![the cutting tool on the lathe exerts a force the cutting tool on the lathe exerts a force](https://www.hobby-machinist.com/data/attachments/139/139822-e8accef148c79af5ae06c588966dc183.jpg)
Depth of cut depends upon cutting speed, rigidity of machine- tool and tool material etc.
![the cutting tool on the lathe exerts a force the cutting tool on the lathe exerts a force](https://www.hobby-machinist.com/data/attachments/139/139823-3a83c9d21522e33bd051ab6b7c91b221.jpg)
It is the advancement of tool in the job in a direction perpendicular to the surface being machined. The rate at which the tool is fed depends upon various factors such as finish required, depth of cut and the rigidity of the machine, e.g., a high rate of feed will get the job done in less time but will give a rough finish and will take more time to drive a slower rate will give a better finish but will take a longer time. It is given in mm per revolution of the job. It is the amount of tool advancement per revolution of job parallel to the surface being machined. The selection of proper cutting conditions (i.e., speed, feed, depth of cut, coolant, etc.) plays a vital role in the economy of production. Where D = Diameter of job in mm, N = Spindle or job speed in R.P.M. Mathematically, cutting speed = (πDN / 1000) metres/minute. While determining the cutting speed for any material, several factors are to be taken into account e.g., the type of job and tool, condition of machine, type of cut required (roughing or finishing), machinability of material, tool materials, presence of hard scale and so on. In lathe, cutting speed means the number of metres measured on the circumference of job that passes the cutting edge of the tool in one minute. Tangential force P x as well as P z and P y increase considerably with increase in flank wear. Radial component P z increases for bigger nose radius resulting in tendency for increase in tool chatter, but tool life and surface finish are improved at higher feeds and depth of cut. The effect of increasing nose radius is similar to as that of reducing the approach angle. The vertical component P z increases slightly as the back rake angle increases from – ve value to + ve value. 1,480 lb F.It controls the direction of chip flow either away from or towards the workpiece depending upon whether it is + ve or – ve. If a = 3 ft, b = 2 ft and z = 600 lb/ft determine the magnitude of that force.
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Now, sum moments using scalar components, the way you normally do…Ĥ.5 Replace the distributed loading with an equivalent resultant force. (Fill in the blank on the cover answer sheet) (5 pts)Ĥ00 kN-m = (50.0 kN)(x) x = 8.0 m Tell me WHY the answer is NOT x = 6.6 m … you know this … HINT: Draw a picture of the resultant force (calculated in question 4.2) at some distance x away from point O. MR = (30)(3) + (40sin30)(5) + 210 = 400 kN-mĤ.4 Replace the force and couple system acting on the beam by only a single equivalent force and find the location, x, where this force acts on the beam measured (in meters) from point O. Remember the difference between projection planes and coordinate direction angles…ġ.4 A force is expressed as a Cartesian vector F = kN-m What is the i component of F1 ? (5 pts) A. What is the magnitude of the resultant force acting on the bracket? (5 pts) A.