*begins*to slide, the frictional force between the chain and the table is maximum and is called

*limiting*frictional force which is equal to

*μ*

_{s}

*N*where

*μ*

_{s}is the coefficient of static friction between the chain and the table and

*N*is the normal force.

*N*= 0.8

*mg*and

*μ*

_{s}

*N =*0.2

*mg*where

*m*is the mass of the chain.

*mg*) balances the frictional force

*μ*

_{s}

*N*].

*μ*

_{s}×0.8

*mg*= 0.2

*mg*from which

*μ*

_{s}= 0.25

*limiting*frictional force is applied on it. If the coefficient of static friction and the coefficient of kinetic friction are respectively 0.5 and 0.4 and the acceleration due to gravity is 10 ms

^{–2}, the acceleration of the wooden block is

^{–2}

^{–2}

^{–2}

^{–2}

*μ*

_{s }

*mg*where

*μ*

_{s}is the coefficient of static friction, and

*mg*is the weight of the wooden block. When the block moves along the horizontal surface, the frictional force acting is reduced to

*kinetic*frictional force and is equal to

*μ*

_{k }

*mg*where

*μ*

_{k}is the coefficient of kinetic friction.

*μ*

_{s }

*mg*–

*μ*

_{k }

*mg*and the acceleration

*a*of the block is given by,

*frictionless*horizontal floor and a block of mass 10 kg is placed on the slab as shown. The coefficient of static friction and the coefficient of kinetic friction between the slab and the block are respectively 0.4 and 0.2 and the acceleration due to gravity is 10 ms

^{–2}. A horizontal force equal to 30 N acts on the block. The acceleration of the slab will be

^{–2}

^{–2}

^{–2}

^{–2}

^{–2}

*μ*

_{s}

*N*) between the block and the slab is given by

*μ*

_{s}

*N*=

*μ*

_{s}

*Mg*= 0.4×10×10 = 40 N

*ogether*, with acceleration

*a*given by

*a =*30 N/(50+10)kg = 0.5 ms

^{–2}.

**aspirants.**

**The following questions are specifically for AP Physics C**

**aspirants:**

*M*= 6 kg is placed on rough horizontal floor. The coefficient of kinetic friction between the floor and the slab is 0.1. A block B of mass

*m*= 4 kg is placed on the slab (Fig.). The coefficient of static friction between the slab and the block is 0.2. When a horizontal force of magnitude

*F*is applied on the slab, the block B

*just begins to slide*along the slab A. What is the value of

*F*? (Take g = 10 ms

^{–2}).

*μ*

_{s}

*mg*where

*μ*

_{s}is the coefficient of static friction between the slab and the block.

*F*is applied on the slab, the entire system containing the slab and the block must move with an acceleration

*a*so that the inertial force (=

*ma*) on the block just balances the frictional force.

*μ*

_{s}

*mg = ma*so that

*a = μ*

_{s}

*g =*0.2×10 = 2 ms

^{–2}

*a*(

*M+m*).

*μ*

_{k}(M+

*m*)

*g*] between the floor and the slab.

*F*=

*μ*

_{k}(M+

*m*)

*g*+

*a*(

*M+m*)

*F*= (M+

*m*)(

*μ*

_{k}

*g + a*) = 10(1+2) = 30 N.

*M*= 80 kg tries to pull down a tree using a rope. He exerts the pulling force at an angle of 30º with the horizontal (Fig.). If the coefficient of static friction between the man’s feet and the ground is 0.6 and the gravitational acceleration is 10 ms

^{–2}, the maximum pulling force he can exert (before he slips) is approximately

*F*=

*T*.

*T*which acts upwards reduces the normal force

*N*and we have

*N*=

*Mg – T*sin 30º = 80×10 –

*T*/2

*μ*

_{s}

*N*= 0.6×(800 –

*T/*2).

*T*=

*T*cos 30º =

*T*×(√3)/2

*T/*2) = T ×(√3)/2

*T*= 1.732

*T*so that

*T*= 960/2.332 = 400 N, approximately