If P1 and P2 are the tight and slack side tensions in the belt, then the initial tension Pi (neglecting centrifugal tension) will be equal to (Where, Pc is centrifugal tension)

$${{\text{P}}_1} - {{\text{P}}_2}$$

$${{\text{P}}_1} + {{\text{P}}_2}$$

$$\frac{1}{2}\left( {{{\text{P}}_1} + {{\text{P}}_2}} \right)$$

$$\frac{1}{2}\left( {{{\text{P}}_1} + {{\text{P}}_2}} \right) + {{\text{P}}_{\text{c}}}$$

Correct Answer: $$\frac{1}{2}\left( {{{\text{P}}_1} + {{\text{P}}_2}} \right)$$

The ratio of driving tensions for flat belts, neglecting centrifugal tension, is (where T₁ and T₂ = Tensions on the tight and slack sides of belt respectively, $$\mu $$ = Coefficient of friction, between the belt and pulley and $$\theta $$ = Angle of contact)

$$\frac{{{{\text{T}}_1}}}{{{{\text{T}}_2}}} = \mu \theta $$

$$\log \frac{{{{\text{T}}_1}}}{{{{\text{T}}_2}}} = \mu \theta $$

$$\frac{{{{\text{T}}_1}}}{{{{\text{T}}_2}}} = {\text{e}}\mu \theta $$

$$\frac{{{{\text{T}}_1}}}{{{{\text{T}}_2}}} = \log \mu \theta $$

If T₁ and T₂ are the tensions on the tight and slack sides of the belt respectively and T_C is the centrifugal tension, then initial tension in the belt is equal to

T1 - T2 + TC

T1 + T2 + TC

$$\frac{{{{\text{T}}_1} - {{\text{T}}_2} + {{\text{T}}_{\text{C}}}}}{2}$$

$$\frac{{{{\text{T}}_1} + {{\text{T}}_2} + {{\text{T}}_{\text{C}}}}}{2}$$

If P₁ and P₂ are the tight and slack side tensions in the belt, then the initial tension P_i (according to Barth) will be equal to (Where, P_c is centrifugal tension)

$${\left( {\frac{{\sqrt {{{\text{P}}_1}} + \sqrt {{{\text{P}}_2}} }}{2}} \right)^2}$$

$${{\text{P}}_1} + {{\text{P}}_2}$$

$$\frac{1}{2}\left( {{{\text{P}}_1} + {{\text{P}}_2}} \right)$$

$$\frac{1}{2}\left( {{{\text{P}}_1} + {{\text{P}}_2}} \right) + {{\text{P}}_{\text{c}}}$$

Power is transmitted by an electric motor to a machine by using a belt drive. The tensions on the tight and slack side of the belt are 2200 N and 1000 N respectively and diameter of the pulley is 600 mm. If speed of the motor is 1500 r.p.m, find the power transmitted.

46.548 kW

56.548 kW

66.548 kW

76.548 kW

The tension on the slack side of the belt is __________ the tension on the tight side of the belt.

Equal to

Less than

Greater than

None of these

The ratio of belt tensions $$\frac{{{{\text{p}}_1}}}{{{{\text{p}}_2}}}$$ considering centrifugal force in flat belt is given by where,
m = mass of belt per meter (kg/m)
v = belt velocity (m/s)
f = coefficient of friction
$$\alpha $$ = angle of wrap (radians)

$$\frac{{{{\text{p}}_1} - {\text{m}}{{\text{v}}^2}}}{{{{\text{p}}_2} - {\text{m}}{{\text{v}}^2}}} = {{\text{e}}^{{\text{f}}\alpha }}$$

$$\frac{{{{\text{p}}_1}}}{{{{\text{p}}_2}}} = {{\text{e}}^{{\text{f}}\alpha }}$$

$$\frac{{{{\text{p}}_1}}}{{{{\text{p}}_2}}} = {{\text{e}}^{ - {\text{f}}\alpha }}$$

$$\frac{{{{\text{p}}_1} - {\text{m}}{{\text{v}}^2}}}{{{{\text{p}}_2} - {\text{m}}{{\text{v}}^2}}} = {{\text{e}}^{ - {\text{f}}\alpha }}$$

In a band and block brake, the ratio of tensions on tight side and slack side of the band is (where $$\mu $$ = Coefficient of friction between the blocks and the drum, $$\theta $$ = Semi-angle of each block subtending at the center of drum and n = Number of blocks)

$$\frac{{{{\text{T}}_1}}}{{{{\text{T}}_2}}} = \mu \theta {\text{n}}$$

$$\frac{{{{\text{T}}_1}}}{{{{\text{T}}_2}}} = {\left( {\frac{{1 - \mu \tan \theta }}{{1 + \mu \tan \theta }}} \right)^{\text{n}}}$$

$$\frac{{{{\text{T}}_1}}}{{{{\text{T}}_2}}} = {\left( {\mu \theta } \right)^{\text{n}}}$$

$$\frac{{{{\text{T}}_1}}}{{{{\text{T}}_2}}} = {\left( {\frac{{1 + \mu \tan \theta }}{{1 - \mu \tan \theta }}} \right)^{\text{n}}}$$

A counter shaft of engineering workshop is being driven by electric motor with the help of canvas belt and pulley arrangement such that the speed of belt is 600 m/min and it transmits 50 KW. What will be the difference between tensions on either side of the belt?

10000

100

5000

In a band and block brake, the ratio of tensions on the tight and slack sides of band is given by (where $$\mu $$ = Coefficient of friction between the blocks and the drum, $$\theta $$ = Semi-angle of each block subtending at the centre of drum and n = Number of blocks)

$$\frac{{{{\text{T}}_1}}}{{{{\text{T}}_{\text{2}}}}} = \mu \theta \times {\text{n}}$$

$$\frac{{{{\text{T}}_1}}}{{{{\text{T}}_2}}} = {\left( {\mu \theta } \right)^{\text{n}}}$$

$$\frac{{{{\text{T}}_1}}}{{{{\text{T}}_2}}} = {\left( {\frac{{1 - \mu \tan \theta }}{{1 + \mu \tan \theta }}} \right)^{\text{n}}}$$

$$\frac{{{{\text{T}}_1}}}{{{{\text{T}}_2}}} = {\left( {\frac{{1 + \mu \tan \theta }}{{1 - \mu \tan \theta }}} \right)^{\text{n}}}$$

When the belt is transmitting maximum power, the belt speed should be
(Where, m = mass of belt per meter (kg/m) and Pmax = maximum permissible tension in belt (N))

$$\sqrt {\frac{{{\text{P}}\max }}{{2{\text{m}}}}} $$

$$\sqrt {\frac{{{\text{P}}\max }}{{3{\text{m}}}}} $$

$$\sqrt {\frac{{{\text{P}}\max }}{{\text{m}}}} $$

$$\sqrt {\frac{{{\text{3m}}}}{{{\text{P}}\max }}} $$