NAME & ID
DATE
MTE 119 – STATICS
TUTORIAL 1
Mechatronics Engineering
PAGE
1
4
PROBLEM 2-11,12: (TEXT BOOK)
11) The force acting on the gear tooth is F = 20 lb .
Resolve this force into two components acting along
the lines aa and bb.
12) The component of force F acting along line aa is
required to be 30 lb . Determine the magnitude of
F and its component along line bb.
Goal: To resolve the force F along aa and bb axes.
Drawing:
Plan: Use the law of sine to resolve the forces
Given:
11) F=20 lb
12) F along aa =30 lb
Execute:
11)
12)
60o
20 lb
80o
60
40o
Fb
F
40o
80o
Fb
o
Fa
Fa
20
=
Fa = 30.6 lb
sin 40 sin 80
Fb
20
=
Fa = 26.9 lb
sin 40 sin 60
Fa =30 lb
30
F
=
Fa = 19.6 lb
sin 80 sin 40
Fb
30
=
Fb = 26.4 lb
sin 80 sin 60
ANS
NAME & ID
DATE
MTE 119 – STATICS
TUTORIAL 1
Mechatronics Engineering
PAGE
2
4
PROBLEM 2-38: (TEXT BOOK)
Determine the magnitude and direction measured
counterclockwise from the positive x axis of the
resultant force of the three forces acting on the ring
A. Take F = 500 N and θ = 20
Goal: To determine the magnitude and direction of the resultant force acting at the ring
Drawing:
Plan:
Resolve the forces in to components and sum the forces to get the resultant.
Knowing the components of resultant, its direction can be calculated
Given: F = 500 N and θ = 20
Execute:
y
y
F
F3=600N F3y 1y F1=500N
20o
35
4
F3x
F1x
F2=400N
F2y
30
o
F2x
x
θ
1029.8 N
FR
37.4 N
x
Using Scalar notation, summing the force components algebraically,
+
4
→ FRx = ∑ Fx : FRx = 500sin 20 + 400 cos 30 − 600   = 37.42 N →
5
3
+ ↑ FRy = ∑ Fy : FRy = 500 cos 20 + 400sin 30 + 600   = 1029.8 N ↑
5
The magnitude of the resultant force FR is:
FR = FR2x + FR2y = 37.422 + 1029.32 = 1030.5 N = 1.03 kN
ANS
The directional angle θ measured counter clockwise from positive x axis is:
 1029.8 
o
 = 87.9
 37.42 
θ = tan −1 
ANS
NAME & ID
DATE
MTE 119 – STATICS
TUTORIAL 1
Mechatronics Engineering
PAGE
3
4
PROBLEM 2-50: (TEXT BOOK)
Determine the magnitude and direction, measured
counterclockwise from the positive x axis, of the
resultant force acting on the ring at O, if FA = 750 N
and θ = 45 .
Goal: To determine the magnitude and direction of the resultant force acting at the ring
Drawing:
Plan
Resolve the forces in to components and sum the forces to get the resultant.
Knowing the components of resultant, its direction can be calculated
Given: FA = 750 N , θ = 45
Execute:
y
FAy
FA=750N
45o
y
FAx
FBx
30o
FB=800N
θ
x
FR
130.33 N
FBy
1223.15 N
x
Using Scalar notation, summing the force components algebraically,
+
→ FRx = ∑ Fx : FRx = 750sin 45 + 800 cos 30 = 1223.15 N →
+ ↑ FRy = ∑ Fy : FRy = 750 cos 45 − 800sin 30 = 130.33 N ↑
The magnitude of the resultant force FR is:
FR = FR2x + FR2y = 1223.152 + 130.332 = 1230 N = 1.23 kN
ANS
The directional angle θ measured counter clockwise from positive x axis is:
 130.33 
o
 = 6.08
 1223.15 
θ = tan −1 
ANS
NAME & ID
DATE
MTE 119 – STATICS
TUTORIAL 1
Mechatronics Engineering
PAGE
4
4
PROBLEM 2-79: (TEXT BOOK)
The bolt is subjected to the force F, which
has components acting along the x,y,z axes
as shown. If the magnitude of F is 80 N and
α=60o and γ=45o, determine the magnitude of
its components.
Goal: Compute the magnitudes of the force F
Drawing:
Given: F =80 N, α=60o and γ=45o
Plan:
Calculate the angleβ, compute the component forces.
Execute:
angleβ:
cos β = 1 − cos 2 α − cos 2 γ
= 1 − cos 2 60 − cos 2 45
β = 120o
Components:
Fx = 80 cos 60 = 40 N
Fy = 80 cos120 = 40 N
Fz = 80 cos 45 = 56.6 N
ANS