0
1.6kviews
Numerical : 2
1 Answer
0
103views

A shaft carries four masses A,B,C,D of magnitude 200 kg, 300 kg, 400 kg and 200 kg respectively and revolving at radii 80 mm, 70 mm, 60 mm and 80 mm in planes measured from A at 300 mm, 400 mm and 700 mm. The angles between the cranks measured anticlockwise are A to B 45 $^{\circ}$, B to C 70 $^{\circ}$ and C to D 120 $^{\circ}$. The balancing masses are to be 100 mm, between X and Y is 400 mm and between Y and D is 200 mm, If the balancing masses revolve at a radius of 100 mm find their magnitude and angular positions.

Given :

$m_A = 200 \ kg$ $ r_A = 80 \ mm = 0.08 \ m $
$ m_B = 300 \ kg $ $ r_B = 70 \ mm = 0.07 \ m $
$m_C = 400 \ kg $ $ r_C = 60 \ mm = 0.06 \ m $
$m_D = 200 \ kg $ $ r_D = 80 \ mm = 0.08 \ m $
$ r_X = r_Y = 100 \ mm = 0.100 \ m$

enter image description here

Assume the plane X as R.P., the distance of the planes to the right of the plane X are taken as positive while the distance of the plane to the left of the plane X are taken as negative.

Draw Table :

Note : The planes are tabulated in the same order in which they occur, Reading from Left to Right.

Plane Mass(m) Kg Radius(r) m Centrifugal force$\div \omega^{2}$ (mr) kg-m Dist. from plane x(L) m couple$\div\omega^{2} $ (mrL) $kg-m^2$
A 200 0.08 16 -0.1 -1.6
X mx 0.1 0.1 mx 0 0
B 300 0.07 21 0.2 4.2
C 400 0.06 24 0.3 7.2
Y My 0.1 0.1 My 0.4 0.04 My
D 200 0.08 16 0.6 9.6

Couple polygon :

enter image description here

Force polygon :

enter image description here

The angular position of mass Mx is obtained by drawing oMx, Parallel to vector oe. By measurement position of mx is $\theta x =140 ^{\circ}$ in clockwise direction from mass Ma.

Please log in to add an answer.