logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
0 votes

A solid sphere is placed rotating with its center initially at rest in a corner as shown. (a) and (b) . Coefficient of friction between all surfaces and sphere is $\large\frac{1}{3}$ Find the ratio of friction forces $\large\frac{fa}{fb}$ by ground in situations (a) and (b)

 

\[\begin {array} {1 1} (a)\;1 & \quad (b)\;\frac{9}{10} \\ (c)\;\frac{10}{9} & \quad  (d)\;none\;of\;these \end {array}\]
Can you answer this question?
 
 

1 Answer

0 votes
Sphere in (a) is in transnational equilibrium. Sphere in (a)
$ \sum F_Y=0$
$mg= f_1+N_2$
$ \sum F_X=0$
$f_2=N_1$
$f_1=\mu N_1$
$f_2= \mu N_2$
$\mu N_2 =N_1$
$N_2=\large\frac{N_1}{\mu}$
$\therefore mg =\large\frac{N_1}{\mu}$$+\mu N_1$
$ \large\frac{\mu mg}{\mu^2+1}$$=N_1$
$N_2= \large\frac{mg}{\mu^2 +1}$
$f_a= \large\frac{\mu mg}{\mu^2+1}$
$\qquad= \large\frac{3mg}{10}$
For sphere in (b)
Since $f_2$ is acting toward right, the sphere has a tendency to move towards right.
$N_1=0, f_1=0$
$\therefore mg=N_2$
$f_2 = \mu mg=\large\frac{mg}{3}$
$\therefore \large\frac{f_a}{f_b}=\frac{9}{10}$
answered Dec 5, 2013 by meena.p
edited Jun 22, 2014 by lmohan717
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...