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A Single slider crank mechanism:

Given: Crank AB = 20 mm = 0.02 m, C. R. BC = 80 mm = 0.08 m

N = 1000 rpm, ωBA= 2πN/60 = 2π x 1000/60 = 104.7 rad/sec

VBA = ωBA x AB = 104.7 x 0.02 = 2.09 m/s

From velocty diagram: Velocity of C w.r.t. B -

VCB = vector cb = 1.15 m/s

Angular velocity of Connecting rod ‘BC’ ωCB = VCB / CB = 1.15/0.08 = 14.375 rad /sec

Velocity of slider ‘C’

VC= vector ac = 2 m/sec

Numerical Problem-A casting weighing 9 kN hangs freely from a rope which makes 2.5 turns ............

Given: W= T1= 9 kN =9000N, d= 0.3 m, N = 20 rpm , µ= 0.25

(i) Force reqd. by a man 

T2- force reqd. by man

As rope makes 2.5 turns,

Therefore angle of contact ,

θ =2.5x2π = 5 π rad.

We know that,

2.3 log {T1/T2} = µ θ = 0.25 x 5 π = 3.9275

log {T1/T2} = 3.9275/2.3 = 1.71 or T1/T2 = 51

T2 = 9000/51 = 176.47 N

(ii) Power to raise casting : 

For a single slider crank mechanism , state the formulae to calculate by analytical method – Also state the meaning of each term.

i) Velocity of slider:

                                Vp = w.r [sinθ + sin2θ/2n ]

              where,

                  Vp - velocity of slider

w- angular velocity

θ – angle of crank to line of stroke ‘PO’

n- l/r = ratio of length of connecting rod to crank radius.

ii) Acceleration of slider:

               fp = w2 r(cos θ + cos2θ/n)

                 where, fp – acceleration of slider

iii) Angular velocity of connecting rod.:

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