Download SOLUTION Consider the three vectors;with A vertical.Note and more Lecture notes Statics in PDF only on Docsity!
If A , B , and D are given vectors, prove the distributive law for the vector cross product, i.e., A : ( B + D ) = ( A : B ) + ( A : D ).
SOLUTION
Consider the three vectors; with A vertical.
Note obd is perpendicular to A.
Also, these three cross products all lie in the plane obd since they are all perpendicular to A. As noted the magnitude of each cross product is proportional to the length of each side of the triangle.
The three vector cross products also form a closed triangle which is similar to triangle obd. Thus from the figure,
Note also,
= (A * B) + (A * D) (QED)
i j k A (^) x A (^) y A (^) z Bx By Bz
i j k A (^) x A (^) y A (^) z Dx Dy Dz
- [(A (^) y Dz - A (^) z Dy) i - (A (^) x Dz - A (^) z Dx) j + (A (^) x Dy - A (^) y Dx) k
= [(A (^) y Bz - A (^) zBy) i - (A (^) x Bz - A (^) z Bx)] j + (A (^) x By - A (^) y Bx) k
- [A (^) x(By + Dy) - A (^) y(Bx + Dx)] k
- [A (^) x(Bz + Dz) - A (^) z(Bx + Dx)] j
= [A (^) y (Bz + Dz) - A (^) z(By + Dy)] i
A * ( B + D ) = 3
i j k A (^) x A (^) y A (^) z Bx + Dx By + Dy Bz + Dz
D = Dx i + Dy j + Dz k
B = Bx i + By j + Bz k
A = A (^) x i + A (^) y j + A (^) z k
A * (B + D) = (A * B) + (A * D)
o¿b¿d¿
bd = ƒ A * D ƒ = ƒ A ƒ ƒ D ƒ sin u 2
ob = ƒ A * B ƒ = ƒ A ƒ ƒ B ƒ sin u 1
od = ƒ A * ( B + D ) ƒ = ƒ A ƒ ƒ B + D ƒ sin u 3
(QED)
Prove the triple scalar product identity A #^ (B : C) = (A : B) #^ C.
SOLUTION
As shown in the figure
Thus,
But,
Thus,
Since | (A * B) #^ C | represents this same volume then
(QED)
Also,
Thus,
A #^ (B : C) = (A : B) #^ C (QED)
LHS = RHS
= A (^) xByCz - A (^) xBzCy - A (^) yBxCz + A (^) yBzCx + A (^) zBxCy - A (^) zByCx
= Cx(A (^) y Bz - A (^) zBy) - Cy(A (^) xBz - A (^) zBx) + Cz(A (^) xBy - A (^) yBx)
i j k A (^) x A (^) y A (^) z Bx By Bz
3 # (^) (Cx i + Cy j + Cz k )
RHS = (A : B) #^ C
= A (^) xByCz - A (^) xBzCy - A (^) yBxCz + A (^) yBzCx + A (^) zBxCy - A (^) zByCx
= A (^) x (ByCz - BzCy) - A (^) y (BxCz - BzCx) + A (^) z (BxCy - ByCx)
= (A (^) x i + A (^) y j + A (^) z k ) #^3
i j k Bx By Bz Cx Cy Cz
LHS = A #^ (B : C)
A #^ (B : C) = (A : B) #^ C
Volume = | A #^ (B * C) |
|h| = | A #^ u ( B * C )| = (^) ` A #^ a
B * C
| B * C |
b `
Volume of parallelepiped is | B * C ||h|
Area = B(C sin u) = | B * C |
2 m 3 m
4 m
60
F 1 250 N^30
B
F 2 300 N
F 3 500 N
A
4 3
5
Determine the moment of each of the three forces about point A.
SOLUTION
The moment arm measured perpendicular to each force from point A is
Using each force where we have
a
(Clockwise) Ans.
a
(Clockwise) Ans.
a
= -800 N #^ m = 800 N #^ m (Clockwise) Ans.
+ 1 MF 32 A = - 5001 1.60 2
= -1299 N #^ m = 1.30 kN #^ m
+ 1 MF 22 A = - 3001 4.330 2
= -433 N #^ m = 433 N #^ m
+ 1 MF 12 A = - 2501 1.732 2
MA = Fd,
d 3 = 2 sin 53.13° = 1.60 m
d 2 = 5 sin 60° = 4.330 m
d 1 = 2 sin 60° = 1.732 m
Determine the moment of each of the three forces about point B.
SOLUTION
The forces are resolved into horizontal and vertical component as shown in Fig. a. For F 1 ,
a
d Ans.
For F 2 ,
a
d Ans.
Since the line of action of F 3 passes through B , its moment arm about point B is zero. Thus
M (^) B = 0 Ans.
= 600 N #^ m
- MB = 300 sin 60°(0) + 300 cos 60°(4)
= 149.51 N #^ m = 150 N #^ m
- MB = 250 cos 30°(3) - 250 sin 30°(4)
2 m 3 m
4 m
60
F 1 250 N^30
B
F 2 300 N
F 3 500 N
A
4 3
5
The railway crossing gate consists of the 100-kg gate arm having a center of mass at Ga and the 250-kg counterweight having a center of mass at GW. Determine the magnitude and directional sense of the resultant moment produced by the weights about point A.
SOLUTION
= 2084.625 N #^ m = 2.08 kN #^ m ( Counterclockwise ) Ans.
+(MR)A = gFd; (MR)A = 100(9.81)(2.5 + 0.25) - 250(9.81)(0.5 - 0.25)
0.5 m^ 1 m
0.25 m
2.5 m
G (^) W 0.75 m
G (^) a B
A
The force F^ acts on the end of the pipe at B. Determine (a) the moment of this force about point
A , and (b) the magnitude and direction of a horizantal force, applied at C , which produces the
same moment.
Given:
F 70 N
a 0.9 m
b 0.3 m
c 0.7 m
60 deg
Solution:
(a) (^) M (^) A F sin (^) (^) c F cos (^) a M (^) A 73.9 N m
(b) F C ( ) a M A F C
M A
a
F C 82.2 N
Ans.
Ans.
The force F acts on the end of the pipe at B. Determine the angles ( 0° 180°) of
the force that will produce maximum and minimum moments about point A. What are the
magnitudes of these moments?
Given:
F 70 N
a 0.9 m
b 0.3 m
c 0.7 m
Solution:
M A F sin ^ c F cos ^ a
For maximum moment
d M A
d
c F cos a F sin 0
max atan
c
a
max 37.9 deg
M Amax F sin max c F cos max a M Amax 79.812 N m
For minimum moment M A F sin^ c F cos^ a 0
min 180 deg atan
a
c
min 128 deg
M Amin F c sin min F ( ) cos a min M Amin 0 N m
Ans.
Ans.
Ans.
Ans.
The Achilles tendon force is mobilized when the man tries to stand on his toes. As this is done, each of his feet is subjected to a reactive force of If the resultant moment produced by forces and about the ankle joint A is required to be zero, determine the magnitude of F f.
F t N f
N t = 400 N.
F t
SOLUTION
Referring to Fig. a ,
a
F = 618 N Ans.
+(MR)A = ©Fd; 0 = 400(0.1) - F cos 5°(0.065) 65 mm 100 mm
200 mm
A
N (^) f 400 N
F t
5
The total hip replacement is subjected to a force of Determine the moment of this force about the neck at A and the stem at B.
F = 120 N.
SOLUTION
Moment About Point A: The angle between the line of action of the load and the neck axis is
a
(Counterclockwise) Ans.
Moment About Point B: The dimension l can be determined using the law of sines.
Then,
= -4.92 N #^ m = 4.92 N #^ m (Clockwise) Ans.
MB = -120 sin 15° 1 0.1584 2
l sin 150°
sin 10°
l = 158.4 mm = 0.1584 m
= 0.418 N #^ m
a +
MA = 120 sin 5° 1 0.04 2
10 °
120 N 15 °
40 mm
150 °
A^ 15 mm
B
The tower crane is used to hoist a 2-Mg load upward at con- stant velocity. The 1.5-Mg jib BD and 0.5-Mg jib BC have centers of mass at G 1 and G 2 , respectively. Determine the required mass of the counterweight C so that the resultant moment produced by the load and the weight of the tower crane jibs about point A is zero. The center of mass for the counterweight is located at G 3.
SOLUTION
a
MC = 4966.67 kg = 4.97 Mg Ans.
- (MR)A = ©Fd; 0 = MC(9.81)(7.5) + 500(9.81)(4) - 1500(9.81)(9.5) - 2000(9.81)(12.5)
C
G (^2) B (^) D
G 3
A
9.5m
7.5 m
4 m
G 1 12.5 m
23 m
Determine the magnitude of the force F^ that
should be applied at the end of the lever such
that this force creates a clockwise moment M
about point O.
Given:
M 15 N m
60 deg
30 deg
a 50 mm
b 300 mm
Solution:
M F cos ^ ^ ^ a b sin^ ^ F sin ^ ^ ^ b cos^
F
M
cos^ ^ a^ b sin ^ ^ sin ^ ^ ^ b cos^
F 77.6 N
Ans.
