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)

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

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

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

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

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

Evaluate the following:
$$\frac{{\cos 2\theta \cdot \cos 3\theta - \cos 2\theta \cdot \cos 7\theta + \cos \theta \cdot \cos 10\theta }}{{\sin 4\theta \cdot \sin 3\theta - \sin 2\theta \cdot \sin 5\theta + \sin 4\theta \cdot \sin 7\theta }}$$

cot6θ.cot5θ

cos6θ.cos5θ

cos6θ.cot5θ

-cot6θ.cot5θ

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

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

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

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

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

The value of $$\frac{{\sec \theta \left( {1 - \sin \theta } \right)\left( {\sin \theta + \cos \theta } \right)\left( {\sec \theta + \tan \theta } \right)}}{{\sin \theta \left( {1 + \tan \theta } \right) + \cos \theta \left( {1 + \cot \theta } \right)}}$$ is equal to:

2cosθ

cosecθsecθ

2sinθ

sinθcosθ

The layout of a shaft supported on bearings at A & B is shown. Power is supplied by means of a vertical belt on pulley B which is then transmitted to pulley C carrying a horizontal belt. The angle of wrap is 180’ and coefficient of friction is 0.3. Maximum permissible tension in the rope is 3kN. The radius of pulley at B & C is 300mm and 150mm. Calculate the tension in the rope of pulley C.

6778.3N and 7765.3N

5948.15N and 2288.75N

5468.4N ad 8678.3N

None of the listed

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

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

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

$$\frac{{{{\left( {1 + \sec \theta \,{\text{cosec}}\,\theta } \right)}^2}{{\left( {\sec \theta - \tan \theta } \right)}^2}\left( {1 + \sin \theta } \right)}}{{{{\left( {\sin \theta + \sec \theta } \right)}^2} + {{\left( {\cos \theta + {\text{cosec}}\,\theta } \right)}^2}}},$$ 0°

1 + sinθ

sinθ

cosθ

1 - cosθ

The expression $$\frac{{{{\left( {1 - \sin \theta + \cos \theta } \right)}^2}\left( {1 - \cos \theta } \right){{\sec }^3}\theta \,{\text{cose}}{{\text{c}}^2}\theta }}{{\left( {\sec \theta - \tan \theta } \right)\left( {\tan \theta + \cot \theta } \right)}},$$ 0°

2tanθ

cotθ

sinθ

2cosθ

If $$\frac{{\sec \theta - \tan \theta }}{{\sec \theta + \tan \theta }} = \frac{1}{7},\,\theta ,$$ lies in first quadrant, then the value of $$\frac{{{\text{cosec}}\,\theta + {{\cot }^2}\theta }}{{{\text{cosec}}\,\theta - {{\cot }^2}\theta }}$$ is:

$$\frac{{19}}{5}$$

$$\frac{{37}}{{19}}$$

$$\frac{{22}}{3}$$

$$\frac{{37}}{{12}}$$

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)

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

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

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

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

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)

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

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

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

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