Engineering Mechanics

(^682) „„„„„ A Textbook of Engineering Mechanics
R=
222
TTTδθ δ δθ δθ
++ = T δθ ...(ii)
... Neglecting
2
⎛⎞δδθT
⎜⎟
⎝⎠
Now resolving the forces vertically,
μ × R=()cos–cosTT 22 T
⎛⎞ ⎛⎞δθ δθ
+δ ⎜⎟ ⎜⎟
⎝⎠ ⎝⎠ ...(iii)
Since the angle δθ is very small, therefore substituting cos 1
2
⎛⎞δθ
⎜⎟=
⎝⎠
in equation (iii),
μ × R=T + δT – T = δT
∴ R=
δT
μ
...(iv)
Equating the values of R from equations (ii) and (iv),
T δθ=
δT
μ
or
T
T
δ
=μ δθ
Integrating both sides from A to B,
i.e.,
1
2
T
T
T
T
δ
∫ =
0
θ
μ∫δθ
or
1
2
loge
T
T
⎛⎞
⎜⎟⎜⎟
⎝⎠
=μ θ ...(v)
∴
1
2
T
T
⎛⎞
⎜⎟⎜⎟
⎝⎠
=eμθ
The equation (v) may also be expressed in terms of corresponding logarithm to the base 10 i.e.
1
2
2·3 log
T
T
⎛⎞
⎜⎟⎜⎟
⎝⎠
=μ θ
The above expression gives the relation between the tight side and slack side tensions, in
terms of coefficient of friction and the angle of contact.
Notes: 1. In the above expression (θ) is the angle of contact at the smaller pulley.

2. In an open belt drive, the angle of contact,
   θ= (180° – 2α)


3. In a cross-belt drive, the angle of contact,
   θ= (180° + 2α)
   Example 33.8. Find the power transmitted by a belt running over a pulley of 600 mm
   diameter at 200 r.p.m. The coefficient of friction between the belt and pulley is 0·25, angle of lap
   160° and maximum tension in the belt is 2·5 kN.
   Solution. Given: Diameter of pulley (d) = 600 mm = 0·6 m ; Speed of the pulley (N) = 200

r.p.m.; Coefficient of friction (μ) = 0·25; Angle of lap (θ) = 160° 160 2·79 rad

180

π

=°× =

°

and

maximum tension (T 1 ) = 2·5 kN.