SOLVED PROBLEMS
Statics Mechanics Assignment
Darya Mamonai – 09CE37
Mehran University of Engineering and Technology, Jamshoro
Mechanics Assignment Submitted to: Dr. Abdul Sami Quershi
09CE37 2 | P a g e
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:
Mechanics Assignment Submitted to: Dr. Abdul Sami Quershi
09CE37 3 | P a g e
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
By Using Cosine Law:
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 = ?
By Using Cosine Law:
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)
Mechanics Assignment Submitted to: Dr. Abdul Sami Quershi
09CE37 4 | P a g e
Let F1= F2 = F, and also R = F ( one of the force ) , θ = ?
By Using Cosine Law:
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
Mechanics Assignment Submitted to: Dr. Abdul Sami Quershi
09CE37 5 | P a g e
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
Mechanics Assignment Submitted to: Dr. Abdul Sami Quershi
09CE37 6 | P a g e
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=?
Mechanics Assignment Submitted to: Dr. Abdul Sami Quershi
09CE37 7 | P a g e
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
Mechanics Assignment Submitted to: Dr. Abdul Sami Quershi
09CE37 8 | P a g e
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
Mechanics Assignment Submitted to: Dr. Abdul Sami Quershi
09CE37 9 | P a g e
∑ 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
Mechanics Assignment Submitted to: Dr. Abdul Sami Quershi
09CE37 10 | P a g e
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
Mechanics Assignment Submitted to: Dr. Abdul Sami Quershi
09CE37 11 | P a g e
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
Mechanics Assignment Submitted to: Dr. Abdul Sami Quershi
09CE37 12 | P a g e
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
Mechanics Assignment Submitted to: Dr. Abdul Sami Quershi
09CE37 13 | P a g e
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
Mechanics Assignment Submitted to: Dr. Abdul Sami Quershi
09CE37 14 | P a g e
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
Mechanics Assignment Submitted to: Dr. Abdul Sami Quershi
09CE37 15 | P a g e
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
Mechanics Assignment Submitted to: Dr. Abdul Sami Quershi
09CE37 16 | P a g e
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