Engineer Science Banding - Essay Example

Solution # 1st:-

(A)    Frictional Force:  we can calculate the maximum frictional force using

F=mN

Where
- F is the frictional force,
- m is the "coefficient of friction", a number we use for those particular surfaces,
- N is the "Normal reaction", which is the force pressing the surfaces together.

Now,          F = 0.3* mg                                    (N = mg)

                 F = 0.3* 300* 9.8                             (m = 300 kg)

Basically, torque = F * r

Torque = 882 * 0.375

Torque = 330.75 N-m

And in special case of shaft rotation torque can be calculated by using these

Power = torque * angular speed                           

Power = torque * 2pi * rev/min

Now by using table, we get the values

510 = torque * 2 * 3.142 * 1500

Torque = 0.0541 N-m

Solution # 1st:-

(B)    (1)             V (final) = v (initial) + a * t                        F = m * a     

                      400 = 0 + 2.94 * t                                       882 = 300 * a

                     400 / 2.94 = t                                               882 / 300 = a

                      t = 136 sec                                                     a = 2.94

(2)               S = V (avg) * t                                                         V (avg) = v (initial) + v (final) / 2

          S =  200 * 136                                                                    V (avg) = 400 / 2 

       S =  3.33 * 136                                                     V (avg) = 200  rev/min       

       S = 453.33                                                            V (avg) = 3.33 rev/sec

Where,

V is angular speed

S is the number of revolutions

1. no 4 (Solution)

(A)          Work done =  Force * height                                    Torque = force * radius                                                                       

             W = 357.14 * 28                                           250 = force * 700/1000

            W = 12.755 Joule                                           250 = force *  0.7

                                                                                  250 / 0.7 = force

  Force = 357.14 N

• 2as = v (final)2 – v (initial)2

As we know that in motion under gravity a is replaced by g

 Where,      

  a = acceleration

  g = gravity

now,                      2gs = v (final)2 – v (initial)2

                                   2 * 9.8 * 28 = v ( final)2 – 0

                                  192.08 = v (final)2

                             Taking square root on both sides

                            V (final) = Linear velocity = 13.85 m/s

• V (final) = v (initial) + g * t

13.85 = 0 + 9.8 * t

13.85 / 9.8 = t

t = 1.41 sec

1. no 2 (solution)

(a)           V (final) = v (initial) + a * t                               Force = mass * acceleration  

• = 0 + a * 240 F = 1400 * 2   

480 / 240 = a                                         F = 2800 N

a = 2

Now,

             Torque = Force * radius

               Torque = 2800 * 0.75

            Torque = 2100 N-m  

s = 16 rev/sec

s = 960 rev /min

And, when the system cascaded(carries sharing operation)

the table shows

s =  480 rev/min    

So, it means the system efficiency reduces to 50 %.