Question

2 Two pulleys of diameters di&. d2 are at distance 'X' apart are
connected by means of an open belt drive, the length of the belt is
Options (Marks: 4)
(Tap/Click on option)
Option 1
TT/2 (dl+d2)2x+(d1+d2)2/4x
Option 2
D TI/2 (d1-d2)2x+(d1-d2)2/4x
Option 3
T/2 (d1+d2)2x+(d1-d2)2/4x
Option 4
TT/2 (d1-d2)2x+(d1+d2)2/4x​

Answer 1

Explanation:

Magnets are objects that produce magnetic fields and attract metals like iron, nickel and cobalt

Answer 2

The length of the belt for an open belt drive between two pulleys is               $L=\frac{\pi}{2}(d_1+d_2)+2x+\frac{(d_1-d_2)^2}{4x}$

Explanation:

• given the diameters of the two pulleys as $d_1, d_2$ such that $d_1>d_2$ hence let their corresponding radii be $r_1\ and\ r_2$ and the distance between them is $x$.
• then let the angle of lap be $\alpha$ , the length of belt taunt at upper side between both the pulleys be $l$, the arc length of belt on first pulley be $l_1$ and the arc length of belt on second pulley be $l_2$ .  
• then we have  , $\alpha=sin\alpha= \frac{r_1-r_2}{x}$  
• for the arc length on first pulley we have,                                                            $l_1=r_1(\frac{\pi }{2}+\alpha )\\l_1=r_1(\frac{\pi}{2}+\frac{r_1-r_2}{x} )$    -----(a)
• for arc length on second pulley we have,                                                          $l_2=r_2(\frac{\pi}{2}-\alpha ) \\l_2=r_2(\frac{\pi}{2}- \frac{r_1-r_2}{x} )$  ----- (b)
• for the length of taunt upper part we have,                                                              $l=\sqrt{x^2-(r_1-r_2)^2} \\l=x({1-(\frac{r_1-r_2}{x})^2 })^\frac{1}{2}$
• using binomial theorem we get,                                  $l=x(1-\frac{(r_1-r_2)^2}{2x^2} ) \\l=x-\frac{(r_1-r_2)^2}{2x}$   -----(c)
• now the total length of the belt is given by, $L=2(l_1+l+l_2)$  
• from (a)(b) and (c) we get,                                                                                        $L=2(r_1(\frac{\pi}{2}+\frac{r_1-r_2}{x} ) +x-\frac{(r_1-r_2)^2}{2x}+ r_2(\frac{\pi}{2}+\frac{r_1-r_2}{x} )) \\L=\pi(r_1+r_2)+2x+\frac{(r_1-r_2)^2}{x}$                                              substituting the values of diameters we get,                                                           $L=\frac{\pi}{2}(d_1+d_2)+2x+\frac{(d_1-d_2)^2}{4x}$       --------ANSWER