NAME & ID
DATE
MTE 119 – STATICS
HOME WORK 1
Mechatronics Engineering
Problem 1:
If an object is near the surface of the earth, the variation of its weight with distance from the
center of the earth can often be neglected. The acceleration due to gravity at sea level is
g=9.81 m/s2. The radius of the earth is 6370 km. The weight of an object at sea level is mg,
where m is its mass. At what height above the earth does the weight of the object decrease to
0.99mg?
Solution:
Use Newton’s law of gravitational attraction and Newton’s second law:
GmmE
r2
W = ma
W=
a=
GmE
r2
(1)
mE - Mass of Earth
r - Distance between two masses
At sea level:
a=g
GmE = gRE2
r = RE
(2)
RE2
By substituting (2) in (1) a = g 2 Acceleration due to gravity at a distance ‘r’ from the
r
center of earth in terms of ‘g’ (sea level)
Therefore, the weight of the object at a distance ‘r’ from the center of the earth (W=ma):
W = mg
RE2
r2
Calling ‘h’ the height above the surface of the earth: r = RE + h
Therefore,
2
⎛ RE ⎞
W = mg ⎜
⎟ = 0.99mg
R
+
h
⎝ E
⎠
Solving for the radial height ‘h’:
⎛ 1
⎞
h = RE ⎜
− 1⎟
⎝ 0.99 ⎠
h = 32.09 km = 32.1 km
PAGE
1
9
NAME & ID
DATE
MTE 119 – STATICS
HOME WORK 1
Mechatronics Engineering
Problem 2: (2-13 Textbook Page 27)
The 500 lb force acting on the frame is to be resolved
into two components acting along the axis of the
struts AB and AC. If the component of force along
AC is required to be 300 lb, directed from A to C,
determine the magnitude of force acting along AB
and the angle θ of the 500 lb force.
Solution:
Parallelogram law of addition is shown in figure (a).
FAB
500 lb
105o
500 lb
60o
θ
45o
45o+θ
75o
300 lb
300 lb
(a)
Using the law of sines (figure b), we have
sin φ sin 75
=
300
500
sin φ = 0.5796
φ = 35.42o
Thus,
45o + θ + 75o + 35.42o = 180o
θ = 24.58o = 24.6o
FAB
500
=
o
sin(45 + 24.58 ) sin 75o
o
FAB = 485 lb
(b)
φ
FAB
PAGE
2
9
NAME & ID
DATE
MTE 119 – STATICS
HOME WORK 1
Mechatronics Engineering
Problem 3: (2-23 Textbook Page 29)
Determine the magnitude and direction of the
resultant FR = F1 + F2 + F3 of the three forces by first
finding the resultant F ′ = F2 + F3 and then forming
FR = F ′ + F1 .
Solution:
y
F1 = 30 N
F3 = 50 N
θ
’
20
o
F2 = 20 N
F′ =
90o-23.53o-(90o-36.87o)=
13.34o
x
70o
θ
x
φ
F ′ = 47.07 N
F′
( 20 ) + ( 50 )
2
2
− 2 ( 20 )( 50 ) cos 70o = 47.07 N
20
47.07
=
sinθ ′ sin 70o
θ ′ = 23.53o
FR =
( 47.07 ) + ( 30 )
2
FR
2
− 2 ( 47.07 )( 30 ) cos13.34o = 19.18 = 19.2 N
19.18
30
=
o
sin13.34
sin φ
φ = 21.15o
θ = 23.53o − 21.15o = 2.37o
PAGE
3
9
NAME & ID
DATE
MTE 119 – STATICS
HOME WORK 1
Mechatronics Engineering
Problem 4:
JG
While emptying a wheel barrow, a gardener exerts on each handle AB a force P directed
JG
along line CD. Knowing that P must have a 135-N
C A
horizontal component, determine:
JG
D
a. The magnitude of the force P
40°
JG
b. The vertical component of the force P
Note: Express your answers using three
significant figures.
B
Solution:
G
Resolve the force p into rectangular components:
C
G
P
G
Py
40o
G
Px = 135 N
(a)
G
G
Px = P cos 40
G
P =
(b)
G
Px
cos 40
=
135 N
= 176 N
cos 40
G
G
G
Py = Px tan 40 = P sin 40
G
Py = (135 N ) tan 40 = 113 N
D
PAGE
4
9
NAME & ID
DATE
MTE 119 – STATICS
HOME WORK 1
Mechatronics Engineering
Problem 5: (2-47 Textbook Page 40)
Determine the x and y
components of each force acting
on the guesset plate of the bridge
truss. Show that the resultant
force is zero.
Solution:
F1x = −200 lb
F1 y = 0
⎛4⎞
F2 x = 400 ⎜ ⎟ =320 lb
⎝5⎠
⎛3⎞
F2 y = −400 ⎜ ⎟ =-240 lb
⎝5⎠
⎛3⎞
F3 x = 300 ⎜ ⎟ =180 lb
⎝5⎠
⎛4⎞
F3 y = 300 ⎜ ⎟ =240 lb
⎝5⎠
F4 x = −300 lb
F4 y = 0
FRx = F1x + F2 x + F3 x + F4 x
FRx = −200 + 320 + 180 − 300 = 0
FRy = F1 y + F2 y + F3 y + F4 y
FRy = 0 − 240 + 240 + 0 = 0
Thus, FR = 0
PAGE
5
9
NAME & ID
DATE
MTE 119 – STATICS
HOME WORK 1
Mechatronics Engineering
PROBLEM 6: (2-52 TEXTBOOK PAGE 41)
The three concurrent forces acting on
the screw eye produce a resultant
JJG
force FR = 0 . If F2 =
JJG
JJG
2
F1 and F1 is
3
to be 90º from F2 as shown,
determine the required magnitude of
JJG
F3 expressed in terms of F1 and the
angle θ .
SOLUTION:
Express forces in Cartesian Vector notation:
JJG
F1 = F1 cos 60D i + F1 sin 60D j = 0.50 F1 i + 0.8660 F1 j
JJG 2
2
F2 = F1 cos 30D i − F1 sin 30D j = 0.5774 F1 i − 0.3333F1 j
3
3
JJG
F3 = − F3 sin θ i − F3 cos θ j
Find the resultant force:
JJG JJG JJG JJG
FR = F1 + F2 + F3 = 0
( 0.50 F1 + 0.5774 F1 − F3 sin θ ) i + ( 0.8660 F1 − 0.3333F1 − F3 cos θ ) j = 0
Equating i and j components we have:
0.50 F1 + 0.5774 F1 − F3 sin θ = 0
(1)
0.8660 F1 − 0.3333F1 − F3 cos θ = 0
(2)
Solving Equations (1) and (2) yields:
θ = 63.7D , F3 = 1.20 F1
PAGE
6
9
NAME & ID
DATE
MTE 119 – STATICS
HOME WORK 1
Mechatronics Engineering
Problem 7: (2-70 Textbook Page 53)
Determine the magnitude and
coordinate direction angles of the
resultant force and sketch this
vector on the coordinate system.
Solution:
FR = F1 + F2
FR = 350 cos 60i + 350 cos 60 j − 350 cos 45k
⎛4⎞
⎛4⎞
⎛3⎞
+ 250 ⎜ ⎟ cos 30i − 250 ⎜ ⎟ sin 30 j + 250 ⎜ ⎟ k
⎝5⎠
⎝5⎠
⎝5⎠
FR = {348.21i + 75.0 j − 97.487 k } N
FR =
( 348.21) + ( 75.0 ) − ( 97.487 )
2
2
= 369.29 = 369 N
⎛ 348.21 ⎞
o
⎟ = 19.5
369.29
⎝
⎠
α = cos −1 ⎜
⎛ 75.0 ⎞
o
⎟ = 78.3
369.29
⎝
⎠
β = cos −1 ⎜
⎛ −97.487 ⎞
o
⎟ = 105
369.29
⎝
⎠
γ = cos −1 ⎜
2
PAGE
7
9
NAME & ID
DATE
MTE 119 – STATICS
HOME WORK 1
Mechatronics Engineering
Problem 8: (2-73 Textbook Page 53)
The bracket is subjected to the two
forces shown. Express each force
in Cartesian vector form and then
determine the resultant force FR.
Find the magnitude and coordinate
direction angels of the resultant
force.
Solution:
Cartesian Vector notation:
F1 = 250 {cos 35sin 25i + cos 35cos 25 j − sin 35k} N
= {86.55i + 185.60 j − 143.39k} N
= {86.5i + 186 j − 143k} N
F2 = 400 {cos120i + cos 45 j + cos 60k} N
= {−200.0i + 282.84 j + 200.0k} N
= {−200i + 283 j + 200k} N
Resultant force:
FR = F1 + F2
= {( 86.55 − 200.0 ) i + (185.60 + 282.84 ) j + ( −143.39 + 200.0 ) k }
= {−113.45i + 468.44 j + 56.61k } N
= {−113i + 468 j + 56.6k } N
The magnitude of the resultant force is
FR = FR2x + FR2y + FR2z
FR =
( −113.45 ) + ( 468.44 ) + ( 56.61)
= 485.30 = 485N
2
2
2
PAGE
8
9
NAME & ID
Mechatronics Engineering
The coordinate direction angles are,
cos α =
FRx
cos β =
FRy
cos γ =
FRz
FR
FR
FR
=
−113.45
α = 104o
485.30
=
468.44
β = 15.1o
485.30
=
56.61
γ = 83.3o
485.30
DATE
MTE 119 – STATICS
HOME WORK 1
PAGE
9
9