The portion AOB of a mechanism consists of a 400mm steel rod

The portion AOB of a mechanism consists of a 400-mm steel rod OB welded to a gear E of radius 120 mm which may rotate about a horizontal shaft O. It is actuated by a gear D and, at the instant shown, has a clockwise angular velocity of 8 rad/s and a counterclockwise angular acceleration of 40 rad/s^2. Knowing that rod OB has a mass of 3 kg and gear E a mass of 4 kg and a radius of gyration of 85 mm, determine (a) the tangential force exerted by gear D on gear E, and (b) the components of the reaction at shaft O.

Solution

solution:

1)as here gear E is driven by D,hence tangential force exist between them

2)mass moment of inertia of system is given by

I=Ib+Ie

I=Mb*Kb^2+Me*Ke^2

Kb=L/2=400/2=200 mm

Ke=85 mm

hence result

I=3*.2^2+4*.085^2=.1489 kgm2

4)hence torque exerted on gear if its accelaration is

T=I*a

a=40 rad/s2

T=.1489*40=5.956 N m

5)tangential force is given by

T=Fte*Re

Ft=5.956/.12=49.633 N

6)here force o Nd is equal in magnitude and opposite direction to Fte

Ftd=Fte=49.633 N

7)reaction at O is if gear E is spur gear

Fr=Ft*tan20=18.06 N=radial force

R=(Ft^2+Fr^2)^.5=52.818 N

Fy=0

R-Ro=0

R=Ro=52.818 N

T=Fte*Re=5.956 Nm