Question

An inclined blade tillage tool of 25 cm wide and 10 cm long is operating at 25 cm depth in cohesionless soil with density equal to 1.25 \(\text{g}\cdot\text{cm}^{-3}\), and angle of internal friction of 39\(^\circ\). The normal load on the tool and coefficient of soil-metal friction are 1000 N and 0.3, respectively. The soil cutting resistance per unit length of cutting edge is 25 \(\text{N}\cdot\text{mm}^{-1}\). If the tool lift angle is 30\(^\circ\), the specific draft will be [Given: \(\cos 30^\circ = 0.866\)]

• A. 1385 N
• B. 885 N
• C. 760 N
• D. 260 N

Hint:

To save time on calculation:
\(1000 \times 0.5 = 500\).
\(300 \times 0.866 \approx 260\).
Sum = \(760\text{ N}\). This matches the target option perfectly.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The draft of a tillage tool is the force component acting parallel to the direction of travel.
For an inclined blade, normal and frictional forces on the face combine to create horizontal draft resistance.

Step 2: Key Formula or Approach:
The draft contribution \(D\) from the normal force \(N_s\) and frictional force \(F\) is:
\[ D = N_s \sin\alpha + F \cos\alpha \] where:
\(\alpha\) = tool lift angle
\(F = \mu \times N_s\) (frictional force)

Step 3: Detailed Explanation:
Given values:
- Normal load (\(N_s\)) = \(1000\text{ N}\)
- Coefficient of soil-metal friction (\(\mu\)) = \(0.3\)
- Lift angle (\(\alpha\)) = \(30^\circ\)
Calculate the frictional force \(F\):
\[ F = 0.3 \times 1000 = 300\text{ N} \] Now, calculate the draft component:
\[ D = 1000 \sin 30^\circ + 300 \cos 30^\circ \] Using \(\sin 30^\circ = 0.5\) and \(\cos 30^\circ = 0.866\):
\[ D = (1000 \times 0.5) + (300 \times 0.866) \] \[ D = 500 + 259.8 = 759.8\text{ N} \approx 760\text{ N} \]

Step 4: Final Answer:
The correct option is 3, which corresponds to 760 N.