### Problems 2

```7. The turnbuckle T is tightened until the tension in cable OA is 5 kN.

Express the force F acting on point O as a vector. Determine the projection

of F onto the y-axis and onto line OB. Note that OB and OC lies in the x-y
plane.
8.
Determine
the
magnitude
and
direction angles of the resultant force
acting on the bracket.
Resultant

  
R   F  F1  F2




F1  450
cos
45 sin 30i  450
cos
45 cos 30 j  450 sin 45k


 F1 xy
 F1 xy




F1  159.1i  275.57 j  318.2k
Direction angles for 𝑭 𝟐
 x  45
 y  60
 z  90
Direction cosines
cos 2  x  cos 2  y  cos 2  z  1
l
2

 m2  n2  1
cos 2 45  cos 2 60  cos 2  z  1
cos 2  z  0.25 
cos  z  0.5
 z  90

cos  z  0.5 
 z  120




F2  600 cos 45i  600 cos 60 j  600 cos120k




F2  424.26i  300 j  300k




F1  159.1i  275.57 j  318.2k

  
Resultant R   F  F1  F2




F2  424.26i  300 j  300k




R   159.1  424.26 i  275.57  300  j  318.2  300 k




R  265.16i  575.57 j  18.2k
Magnitude of Resultant Force

R  R  265.16 2  575.57 2  18.2 2  633.97 N
Direction Cosines of Resultant Force
Ry
Rx
cos  x 
cos  y 
R
R
265.16
cos  x 
 0.418
633.97
Rz
cos  z 
R
575.57
cos  y 
 0.907
633.97
cos  z 
18.2
 0.029
633.97
Direction angles for 𝑹
 x  arccos(0.418)  65.3
 y  arccos(0.907)  24.9
 z  arccos(0.029)  88.3
9. Determine the parallel and normal components of force 𝐹 in vector form with
respect to a line passing through points A and B.
Cartesian components of 𝑭
5
Fxy  75
5 3
2
2
 64.311 kN
A
Fx  Fxy cos 50  41.34 kN
Fz  75
5 3
2
2
 38.59 kN




F  41.34i  49.265 j  38.59k
(6, 4, 8) m
(3, 2, 5) m

F
Fy  Fxy sin 50  49.265 kN
3
B
z
5
x
50°
y
3
F=75 kN




F  41.34i  49.265 j  38.59k
B
z
Unit vector of line AB







3i  2 j  3k
n AB 
 0.639i  0.426 j  0.639k
2
2
2
3 2 3

n AB
A
(3, 2, 5) m

F
y
Parallel component of 𝑭 to line AB
3
5
(its scalar value) :
 
50°
x
F//  F  n AB






F//  41.34i  49.265 j  38.59k  0.639i  0.426 j  0.639k


(6, 4, 8) m
F=75 kN

F//  41.34 0.639   49.2650.426   38.59 0.639   72.14 kN
Parallel component of 𝑭 to line AB (in vector form):








F//  F// n AB  72.14 0.639i  0.426 j  0.639k  46.14i  30.76 j  46.14k


Normal component of 𝑭 to line AB (in vector form):


 


F  F  F//  4.8i  18.505 j  7.55k
10. An overhead crane is used to reposition the boxcar within a
railroad car-repair shop. If the boxcar begins to move along the rails
when the x-component of the cable tension reaches 3 kN, calculate
the necessary tension T in the cable. Determine the angle xy between
the cable and the vertical x-y plane.
Tx=3 kN, calculate tension T, the angle xy between the cable and the vertical x-y plane.
Unit vector of 𝑻

nT 

 
5i  4 j  k
52  4 2  12




nT  0.77i  0.617 j  0.154k
𝑻 in vector form





T  T 0.77i  0.617 j  0.154k

x-component of 𝑻
0.77T  3

T  3.896 kN (magnitude of 𝑻)
y and z-components of 𝑻
Ty  0.6173.896   2.4 kN




T  3i  2.4 j  0.6k
Tz  0.1543.896   0.6 kN
Txy  Tx2  Ty2  32  2.4 2  3.84 kN
Txy 3.84
cos  xy 

 0.896 
 xy  9.598
T
3.89
11. The spring of constant k = 2.6 kN/m is attached to the disk at point A and
to the end fitting at point B as shown. The spring is unstretched when A and
B are both zero. If the disk is rotated 15° clockwise and the end fitting is
rotated 30 ° counterclockwise, determine a vector expression for the force
which the spring exerts at point A.
12. The rectangular plate is supported by hinges along its side BC and
by the cable AE. If the cable tension is 300 N, determine the projection
onto line BC of the force exerted on the plate by the cable. Note that E is
the midpoint of the horizontal upper edge of the structural support.
If T=300 N, determine the projection onto line BC of the force exerted on the plate by the cable.
𝑪𝒐𝒐𝒓𝒅𝒊𝒏𝒂𝒕𝒆𝒔 𝒐𝒇 𝒑𝒐𝒊𝒏𝒕𝒔 𝑨, 𝑩, 𝑪 𝒂𝒏𝒅 𝑬 𝒘𝒊𝒕𝒉 𝒓𝒆𝒔𝒑𝒆𝒄𝒕 𝒕𝒐 𝒄𝒐𝒐𝒓𝒅𝒊𝒏𝒂𝒕𝒆 𝒔𝒚𝒔𝒕𝒆𝒎
𝑨 400, 0, 0
B (0, 0, 0)
𝑪 0, 1200𝑠𝑖𝑛25, −1200𝑐𝑜𝑠25
C (0, 507.14, 1087.57)
𝑬 0, 1200𝑠𝑖𝑛25, −600𝑐𝑜𝑠25
E (0, 507.14, 543.78)




  400i  507.14 j  543.78k 


T  TnT  300

844.33






T  142.12i  180.19 j  193.21k
y
x
z
Unit vector of 𝒍𝒊𝒏𝒆 𝑩𝑪





nBC   cos 25k  sin 25 j  0.423 j  0.906k
Projection of T onto line BC
TBC

y

z


 



 T  nBC   142.12i  180.19 j  193.21k  0.423 j  0.906k
TBC  251.26 N

nBC

25o
25o
13. The y and z scalar components of a force are 100 N and 200 N,
respectively. If the direction cosine l=cosx of the line of action of the
force is 0.5, write 𝐹 as a vector.
Fy  100 N
Fz  200 N
l 2  m2  n2  1

l  0.5
m 2  n 2  1  0.52

m 2  n 2  0.75
Fy2  Fz2  223.61 N  Fyz 
Fy  F cos  y  Fm
Fz  F cos  z  Fn
F 0.75  223.61

F  258.2 N
Fx  F cos  x  129.1 N




F  129.1 i  100 j  200k
```