Question

A belt drive system given in the figure. Such that distance between the centres $OO'=120cm$ & radii are 20 cm & 80 cm respectively.

Let $\beta$ equal to length of indirect common tangent $AA'$. $[\beta]$ is equal to, where $[x]$ is greatest integer less than or equal to x.

Answer 1

66

Answer 2

Solution:

For a crossed (indirect) belt drive the straight belt (common tangent) length is given by

$$\beta = \sqrt{d^2 - (R + r)^2},$$

with

$$d = 120\text{ cm},\quad r = 20\text{ cm},\quad R = 80\text{ cm}.$$

Thus,

$$\beta = \sqrt{120^2 - (80+20)^2} = \sqrt{14400 - 10000} = \sqrt{4400} = 20\sqrt{11}\text{ cm}.$$

Now, computing the greatest integer function:

$$20\sqrt{11} \approx 20\times3.3166 \approx 66.33,$$

so

$$[\beta] = 66.$$

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Core Explanation (Minimal):

• Use formula for crossed belt drive: $\beta=\sqrt{d^2-(R+r)^2}$.
• Substitute: $\sqrt{14400-10000} = 20\sqrt{11}$.
• Numerically, $20\sqrt{11}\approx66.33$; hence $[\beta]=66$.