Question

For a right-hand offset disc harrow, the longitudinal distance from the hitch point to the centers of the front and rear gangs are 2.5 m and 4.5 m, respectively. The resultant horizontal soil forces in longitudinal direction on the front and rear gangs are 3.0 kN and 3.5 kN, respectively; while the resultant horizontal soil forces in lateral directions are 2.5 kN and 4.0 kN, respectively. Considering the resultant soil forces acting at the centers of front and rear gangs, the amount of offset of the center of cut with respect to the hitch point is _________ m. (Rounded off to 2 decimal places)

Hint:

To calculate the offset, sum the moments of the forces at each gang and divide by the total lateral force to determine the distance of the center of cut from the hitch point.

Answer 1

We are given: Longitudinal distance from hitch to front gang center \( L_1 = 2.5 \, {m} \),
Longitudinal distance from hitch to rear gang center \( L_2 = 4.5 \, {m} \),
Horizontal soil force in longitudinal direction on the front gang \( F_{L1} = 3.0 \, {kN} \),
Horizontal soil force in longitudinal direction on the rear gang \( F_{L2} = 3.5 \, {kN} \),
Horizontal soil force in lateral direction on the front gang \( F_{T1} = 2.5 \, {kN} \),
Horizontal soil force in lateral direction on the rear gang \( F_{T2} = 4.0 \, {kN} \).
Step 1: Calculate the moments about the hitch point for both gangs.
The moment for the front gang is:

\[ M_1 = F_{T1} \times L_1 = 2.5 \times 2.5 = 6.25 \, {kN.m}. \]

The moment for the rear gang is:

\[ M_2 = F_{T2} \times L_2 = 4.0 \times 4.5 = 18.0 \, {kN.m}. \]

Step 2: Calculate the total moment about the hitch point.
The total moment \( M_{{total}} \) is the sum of the moments from both gangs:

\[ M_{{total}} = M_1 + M_2 = 6.25 + 18.0 = 24.25 \, {kN.m}. \]

Step 3: Calculate the total horizontal force in the lateral direction.
The total horizontal force \( F_{{total}} \) is the sum of the lateral forces:

\[ F_{{total}} = F_{T1} + F_{T2} = 2.5 + 4.0 = 6.5 \, {kN}. \]

Step 4: Calculate the offset (distance of the center of cut from the hitch).
The offset of the center of cut \( d \) is calculated by dividing the total moment by the total lateral force:

\[ d = \frac{M_{{total}}}{F_{{total}}} = \frac{24.25}{6.5} \approx 3.73 \, {m}. \]

Thus, the offset of the center of cut with respect to the hitch point is 3.73 m.