Engineering Mechanics

Chapter 34 : Transmission of Power by Gear Trains „„„„„ 711

Speed of wheel B when wheel A is fixed

Since the speed of arm is 150 r.p.m. clockwise, therefore from the fourth row of the table

y= + 150

Moreover, as the wheel A is fixed, therefore

x + y=0

or x=– y = – 150

∴ Speed of wheel B

36

• 150 – – 150
  45

A

B

T

yx

T

==⎛⎞×

⎜⎟

⎝⎠

= 270 r.p.m. Ans.

Speed of wheel B when wheel A makes 300 r.p.m.

Since the wheel A makes 300 r.p.m. anticlockwise, therefore from the fourth row of
the table

x + y= – 300

or x= – 300 – y = – 300 – 150 = – 450

∴ Speed of wheel B

36

• 150 – – 450
  45

A

B

T

yx

T

==⎛⎞×

⎜⎟

⎝⎠

= 510 r.p.m. Ans.

Algebraic method

Let NA= Speed of wheel A,

NB= Speed of wheel B, and

NC= Speed of arm C.

First of all, assume the arm C to be fixed, speed of the wheel A relative to the arm C

=NA – NC

and speed of wheel B relative to arm C

=NB – NC

Since the wheels A and B revolve in opposite directions, therefore

BC

AC

NN

NN

• A
  B

T

T

= ...(i)

Speed of wheel B when wheel A is fixed

When wheel A is fixed, the arm rotates at 150 r.p.m. in the clockwise direction. Therefore

NA=0

and NC= + 150 r.p.m. ...(clockwise)

Substituting this value of NC in equation (i),

• 150
  0–150

NB ––– 36 4

45 5

A

B

T

T

===

∴ NB=

4

150 × + 150 = 270 r.p.m.

5

⎛⎞

⎜⎟⎝⎠ Ans.