 so that the resultant force is

directed vertically obtained and has magnitude of 800 N.

SOLUTION:

The angle formed by F3 is Tan θ3 =3/4 => θ3 = tan-1 (0.75) => θ3= 36.8698

w.r.t X-axis , Φ3 =90-36.8698 => 53.130o

As, R = 800 N vertically Upwards its component

∑ Fx=Rx=0 N & Ry =800 N

Rx=Rcosθ => Rx= 800 cos 90=0

Ry=R sin θ =>Ry =800 sin 90 =800N =∑ Fy

Now Components for F1 are: F1x = F cos θ And F1y= F sin θ

Similarly components for F2 are:

F2x= F2cos 30= 400 X 0.866=346.41

F2x= F2cos 30= 400 X 0.866=346.41

F2y= F2sin 30= 400 X sin30= 200 N

F3x= F3cos 53.130= 600 X Cos 153.130= 360.00 N

F3y= F3 Sin 53.130= 600 X Sin 153.130= 479.99 N

Applying submission


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∑ F x= F1x+ F2x +F3x = F1x +F2x -F3x

0= F1 cos Φ+346.41-360

F1 cos Φ= 13.591 ----------------(1)

∑ F y= F1y+ F2y +F3y

800= F1 sinΦ+200+479.99

F1 sinΦ= 120.01 ------------------ (2)

Divide Equation (2) by (1)

R sin Φ/ R cos Φ 120.01/13.591

tanΦ=8.83 => Φ =83.53

For θ

θ=90-Φ => θ=90-83.53

θ=6.47o

Now for F1

As From Equation (1)

F1 cosΦ=13.59

F1 cos 83.53 =13.59

F1 = 13.59/0.11268

F1=120.60+37

F1=157.6 N

Problem#11:

As F1 =F2 and equal to 30 lb , determine he angle θ & Φ, sothat the resultant force is directed along the +ve x-axis and has magnitude of FR= 20 lb

SOLUTION:

First we have to find angle between forces, by using cosine law

FR =√ F12 +F2

2 +2F1 F2


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20==√ F2 +F2 +2(F)(F) cosα Since F1=F2=F

20=√2 F2 +2F2 cosα

20=√2 F2 +(1+ cosα)

Squaring both sides

(20)2= 2(30)2 (1+ cosα)

400=1800(1+ cosα) =>400/1800=1+ cosα

0.222=1+ cosα => cosα= 0.222-1

α= cos -1 (-0.778)

α= 141.077

Now

R/ Sinα = F1/ SinΦ

20/Sin 141.07 = 30 / SinΦ

SinΦ= 30 X Sin 141.07/20

Φ= 70.4600

And Similarly , Φ = 70.46 = θ

R/ Sinα = F2/ SinΦ

30/ Sin 141.07 = 30/ Sinθ

θ = 70.4600

ALTERNATE

Since forces have equal magnitude hence their respective angles will also be the same

θ= 141.077/2 = Φ

θ = Φ = 70.5380


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Problem#12:

Resolve the 200 lb forces shown acting on the poin as drawn in the figure

Into its component in

(a) x and y direction (b) x’ and y ‘ direction (c) x’ and y ‘ direction (d) x and y ‘ direction

SOLUTION:

(a) x and y direction

As component of F

Will be Fx & Fy

Fx =F cosθ → 200 cos 40 → 153.20 lb

Fy = F sin θ →200 sin 40 → 128.55 lb

(b) x’ and y ‘ direction

Θx’ = 40+30=70 , Θy =90-70=20

Fx’ =F cos θ →200 cos 70 → 68.404 lb

Fy’ = F sin θx → 200 sin 70 → 4187.93 lb

(c) x’ and y ‘ direction

Θx =30+40=70 , Θy =120-70=50


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By resolving into Components

Fx= F cos Θx → 200 cos 70 → 68.404 lb

Fy = F Sin Θx → 200 Sin 70 → 153.20 lb

(d) x and y ‘ direction

Θx =40 , Θy =20

Fx =F cos Θx → 200 cos 40 → 153.20 lb

Fy =F sin Θx →200 sin 20 → 128.55 lb

Problem# 13: Three forces are shown in figure by Its direction given in tangent Form, compute

the magnitude of Remaining forces

SOLUTION:

Hint: By resolving Forces along the direction A,B,C,D & FG

Tanθpx=3/4 → θpx=tan-1 3/4 → 36.869

Tanθx=8/15 → tan-1 (8/15) → 28.07

Tanθqx=12/5 → θqx= tan-1 (12/5) → 67.38

From figure,

θ1= θqx – θx → 67.38-28.07 → 39.31

since from figure

θα=90-θqx → 90-67.83 → 22.62

θb=90-θpx → 90-36.869 → 53.131

Now for θ2

θ2= θα+ θb → 22.62+53.131 → 75.751

Again for θ3

As ∆ABC=180


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180(θ1 +θ2) = θ3

θ3= 180(39.31+75.751) → 64.939

Now for Fq

By using sine law

F/sin 75.751=q/sin 64.939 → q=170/0.9692 X 0.9058 → q= 158.88 lb

Similarly for P

F/sin 75.751=P/sin39.31 → 170/0.9692=P/0.633 → p=111.12 lb

Problem#14: Resolve the forces F having magnitude equal to 130 lb into its components normal &

perpendicular (tangential) to the incline as shown in fig

SOLUTION:

The angle which slope makes with X-axis is given as

Tanθ=3/4 → θ=tan-1 (3/4) → θ=36.869

Tanθ= 15/12 → θ =tan-1(15/12) → θ= 22.6198

As parallel lines have same angle w.r.t same reference angle

ΘT =θ+θ1 → 36.869+22.6198 → 59.488

ΘN=90- ΘT → 30.512

Now

FT=F cos θT→FT =130*cos 59.488→ 66.003 lb

Similarly

FN =F cos θN=130*cos30.512→ 111.99 lb


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Problem#15: Resolve the force 130 lb as shown in fig : into two non rectangular component one having a

line of action along AB & the other parallel to θ

SOLUTION:

By drawing fig according to given problem

TanθAB =12/5 → θAB=tan-1

(12/5) → θAB =67.38

Now θ1

Tanθ1 =3/4

Θ1=tan-1

(3/4) →Θ1=36.869

ΘCD=180- θ1+(67.3) → ΘCD=180- 36.869-67.3 → ΘCD=75.811

Now

FCDx = F cos θCD → FCD=130*cos75.811 → FCD=31.865 lb

FeDy =F sin θCD FeDy=130 sin 75.811 → FCDy=126.03 lb

Problem#16: In the given figure (1) the x-component of P is 893 lb . Determine P and its y-component

SOLUTION:

From figure

Θ1=tan-1

P/B = tan-1

= (1/2) → 26.56

Θ2=tan-1

(4/3) → Θ2=53013

Now

Θx= Θ2 –Θ1 → 53.13-26.56 → 26.57

Now, as we know that

Px= P cosθx

Px/ cosθx =P → P = 893/ cos 26.57 → P=998.44 lb

Now for Py

Py= P sinθx → Py=998.44 sin 26.57 → Py=446.59 lb


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Problem#17: A body in an inclined as shown in figure is subjected to vertical and horizantol forces – find

the component of each force along x-y axis oriented parallel and perpendicular to the inclined

SOLUTION: For inclined angle

Tanθ=3/4 → Θ=tan-1

(3/4) → Θ= 36.86

From figure

Θpx=90- θ → Θpx=90- 36.86 → Θpx= 53.14

Now component of p will be

Px=P cos Θpx → 1200*cos 53.14 → Px=719.83 lb

Py=P sin Θpy → 1200*sin 53.14 → Py=960.12 lb

Now for component of F, From figure Θxf =36.86

Fx=F cos Θxf →Fx=1000 cos 36.86 → Fx=800.10 lb

Fy=F sin Θxf →Fy=1000 sin Θxf → Fy=599.86 lb

Problem#18: A body is resting on inclined which is making an angle of 30o with the x – axis the body is

subjected to a horizontal force making an angle of 20o with the x-axis , compute the component of this force

oriented perpendicular and parallel to an inclined Assume the magnitude of force is 113 lb .

SOLUTION:

From figure

Θxf =θ1+θ2 → Θx=30+20 → Θx=50

Fx= F cos Θx →Fx =113*cos 50 → Fx=72.634 lb

Fy= 113* sin 50 →Fy =113*sin 50 → Fy=86.56 lb


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Problem#16: The ring shown in the figure is subjected to two forces F1 and F2 . The resultant of two applied

forces is 1128 N directed vertically up word (A) Determine the magnitude of F1 &F2 (B) The magnitude when θ=20

of F1 &F2 when F2 is minimum

SOLUTION:

Φ=180-θ-20 → 180-50-20 → 110o

Now by using sine law

R/sin Φ = F1/ sin Φ

1128/sin 110 = F1/ sin 50 → 1128*sin 20/ sin 110 = F 2

F1=919.55 N

For F

R/ sin Φ= F 2/ sin 20 → 1128*sin 20/sin 110 =F2

F2=410.55 N

(A) When F2 is minimum, Since F2 is minimum at θ= 90o

Θ= 180(90+20) → Θ=70

Now

R/ sin θ= F1/ sin 90

1128/ sin 70 = F1/1

F1=1200.39 N

Now

R/ sin θ= F2/ sin 20 → 1128/ sin 70 = F2/ sin 20

F2=410.55 N

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