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?

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.

A

6778.3N and 7765.3N

B

5948.15N and 2288.75N

C

5468.4N ad 8678.3N

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. 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 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

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

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

Two persons are holding a rope of negligible mass horizontally. A 20 kg mass is attached to the rope at the mid-point, as a result, the rope deviates from the horizontal direction. The tension required to completely straighten the rope is (g = 10 m/s²)

A

200 N

B

20 N

C

10 N

D

infinitely large

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

The radius of a friction circle for a shaft rotating inside a bearing is (where r = Radius of shaft and $$\tan \varphi $$ = Coefficient of friction between the shaft and bearing)

A

$${\text{r}}\sin \varphi $$

B

$${\text{r}}\cos \varphi $$

C

$${\text{r}}\tan \varphi $$

D

$$\frac{{\text{r}}}{2}\cos \varphi $$

Figure shows a two pulley system for hoisting a load of 10 kN. The coefficient of friction between each pulley and the rope is 0.2. The vertical and horizontal distances between the centers of the pulleys are 25m and 16m respectively. The tensions T₁ and T₂ respectively in kN are

A

6.00, 5.38

B

12.37, 11.06

C

18.74, 16.73

D

25.11, 22.41