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.

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 torque supplied to the shaft.

A

453.5N-m

B

549.3N-m

C

657.3N-m

D

None of the listed

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. If allowable shear stress in the shaft is 70N/mm² and torsional and bending moments are M=1185000N-mm and m=330000N-mm, find the diameter of the shaft.

A

36.8mm

B

39.7mm

C

44.7mm

D

40.3mm

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. If bending moment on point B in horizontal plate is M and in vertical plane is m, then the net bending moment at point B is?

A

M

B

m

C

M+m

D

√M²+m²

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)

A

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

B

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

C

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

D

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

Consider the following statements in connection with thrust bearings: 1) Cylindrical thrust bearings have higher coefficient of friction than ball thrust bearings. 2) Taper rollers cannot be employed for thrust bearings. 3) Double-row thrust ball bearing is not possible 4) Lower race, outer race and retainer are readily separable in thrust bearings. Which of the above statements are correct?

A

1 and 2

B

2 and 3

C

3 and 4

D

1 and 4

Which of the following is NOT true about Frictional force? A. Friction is the force which opposes the relative motion of two surfaces in contact. B. The force of friction that acts when a body is moving (sliding) on a surface is called sliding friction. C. Friction in machines wastes energy and also causes wear and tear. D. Rolling friction is much more than sliding friction, the use of ball bearings in a machine considerably reduces friction.

A

D

B

A

C

D

B

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))

A

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

B

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

C

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

D

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

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))

A

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

B

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

C

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

D

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

If the angle of wrap on smaller pulley of diameter 250 mm is 1200 and diameter of larger pulley is twice the diameter of smaller pulley, then the centre distance between the pulleys for an open belt drive is

A

1000 mm

B

750 mm

C

500 mm

D

250 mm

A direct rope haulage pulls 8 tubs loaded with coal through an incline of length 500 m having an inclination of 1 in 6. Consider the following additional data.
Capacity of tub = 1.0 tonne.
Tare weight of tub = 500 kg
Hauling speed = 9 km per hour
Coefficient of friction between wheel and rail = $$\frac{1}{{60}}$$
Coefficient of friction between rope and drum = $$\frac{1}{{10}}$$
Mass of rope per meter = 1.5 kg
The minimum power required to haul the tubs in kW is

A

345.50

B

348.60

C

350.10

D

365.50

E

58.86

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