A rough vertical board has an acceleration 'a' such that a 2kg block pressing against it does not fall. What should be the coefficient of friction between the block and the board ?

Question

A rough vertical board has an acceleration 'a' such that a 2kg block pressing against it does not fall. What should be the coefficient of friction between the block and the board ? Question

(A) >g/a (B) < g/a (C) =g/a (D) >a/g

Tardigrade Admin · Accepted Answer

In order to balance the block, upward friction force between board and block will balance the weight of the block. This means that weight should be less than maximmum value of friction. ⇒ mg < μ R where R represents normal reaction from the board.

We will go in accelerated board's frame to solve this problem. Therefore, by concept of pseudo force R=ma (see FBD for details). ∴ from concept of static friction mg < μ ma ⇒ μ >(g/a)

Q. A rough vertical board has an acceleration $^{'} a^{'}$ such that a $2 k g$ block pressing against it does not fall. What should be the coefficient of friction between the block and the board ?

$> g / a$

$< g / a$

$= g / a$

$> a / g$

Q. A rough vertical board has an acceleration ′a′ such that a 2kg block pressing against it does not fall. What should be the coefficient of friction between the block and the board ?

>g/a

<g/a

=g/a

>a/g

Q. A rough vertical board has an acceleration $^{'} a^{'}$ such that a $2 k g$ block pressing against it does not fall. What should be the coefficient of friction between the block and the board ?

$> g / a$

$< g / a$

$= g / a$

$> a / g$