The front wall of a drawer in a cabinet is a provider with two symmetrical handles. The distance between the handles is l and the length (i.e. depth) of the drawer is a . The maximum value of the coefficient of friction between drawer and cabinet, for which drawer can be pulled out by applying a force on one handle perpendicular to the face of the drawer is: (Neglect the weight of the drawer)

Question

The front wall of a drawer in a cabinet is a provider with two symmetrical handles. The distance between the handles is l and the length (i.e. depth) of the drawer is a . The maximum value of the coefficient of friction between drawer and cabinet, for which drawer can be pulled out by applying a force on one handle perpendicular to the face of the drawer is: (Neglect the weight of the drawer) (A) ((a/l)) (B) (2 a/l) (C) (a/2 l) (D) None of these

Tardigrade Admin · Accepted Answer

If a force is applied only on one handle, as shown in figure, the drawer will have a tendency to rotate clockwise. That is, the right and left corners (shown in figure), will press against the side of the cabinet and produce some friction. Balancing force, F=2μ N …….(i) Balancing torque (about right side corner): Fx+Na=μ N(.2x+l.) ……..(ii) Solving gives, μ =(a/l)

$(\frac{a}{l})$

$\frac{2 a}{l}$

$\frac{a}{2 l}$

None of these

(la​)

l2a​

2la​

None of these

$(\frac{a}{l})$

$\frac{2 a}{l}$

$\frac{a}{2 l}$