The Crank AB Turns Uniformly At 180 rpm. In The Clockwise Direction & The Blocks At D & E Are Working In Frictionless Guides. Draw The Velocity Diagram For The Mechanism & Find The Velocities Of The Sliders D & E In Their Guides. Also Determine The Turning Moment At A If a Force Of 500N. Acts On D In The Direction Of Arrow X & a Force Of 750N. On E In The Direction Of Arrow Y.

Subject: Kinematics of Machinery

Topic: Velocity Analysis of Mechanism (Mechanism up-to 6 links)

Difficulty: High

Given: $N_{BA}$ = 180 rpm. or $\omega_{BA}$ = $2\pi\times\frac{180}{60}=18.85$ rad/s

Since AB= 0.45m, therefore velocity of B with respect to A or velocity of B (because A is a fixed point)

$v_{BA}=v_{B}=\omega_{BA}\times AB=18.85\times 0.45=8.5 m/s$

Velocities of the sliders D and E

First of all the space diagram , to some suitable scale , as shown in fig . (a) now the velocity diagram, as shown in fig.(b) is drawn as discussed below:

Vector ab = $v_{B}$ = 8.5 m/s

By measurement, we find that

Velocity of slider D, $v_{D}$ = vector ad = 9.5 m/s

Velocity of slider E, $v_{E}$ = vector ae =1.7 m/s

Turning moment at A

Let $T_{A}$ = Turning moment at A (or at the crank-shaft)

We know that force D, $F_{D}$ =500N

Force at E, $F_{E}$ = 750N

Power input = $F_{D}\times v_{D}-F_{E}\times v_{E}$.............(-ve sign indicates that FE opposes the motion)

= $500\times 95-750\times T_{A}\times 18.85T_{A}Nm/s$

Power output = $T_{A},\omega_{BA}= T_{A}\times 18.85T_{A}Nm/s$

Neglecting losses, power input is equal to power output

3475 = 18.85 $T_{A}$ or $T_{A}$= 184.3 N-m.