Common Data:
In orthogonal turning of a bar of $100 m m$ diameter with a feed of $0.25 m m / r e v$ , depth of cut of $4 m m$ and cutting velocity of $90 m / m i n$ , it is observed that the main (tangential) cutting force is perpendicular to the friction force acting at the chip-tool interface. The main (tangential) cutting force is $1500 N$ .

The normal force acting the chip-tool interface in N is

1000

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1500

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

2000

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2500

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $1500$
Normal force,
$N = F_{C}$
$N = 1500 N$

Suggest Corrections

Similar questions

Q. Common Data:
In orthogonal turning of a bar of

100 m m

diameter with a feed of

0.25 m m / r e v

, depth of cut of

4 m m

and cutting velocity of

90 m / m i n

, it is observed that the main (tangential) cutting force is perpendicular to the friction force acting at the chip-tool interface. The main (tangential) cutting force is

1500 N

.

The orthogonal rake angle of the cutting toll in degree is

Common Data:
In an orthogonal machining operation:
Uncut thickness $= 0.5 m m$
Cutting speed $= 20 m / m i n$
Rake angle $= 15^{o}$
Width of cut $= 5 m m$
Chip thickness $= 07 m m$
Thrust force $= 200 N$
Cutting force $= 1200 N$
Assume Merchant's theory.

The coefficient of friction at the tool-chip interface is