Numerical - Axial force on worm gear

Q.  If pressure angle is 22o and coefficient of friction is 0.05. What is the axial force acting on the worm gear designated as 1/40/10/5, if tangential force of 1200 N acts on it?
- Published on 19 Oct 15

a. 1241.2 N
b. 2618.4 N
c. 7755.1 N
d. 7847.4 N

ANSWER: 7755.1 N

Related Content

 

    Discussion

  • Sravanthi   -Posted on 01 Oct 15
    Given: Pressure angle = 22o, coefficient of friction = 0.05, Ft = 1200 N, worm gear 1/40/10/5

    Formula: Axial force = (Ft) / tan (Φv + λ)

    Φv = tan-1 (μ / cos Φn)
    = 3.086o

    λ = tan-1 (zw / q)
    = 5.71o

    Substituting the given values in formula,

    Axial force = 7755.10

➨ Post your comment / Share knowledge


Enter the code shown above:

(Note: If you cannot read the numbers in the above image, reload the page to generate a new one.)
▲