3. Dynamic balance of tractor      return

 

3-1. Pulling operation (2-dimensional case)

 

 

Figure 3-1 Pulling operation

 

From Figure 3-1, where H is the thrust force, and P is the pulling force, the following three equations can be established:

 

                                                  (1)

                          (2)

                  (3)

     From Eq.(1):

                            (4)

 

 

 

that is, the thrust force H has a component along the x-axis which is equal and opposite to the horizontal component of the pulling force P.

 

     Also, from [Eq. (2) × Cx + Eq.(3) ]

 

             

And:

                    (5)

Substituting Eq.(2) into Eq.(5):

                    (6)

              Equations (4), (5) and (6) are general solutions for two-dimensional cases of pulling operation.

              Considering the following values defining the relative positions of points:

 

              Gx - Cx = a,       Bz = h,       Ex - Cx = l,       Cx - Bx = b,

              Bx − Ex = −(l + i),

 

and ( Px < 0, Pz < 0 ), substitution into the equations will lead into,

 

                            (7)

                            (8)

 

     At stationary condition, the values of R1z and R2z are Wz(b/l) and Wz(a/l) respectively. If these are compared with those of (7) and (8):

                                 (9)

                                (10)

such that, with regards to the weight transfer, the supporting force on the rear wheel increases by:

             

due to the horizontal component of pull and by:

due to the vertical component of pull.

     Likewise, the supporting force on the rear wheel decreases by:

               

     Also, the direction of pull with respect to the horizontal is given by angle b, where:

                                                                       (11)

and, if the distance between point C and the point of application of the pull is d, then, with:

             

therefore:

             

such that Eqs.(7) and (8) becomes:

              　　    (12)

              　　　　　  (13)

such that the supporting force on the rear wheel increases by:

             

and the supporting force on the front wheel decreases by:

             

Sample problem:

     If Pz = 0, how much is the weight transfer? Also, determine the condition for the turnover of the tractor.

 

Solutions:

     With Pz = 0 in Eq.(9) and (10):

             

             

 

     Also, when the value of the supporting force on the front wheel becomes negative, then the tractor would turnover to the rear:

             

 Therefore,

             

which means that when the weight Wz is small, or when the center of gravity G is very near to the rear, or when the point of application of the pulling force is high, the tractor would easily turnover.

 

Sample problem:

     The pulling force P = 400 kgf is applied 50cm above axis of the rear wheel at an angle of 30 degrees.Determine the loads to the front and rear wheels and the weight transfer. The tractor weight is 800 kgf, the center of gravity is 80 cm from the axis of the rear wheel and the distance between the axes of the front and rear wheels is l = 180 cm.

 

Solutions:

     For the front and rear wheels:

             

             

 

     The load to the front wheel was reduced by 96.2 kgf and the load to the rear wheel was increased by 296.2 kgf.