Download Solution Hw2 - Mechanical and Mechatronics Engineering

NAME & ID
DATE
Mechatronics Engineering
MTE 119 – STATICS
HOMEWORK 2
SOLUTIONS
Problem 2-95:
The door is held opened by means of two
chains.
Given:
Tension in AB: FA=300 N
Tension in CD: FC= 250 N
Goal:
Express each of these forces in Cartesian
vector form
Solution:
Plan:
K
K
First determine the position vectors rAB , rAC and then calculate the respective unit vectors.
The force vector will be given by the magnitude of the force multiplied by the unit vector.
Coordinates of points A and C:
(
C = (− 2.5
A = 0 − (1 + 1.5 cos(30 D )) 1.5 sin(30 D )
m
− (1 + 1.5 cos(30 D )) 1.5 sin(30 D )
A = (− 0.5 0 0 )
Position
)
)
m
m
Vector
{
= {(0.5 − 2.5)iˆ
} {
(0 − 0.75)kˆ} = {2iˆ
}
K
rAB = (0 − 0)iˆ (0 − (−2.299)) ˆj (0 − 0.75)kˆ = 2.299 ˆj − 0.75kˆ
K
rAC
(0 − (−2.299)) ˆj
m
}
+ 2.299 ˆj − 0.75kˆ
m
PAGE
1
9
NAME & ID
DATE
MTE 119 – STATICS
HOMEWORK 2
SOLUTIONS
Mechatronics Engineering
Unit
Vector
K
r AB
K
u AB = K
r AB
K
r AC
K
u AC = K
r AC
{
2.299 ˆj
=
− 0.31kˆ
2.299 + 0.75
2iˆ 2.299 ˆj − 0.75kˆ
2
=
} = {0.95 ˆj
− 0.75kˆ
2
{
2 + 2.299 + 0.75
2
2
} = {0.64iˆ
}
}
}
+ 0.73 ˆj − 0.24kˆ
2
Force Vector
K
K
F A = F A u AB = 285 ˆj − 93.0kˆ
N
K
K
F C = F C u AC = 159iˆ + 183 ˆj − 59.7kˆ
{
{
}
N
Problem 2-107:
The cable, attached to the shear-leg derrick, exerts a
force on the derrick of F = 350 lb.
Goal:
Express this force as a Cartesian vector.
Solution:
Plan:
First find the coordinates of points A, B and then find the position vector and the unit vector.
Multiply the force magnitude by the unit vector to get the force vector.
A = (0 0 35)
(
B = 50 sin 30 D
ft
50 cos 30 D
0
)
ft
PAGE
2
9
NAME & ID
DATE
Mechatronics Engineering
Position
{
and
unit
MTE 119 – STATICS
HOMEWORK 2
SOLUTIONS
vector
K
rAB = ( 25 − 0)iˆ ( 43.301 − 0) ˆj (0 − 35) kˆ
{
} {
}
{
ft = 25iˆ + 43.301 ˆj − 35kˆ
K
rAB
25iˆ + 43.301 ˆj − 35kˆ
K
= 0.409iˆ + 0.709 ˆj − 0.573kˆ
u AB = K =
2
2
2
rAB
25 + 43.301 + 35
Force Vector
K
KK
F = F u AB = 143iˆ + 248 ˆj − 201kˆ
{
}
}
ft
}
lb
Problem 2-114:
The force F = {25i – 50j + 10k} N acts at the
end A of the pipe assembly.
Goal:
Determine the magnitude of the components F1
and F2 which act along the axis of AB and
perpendicular to it.
Solution:
Plan:
K
K
First find the unit vector u AB . The magnitude of the projected component of F along AB axis
can be found by the dot product of the force and the unit vector along the axis.
AB = −4 ˆj − 6kˆ
− 4 ˆj − 6kˆ
K
u AB =
= −0.554 ˆj − 0.832kˆ
2
2
4 +6
K
K K
F1 = F ⋅ u AB = (25iˆ − 50 ˆj + 10kˆ) ⋅ (−0.554 ˆj − 0.832kˆ) = 19.414 N = 19.4 N
PAGE
3
9
NAME & ID
DATE
Mechatronics Engineering
K K K
F = F1 + F2
K
2
F = F1 + F2
2
⇒ F2 =
2
F − F1
2
MTE 119 – STATICS
HOMEWORK 2
SOLUTIONS
= 25 2 + 50 2 + 10 2 − 19.414 2 = 53.4 N
Problem 2-139: Determine the magnitudes of the projected components of the force
F = {60i + 12j – 40k} N in the direction of the cables AB and AC.
Solution:
Plan:
K
Find the unit vectors and use dot product to find the projected component of F in AB and
AC direction.
3iˆ + 0.75 ˆj − kˆ
K
u BA =
= 0.923iˆ + 0.23 ˆj − 0.307 kˆ
2
2
2
3 + 0.75 + 1
3iˆ − ˆj + 1.5kˆ
K
uCA =
= 0.857iˆ + 0.285 ˆj − 0.428kˆ
32 + 12 + 1.52
K K
FBA = F ⋅ u BA = 60(0.923) + 12(0.23) − 40(−0.307) = 70.46 = 70.5 N
K K
FBA = F ⋅ u BA = 60(0.857) + 12(0.285) − 40(−0.428) = 65.14 = 65.1 N
PAGE
4
9
NAME & ID
DATE
Mechatronics Engineering
MTE 119 – STATICS
HOMEWORK 2
SOLUTIONS
Problem 5:
The system of cables suspends
a 1000 lb bank of lights above
a movie set.
Goal:
Determine the tensions in
cables AB, CD and CE.
Solution:
Plan:
Isolate juncture A, and solve the equilibrium equations. Repeat for the cable juncture
C.
For Junction A:
K
TAB = TAB (cos(180 − β )iˆ + sin(180 − β ) ˆj )
K
TAC = TAC (cos(α )iˆ + sin(α ) ˆj )
α = 30D , β = 45D
∑F
∑F
x
= 0 ⇒ −TAB cos( β ) + TAC cos(α ) = 0
y
= 0 ⇒ TAB sin( β ) + TAC sin(α ) − W = 0
⇒ TAC = 732.05
lb,
TAB = 896.5
lb
PAGE
5
9
NAME & ID
DATE
Mechatronics Engineering
MTE 119 – STATICS
HOMEWORK 2
SOLUTIONS
For junction C:
K
TCA = TCA (cos(180 + α )iˆ + sin(180 + α ) ˆj )
K
K
TCE = TCE iˆ, TCD = TCD ˆj
α = 30 D
∑F
∑F
x
= 0 ⇒ TCE − TCA cos(α ) = 0
y
= 0 ⇒ TCD − TCA sin(α ) = 0
TCA = 732.05 lb
⇒ TCE = 634
lb,
TCD = 366
lb
Problem 3-14:
The un-stretched length of spring AB is
2m.
Goal:
If the block is held in the equilibrium
position shown, determine the mass of the
block at D.
Solution:
Plan:
Find the new length of AB due to the load and use F = kx to find the force in spring AB.
Write the equilibrium conditions for point A and solve for the unknowns.
PAGE
6
9
NAME & ID
DATE
Mechatronics Engineering
MTE 119 – STATICS
HOMEWORK 2
SOLUTIONS
Free Body Diagram:
y
FAC
FAB
5
45o
3
4
x
W
FAB = k AB x = 30(5 − 2) = 90
N
4
= 0 ⇒ FAB ( ) − FAC cos 45D = 0 ⇒ FAC = 101.82 N
5
3
∑ Fy = 0 ⇒ W − FAB ( 5 ) − FAC sin 45D = 0 ⇒ W = 126.0 N
126.0
m=
= 12.8 kg
9.81
∑F
x
Problem 3-19:
The cords BCA and CD can each support a
maximum load of 100 lb.
Goal:
Determine the maximum weight of the crate that can
be hoisted at constant velocity, and the angle θ for
equilibrium.
PAGE
7
9
NAME & ID
DATE
Mechatronics Engineering
MTE 119 – STATICS
HOMEWORK 2
SOLUTIONS
Solution:
Plan:
Draw the free body diagram. Using your physical sense it is clear that the maximum load is in
the cord CD. Assign 100 lb to FCD and find the maximum weight.
Free Body Diagram:
y
FCD
α
5
12
x
13
W
W
5
= 0 ⇒ 100 cosθ = W ( )
13
12
∑ Fy = 0 ⇒ 100 sin θ = W (13 ) + W
25
W
⇒ tan θ = 13 = 5 ⇒ θ = 78.7D
5
W
13
∑F
x
W = 51.0
lb
PAGE
8
9
NAME & ID
DATE
Mechatronics Engineering
MTE 119 – STATICS
HOMEWORK 2
SOLUTIONS
Problem 3-27:
The lift sling is used to hoist a container having a
mass of 500 kg. Determine the force in each of the
cables AB and AC as a function of θ.
Determine:
If the maximum tension allowed in each cable is 5
kN, determine the shortest lengths of cables AB
and AC that can be used for the lift. The center of
the gravity of the container is located at G.
Solution:
Plan:
Draw the free body diagram and write the equilibrium equations in x and y directions and
solve for F.
∑ Fx
y
FAB = FAC = F
F1=500(9.81)=4905
α
FAB
= 0 ⇒ FAC cos θ − FAB cos θ = 0
x
FAC
∑ Fy
F =
If the maximum allowable tension is 5 kN:
2452.5
= 5000 ⇒ θ = 29.37 D
sin θ
1.5
from the geometry : l =
cos θ
1.5
l =
= 1.72 m
cos 29.37 D
= 0 ⇒ 4905 − 2F sin θ = 0
2452.5
2.45
N =
kN
sin θ
sin θ
PAGE
9
9