Numerical - Frictional torque, given radial load & diameter

Q.  What is the frictional torque acting on a cylindrical roller bearing of 50 mm diameter which is subjected to radial load of 30 kN and has coefficient of friction 0.0011?
- Published on 23 Sep 15

a. 109 kN-mm
b. 900 kN-mm
c. 825 N-mm
d. 1650 N-mm

ANSWER: 825 N-mm

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    Discussion

  • Sravanthi   -Posted on 01 Oct 15
    Given: Radial load ( Fr ) = 30 kN, diameter of roller bearing(d) = 50 mm, co-efficient of friction (f) = 0.0011

    • When two objects in contact move together, torque caused by frictional force is called as frictional torque.

    Formula: Tf = f · Fr · d/2

    Substituting the given values we get,

    Frictional torque (Tf) = f x Fr x d/2

    = (0.0011 x 30000 x 50 / 2)

    Frictional torque = 825 N-mm

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