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SOLUTION

VIEWS: 46 PAGES: 7

									                  SOLUTIONS OF ASSIGNMENT NO. (2)

                        COPLANER FORCE SYSTEMS
                                                                                        y


 1. Determine the magnitude and the direction of
                                                                                            400N
    the resultant force for the shown system .
     (Ans. FR = 250 N , θ=36.9º )                                                                       100N       x
                                                                                  45°               5
                                                                                            3
                                                                                                4           250N
         SOLUTION
                                   y                                    100 2 N



                                       400N
               1
    100 2 (       ) = 100 N
                2                                  100N         x
                             45°               5
                                       3                             250 (4/5) = 200 N
                                           4             250N

                   100 2 N




                1
     100 2 (       ) = 100 N           250 (3/5) = 150 N
                 2

                                                                            y
FRx = 200 + 100 − 100
    = 200 N →
                                                                    150 N
                                                                                FR
FRy = 400 − 150 − 100
                                                                                θ                       x
    = 150 N ↑
                                                                                200 N
                                                            150
FR = 200 2 + 1502        ,             θ = tan − 1        (     )
                                                            200
   = 25O N                                     = 36.9o

                                                     1
                                                                                 F             y
 2. The resultant force of the forces 50N , 75N ,
                                                                                  30º                x
     and F       is directed along the axis BC .                                               C
     Determine the magnitudes of the force F and
     the resultant force.                                                                30º   30º


         (Ans. F =89.95 N , FR=80.8 N )
                                                                                                         75N
                                                                                               50N
                                                                        B
                     SOLUTION
                                                          FR
                                                               x′
               y′           F

                                                y

                                    30º         C         x



                                          30º   30º                 50 sin 30o
                      50 cos 30o


                                                              75N
                                                                            75 sin 60o
                                                50N
                        B
                                75 cos 60o




Assuming the direction of the resultant force FR to be the x ′ - axis so that :


FRx ' = ∑ Fx '                  ,      FRy '    =0

− FR = − 50 cos 30o − 75 cos 60o                              ∴ FR = 80.8 N


   0 = F − 50 sin 30o − 75 sin 60o                         ∴ F = 89.95 N


                                                      2
                                                                                  y

   3. If the resultant force of the forces 200 N , 260
                                                                                      200N
       N ,and F is directed to the positive          x'-axis                                          x'
       and of 800 N in magnitude , determine the                    260N
                                                                                                           F
       magnitude of the force F and its orientation                         13
                                                                        5             60º    θ
       θ.                                                                                                  x
       ( Ans. F = 938.2 N , θ = 23.88º )
                 y'
                    200 sin 60o                        SOLUTION
                o
   260 sin 52.6

                                 y
                                                                                             FR = 800 N
                                     200N          200 cos 60o              x'

                    260N
                                                                        F cos θ
                                                                    F
                            13
                        5            60º       θ
                                                                    x
                o
 260 cos 52.6


                                                   F sin θ


Assuming the direction of the resultant force FR = 800 N to be the x ′ - axis so
that :

FRx' = ∑ Fx' = 800 N                       ,        FRy'       =0
F cos θ + 200 cos 60 o − 260 cos 52.6 o = 800
∴ F cos θ = 857.9 ....................................( 1 )


0 = 200 sin 60 o + 260 sin 52.6 o − F sin θ
∴ F sin θ = 379.8                 .................. ( 2 )
                                        379.8
dividing ( 2 ) by ( 1 ) : θ = tan − 1 (           ) = 23.88 o
                                        857.9
Substituti ng in ( 1 ) : F = 938.2 N


                                                      3
                                                                                       F
    4. The shown forces are acting on the bearing which is                                       3kN
       designed to       a minimum magnitude of the resultant
                                                                                           30º
       force. Determine the magnitude of the force F and the
       magnitude and direction of the minimum resultant                                                      4kN
       force.
         ( F = 4.6 KN , FR = 1.96 KN , θ = 30º )

            y'
                         3 cos 30 o
                                                                   x'
                                      3kN                     o
                                                   3 sin 30
                     F                                                4 cos 30 o
                                30º

                                                                                           FR
                                                   4kN
                                            30o                                  FRy
                                                                                                 ψ
                                                                                                           FRx
                                                                                           θ         30O


                                                     4 sin 30 o


Assuming the direction of the force F to be the y' axis so that :

FRx' = ∑ F x'               ,                     FRy' = ∑ F y'

4 cos 30 o − 3 sin 30 o = FR cosψ , ∴ FR cosψ = 1.96
                 1.964
∴     FR =                                  .......... .......... .......... ... ( 1 ) ,
                 cosψ
F − 4 sin 30 o − 3 cos 30 o = FR sinψ
∴     F − 4.6 = FR sinψ                                  .......... ........ ( 2 )

From (1) FR is minimum at
cosψ = 1 , i .e . at ψ = 0 o , then FR min = 1.96 N , in ( 1 ) ∴ F = 4.6 N
θ =ψ + 30 o = 0 + 30 o = 30 o

                                                         4
                                                                     27º                  17º
    5. I f the resultant force of the shown force
       system , acting on the cable-post , is                         40o
                                                          130N                                       50N
       directed vertically downward , determine                             70N
       the magnitudes of the force F and the                F

       resultant force.
      ( Ans. F = 88.79 N , R= 86.56 N )



                                                    50 cos 57o
                 y'
                                               x'                                    FR
   130 sin 13o                                                       y'                         x'
                                                                                    ψ
                                   40o                                              40o
                      27º                17º

                       40o   F
          130N                                      50N
                             70N
          o
130 cos 13             50o

          70 cos 50o                            o
                                                        50 sin 57o
                                   70 sin 50


                                                                     y'                         x'
  Assuming the line of action of the force F
  to be the x' – axis:
  FRx' = ∑ Fx' :                                                                    40o

  FR cosψ = F + 50 cos 57 o − 70 cos 50 o                 ψ =230o
            − 130 cos 13 o
  ∴ FR cosψ = F − 144.4 ..........( 1 )
  FRy' = ∑ F y' :
                                                                              FR
  FR sinψ = 130 sin 13 o − 50 sin 57 o
                                                          Problem(5):              ψ = 230 o
            − 70 sin 50 o
  ∴ FR sinψ = − 66.31 ...............( 2 )
                                                    5
In problem (5), if the resultant force is vertically downward , the angle
ψ = 360 o − 130 o = 230 o , then in (1) & (2) :
               66.31
∴ FR = −             o
                       = 86.56 N ,
             sin 230


F = 144.4 + 86.56 cos 230 o = 88.79 N

   6. If the resultant force of the shown force system , acting on the cable-post , is 73 N in
      magnitude      , determine the magnitudes of the force F and the direction of the resultant
      force. Show that the problem has two solutions of different results.
    ( Ans.1 : F = 174.84 N , θ1 = - 25.28º ) , ( Ans.2 : F = 114 N , θ2 = - 74.72º )


                                         SOLUTION


Using the same equations          (1) & (2) of the previous problem , and substituting
FR= 73N , then from (2) :


             66.31
sinψ = −                    ⇒ the equation has two solutions :
              73
∴ ψ 1 = 245.28 o         , ψ 2 = 294.72 o


in ( 1 ) :           ∴ F1 = 114 N          ,   F2 = 174.84 N


θ = 4O o + ψ         ∴θ 1 = 40 o + 245.28 o = 285.28 o = − 74.72 o


                       θ 2 = 40 o + 294.72 o = 334.72 o = − 25.28 o




                                               6
    7. If the resultant force of the shown force system , acting on the cable-post , is to be of a
        minimum magnitude , determine the magnitudes of the force F and the magnitude and
        direction of the resultant force.
       ( Ans. F = 144.43N, FR= 66.3 N , θ1 = - 50º )




                                            SOLUTION


Using the same equations          (1) & (2) of the previous problem , and substituting
, then from (2) :


         66.31
FR = −                     ,
         sinψ
FR is minimum if sinψ = − 1 , ψ = − 90 o


∴ FR min = 66.3 N


θ = 4 O o + ψ = 40 o − 90 o = − 50 o




                                                 7

								
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