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Q3:A ball bearing designated as 6214 runs at N=900 rpm. Co=45 kN; C= 60,5 kN. Radial force and axial force acting on the ball bearing is Fr=4,2 kN and Fa=3,4 kN, respectivelly. Find the life of bearing in terms of number of revolution and hours. $F= X Fr + Y Fa$ $L=(\frac{C}{F})^E$; $L=Lh.N.60$ Type of bearing Specifications $\frac{W_A}{W_R} \le e$ $\frac{W_A}{W_R} > e$ $e$ $X$ $Y$ $X$ $Y$ Deep groove ball bearing $\frac{W_A}{C_0} = 0.025$ $\frac{F_A}{C_0}$ $2.0$ $0.22$ $-0.04$ $1.8$ $0.24$ $-0.07$ $1.6$ $0.27$ $-0.13$ $1$ $0$ $0.56$ $1.4$ $0.31$ $-0.25$ $1.2$ $0.37$ $-0.50$ $1.0$ $0.44$

Name: q3a ball bearing designated as 6214 runs at n900 rpm co45 kn c 605 kn radial force and axial force acting on the ball bearing is fr42 kn and fa34 kn respectivelly find the life of bearing in 10808
Uploaded: 2025-06-21T10:49:11-08:00
Duration: 16 min 15 s
Channel: Mark Scythian
Description: q3a ball bearing designated as 6214 runs at n900 rpm co45 kn c 605 kn radial force and axial force acting on the ball bearing is fr42 kn and fa34 kn respectivelly find the life of bearing in 10808

          Q3:A ball bearing designated as 6214 runs at N=900 rpm. Co=45 kN; C= 60,5 kN. Radial force and axial force acting on the ball bearing is Fr=4,2 kN and Fa=3,4 kN, respectivelly. Find the life of bearing in terms of number of revolution and hours.
$F= X Fr + Y Fa$
$L=(\frac{C}{F})^E$; $L=Lh.N.60$
Type of bearing
Specifications
$\frac{W_A}{W_R} \le e$
$\frac{W_A}{W_R} > e$
$e$
$X$
$Y$
$X$
$Y$
Deep groove ball bearing
$\frac{W_A}{C_0} = 0.025$
$\frac{F_A}{C_0}$
$2.0$
$0.22$
$-0.04$
$1.8$
$0.24$
$-0.07$
$1.6$
$0.27$
$-0.13$
$1$
$0$
$0.56$
$1.4$
$0.31$
$-0.25$
$1.2$
$0.37$
$-0.50$
$1.0$
$0.44$

q3a ball bearing designated as 6214 runs at n900 rpm co45 kn c 605 kn radial force and axial force acting on the ball bearing is fr42 kn and fa34 kn respectivelly find the life of bearing in 10808

Added by Holly F.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Mark Scythian

06/21/2025

Step 1

5$ kN $F_r = 4.2$ kN $F_a = 3.4$ kN Step 2: Calculate the ratio $\frac{F_a}{C_0}$. $\frac{F_a}{C_0} = \frac{3.4 \text{ kN}}{45 \text{ kN}} = 0.0755...$ Step 3: Determine the values of $X$, $Y$, and $e$ from the table. The table provides values for Show more…

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Q3:A ball bearing designated as 6214 runs at N=900 rpm. Co=45 kN; C= 60,5 kN. Radial force and axial force acting on the ball bearing is Fr=4,2 kN and Fa=3,4 kN, respectivelly. Find the life of bearing in terms of number of revolution and hours. $F= X Fr + Y Fa$ $L=(\frac{C}{F})^E$; $L=Lh.N.60$ Type of bearing Specifications $\frac{W_A}{W_R} \le e$ $\frac{W_A}{W_R} > e$ $e$ $X$ $Y$ $X$ $Y$ Deep groove ball bearing $\frac{W_A}{C_0} = 0.025$ $\frac{F_A}{C_0}$ $2.0$ $0.22$ $-0.04$ $1.8$ $0.24$ $-0.07$ $1.6$ $0.27$ $-0.13$ $1$ $0$ $0.56$ $1.4$ $0.31$ $-0.25$ $1.2$ $0.37$ $-0.50$ $1.0$ $0.44$