**1 Answer**

written 2.1 years ago by |

If the chip is isolated as a free body it can be seen that the chip is equilibrium under the action of only two forces R and R as shown in figure.

Force R is the force between the chip and the tool face and is a resultant of normal and tangential component of forces N and F on the tool face.

Similarly, R’ is the force between the chip and work piece and is a resultant of normal and tangential component’s Ns and Fs on the shear plane.

For equilibrium R and R’ must be equal and opposite. Actually they may not be col-linear but the small couple due to this can be neglected.

If the force R is assumed to act at the top tip instead of at its actual point of application on the tool face a compact force diagram shown in figure can be obtained. This type of plane was first suggested by Merchant (1942) and is known as Merchants diagram.

It is widely used foe analysis of the orthogonal cutting process.

In this plot, circle AD is drawn from the top of the tool with AD=R.

The resultant force R can then be resolved in any desired direction.

The three most convenient directions as shown in the figure are:

1] Along and at right angles to tool F and N.

2] Along and at right angles to the shear plane Fs and Ns.

3] Horizontal and vertical directions Fp and Fq.

Force Fp is the main cutting force on the tool and act in the direction of tool travel. Fq is the force at right angles to the cutting force.

Fp and Pq can be easily measured with the help of a two component force dynamo-meter.

Once Fp and Fq are known all other force components can easily be calculated from the force circle.

**NOTE:**

$1. F=F_{p} \cdot \sin \gamma+F_{q} \cdot \cos \gamma$

$2. N=F_{p} \cdot \cos \gamma-F_{q} \cdot \sin \gamma$

$3. F_{s}={F}_{p} \cdot \cos \emptyset-F_{q} \cdot \sin \emptyset$

$4. N_{s}=F_{p} \cdot \sin \emptyset+F_{q} \cdot \cos \emptyset=F_{s} \cdot \tan (\emptyset+\beta-\gamma)$