CARAGA STATE UNIVERSITY

CABADBARANCAMPUS
T.Curato St.,Cabadbaran City, Philippines
Tel. Nos. (085)3431020 / 281-2032
csucc.carsu.edu.ph

COLLEGE OF ENGINEERING AND INFORMATION TECHNOLOGY

Part 1:
1. Dynamics
2. Statics
3. Particle
4. Rigid bodies
5. Concentrated force
6. Scalar
7. Vector
8. Free Body Diagram

Part 2:
1. First Law at rest, having a constant velocity tends to remain in this state provided the
particle is not subjected to an unbalanced load.

Second Law unbalanced force experiences an accelaration has same direction as the
force and magnitude that is directly proportional to the force.

Third Law the mutual force of action between two particles are equal, opposite and
collinear.

2. Equilibrium- is the condition of a system in which all competing influences are balanced.
 Part 3:
Problem #1
Using triangle method:
= 180 - 60 - 45
FR = 4002 + 2502 -2(400)(250)(cos75)
FR = 413.20 N
sin sin 75
400
= 413.20
= 35.760
= 60 - 35.76 = 24.240

Problem #2
S1 = 22 + 1.52 = 2.5 m
S2 = 22 + 22 = 22 m
F = sk
= (22 - 2.5)(100)
F = 32.84 N
2
= tan ( 2 ) = 45 0

Fy = 0 let W weight of a cylinder

Problem #3
FAB = FAB i
FAC = -FAC j
 (20) i+(20) j+(10)k
FAD = FAD ( (2 0) 2+(2 0)2+(1 0) 2
2 2 1
FAD = - 3 FADi + 3 FADj + + 3 FADk
W = [-m(9.81)k]
Since,
F = 0
FAB + FAC + FAD - W = 0
2 2 1
FABi + (-FACj) + (- FADi + FADj + + F k) + [-m(9.81)k]
3 3 3 AD
2 2 1
(FAB - F )i + (-FAC + F )+( F - 9.81m)k
3 AD 3 AD 3 AD
By i, j and k components:
2
FAB - F =0
3 AD
2
FAC + F =0
3 AD
1
F - 9.81m = 0
3 AD
Since the FAD = 0, then the mass of the crate is:
1
(3000) - 9.81m = 0
3
m = 101.94 kg

Problem #4
Joint C:
+Fy = 0
sin30(FCB) - 1.5 = 0
FCB = 3 kN (T)
+Fx = 0
FCD + 3(cos30) = 0
FCD = -2.6 = 2.6 kN (C)
Joint D:
+Fx = 0
FDE 2 = 0
FDE = 2 kN (T)
 +Fy = 0
FDB + 2.6 = 0
FDB = -2.6 = 2.6 kN (C)
Joint B:
+Fy = 0
FBE(cos30) + 2(cos30) = 0
FBE = -2 = 2 kN (C)
+Fx = 0
(2+2)(sin30) + 3 FBA
FDB = 5 kN (T)
Problem #5
+Fx = 0
400 Ax = 0
Ax = 400 N
MC = 0
1200(8) + 400(3) Dy(12) = 0
Dy = 900 N
+Fy = 0
Ay -1200 +900 = 0
Ay = 300 N
At point G:
MG = 0
-300(4) 400(3) FBC(3) = 0
FBC = 800 (T)
At point C:
MC = 0
300(8) +FGE(3) = 0
FGE = -800 N = 800 N (C)
+Fy = 0
3
300 ( 5 )FGC = 0
FGC 500 N (T)

Problem #6
At member AC:
 MA = 0
-Cx(1.5sin60) + Cy(1.5cos60) + 600(0.75) = 0
At member CB:
MB = 0
-500(1) + Cx(1) + Cy(1) = 0
Therefore:
Cx = 402.63 N
Cy = 97.37 N
At member AC:
+Fx = 0
-Ax + 600(sin60) 402.63 = 0
Ax = 117 N
+Fy = 0
Ay 600(cos60) 97.37 = 0
Ay = 397.37 N
At member CB:
+Fx = 0
402.63 500 + Bx = 0
Bx = 97.37 N
+Fy = 0
-By +97.37 = 0
By = 97.37 N

Problem #7
Fy = 2T - 700 = 0
T = 350 lb
Member ABC:
MA = 0
By(4) - 700(8) - 100(4) - (350sin60)(4) = 0
By = 1803.11 lb
Fy = 0
-Ay - 350sin60 -100 -700 + 1803.11 = 0
Ay = 700 lb
Fx = 0
Ax - 350cos60 - Bx + 350 - 350 = 0
Ax - Bx = 175 lb
 Member DB:
MD = 0
-40(2) - 1803.11(4) + Bx = 0
Bx = 1823.11 lb
Fx = 0
-Dx + 1823.11 = 0
Dx + 1823.11 lb
FY = 0
Dy - 40 1803.11 = 0
Dy = 1843.11 lb
Since, Ax - Bx = 175
Ax = 175 + 1823.11
Ax = 1998.11 lb

Problem #8

Problem #9