Question

210

For the mechanism Illustrated in Figure 3, draw free body diagrams of the links. Assume all links have mass and that the centers of mass of all links are in the middle of the links. Assume the center of mass of link 3 is at C. Write out the equations need to determine all forces and moments. Assume that there is an applied torque at O₂ driving the mechanism. Do not solve!

α₂=0, ω₂=60 rad/sec cw, θ₂ = 210 deg, R_AO2=150mm, R_BA=300mm, R_O4O2=75mm, R_BO4=300mm, R_DA=150mm, R_CD=100mm

Answer 1

Link 2:

30 Ng

(a)

Fr = 0

$\sum F_{y}=N_{2}-W-F=0$ (b)

(c)

02-A k cos301 x *-* sin300-0 N2 -

k cos30

Here, in place of force you can also take two force components in the radial and tangential directions to link 2, which will serve the same purpose. This method is applied in case of link 3 and 4 below.

Link 3:

4R

$\sum F_{x}=-T_{4R}sin\theta-T_{4T}cos\theta=0$ (d)

$\sum F_{y}=F_{4R}cos \theta+F-W_{3}=0$ (e)

(f)

M.=F*100-F,T (X) = 0

where X is the perpendicular distance of from point 'C'. To find it's value follow the method given in the following figure.

AT

N G ces e 150 (450-100pine) DL 100 (151 ) 1501o0 ne

Here, $X=BC"=BC'-C'C"$

r. (150)- (150 - 100sine) cost

And the value of $\theta$ can be found geometrically pretty easily.

Note: Don't forget to take sign of
AT
and $F_{4R}$ into account and change the directions as required.

Link 4:

(g)

TCOS sin

$\sum F_{y}=F_{4T}sin\theta+N_{4}-F_{4R}cos\theta-W_{4}=0$ (h)

$M_{N_{4}}=300*F_{4T}-150*W_{4}sin\theta=0$ (i)

Equations (d) and equation (g) are basically the same.