Question

The bearings of a shaft at A and B are 5 meters apart. The shaft carries three eccentric masses C, D and E which are 160 kg, 170 kg and 85 kg respectively. The respective eccentricity of each mass measured from the axis of rotation, is 0.5 cm, 0.3 cm and 0.6 cm and the distance from A is 1.3m, 3m and 4m respectively.

The Net dynamic force at A for this arrangement when the shaft runs at 100 rpm.

• A. 63.9 N
• B. 74.36 N
• C. 54.21 N
• D. 42.36 N

Answer 1

The correct option is A 63.9 N

Planes	m(kg)	r(cm)	L(cm)	$\theta$(degree)	mr	mrL
C	160	0.5	1.3	0	80	$104\to M_C$
D	170	0.3	3	${\theta }_1$	51	$153\to M_D$
E	85	0.3	4	${\theta }_2$	51	$204\to M_E$
A	----	0.6			$R_A$	0

$N=100\text{rpm}\Rightarrow \omega =\frac{2\pi \times 100}{60}=10.47\text{rad/sec}$
$R_B=0\text{(given)}$

${\text{Since there is no dynamic reaction at 'B' and there is dynamic reaction at "A", it means the system is moment balanced. Moment of unknown dynamic reaction}}^{\prime }{R}_A^{\prime }\text{wrt its plane is zero.}$

$\sum M_A=0$

$\text{We can say that the three moment vectors}{M}_C,M_D\text{and}{M}_E\text{are in equilibrium. Resultant of}{M}_C,M_D\text{is balanced by}{M}_E\text{i.e.,}R=M_E\text{by magnitude but opposite in direction.}$


$\cos \theta =\frac{{204}^4+{104}^2-{153}^2}{2\times 204\times 104}=0.684$

$\theta ={46.84}^{\circ }$

$\cos \varphi =\frac{{204}^2+{15}^3-{104}^2}{2\times 204\times 153}=0.8684$

$\varphi ={29.7267}^{\circ }$

${\theta }_1=46.84+29.7267={76.5667}^{\circ }$

${\theta }_2=180+46.84={226.84}^{\circ }$

To get the dynamic reactions at 'A'

$\sum F_x=(m_C{r}_C+m_D{r}_D\cos {\theta }_1+m_E{r}_E\cos {\theta }_2){\omega }^2$

$=(\frac{80}{100}+\frac{51}{100}\cos 76.5667+\frac{51}{100}\cos 226.84)\times {10.47}^2=62.443\text{N}$

$\sum F_y=(m_D{r}_D\sin {\theta }_1+m_E{r}_E\sin {\theta }_2){\omega }^2$

$=(\frac{51}{100}\sin 76.5667+\frac{51}{100}\sin 226.84)\times {10.47}^2=13.6\text{N}$

Net dynamic reaction at 'A',

$R_A=\sqrt{{62.443}^2+{13.6}^2=63.9}\text{N}$