Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
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$
$f_1=\mu N_1$
$f_2= \mu N_2$
$\mu N_2 =N_1$
$\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
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App