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Questions involving friction were discussed on this site earlier. You can access them by clicking on the label ‘friction’ below this post or by trying a search for ‘friction’, using the search box provided on this page.

Today we will discuss some more multiple choice practice questions on friction:

(1) A uniform chain is placed on a rough horizontal table so that one end of the chain hangs down over the edge of the table. When 20% of the length of the chain hangs over the edge, it starts sliding. What is the coefficient of static friction between the chain and the table?

(a) 0.1

(b) 0.15

(c) 0.25

(d) 0.35

(e) 0.5

When the chain just

*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.
Here

*N*= 0.8*mg*and*μ*_{s}*N =*0.2*mg*where*m*is the mass of the chain.
[Note that the weight of 20% of the length of the chain (= 0.2

*mg*) balances the frictional force*μ*_{s}*N*].
Therefore,

*μ*_{s}×0.8*mg*= 0.2*mg*from which*μ*_{s}= 0.25
(2) A wooden block is placed on a horizontal surface and a horizontal force equal to the

*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
(a) 1 ms

^{–2}
(b) 1.1 ms

^{–2}
(c) 1.4 ms

^{–2}
(d) 1.5 ms

^{–2}
(e) zero

The limiting (or, maximum) frictional force is

*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.
[If the frictional force were not reduced, the block would have moved with uniform velocity since there would be no net force].

The net force acting on the block is

*μ*_{s }*mg*–*μ*_{k }*mg*and the acceleration*a*of the block is given by,
(3) A slab of mass 50 kg is placed on

*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
(a) 0.1 ms

^{–2}
(b) 0.2 ms

^{–2}
(c) 0.3 ms

^{–2}
(d) 0.5 ms

^{–2}
(e) 1.2 ms

^{–2}
The limiting frictional force (

*μ*_{s}*N*) between the block and the slab is given by*μ*

_{s}

*N*=

*μ*

_{s}

*Mg*= 0.4×10×10 = 40 N

The block will not slide along the slab since the applied force (= 30 N) is less than the limiting frictional force. So the slab and the block will move t

*ogether*, with acceleration*a*given by*a =*30 N/(50+10)kg = 0.5 ms

^{–2}.

[Since there is no relative motion between the slab and the block, the coefficient of kinetic friction does not play any role in the above problem. It just serves the purpose of a distraction].

The above questions may be ‘enjoyed’ by AP Physics B & C

--> **aspirants.****The following questions are specifically for AP Physics C****aspirants:**
(4) A slab A of mass

*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}).
(a) 40 N

(b) 30 N

(c) 15 N

(d) 10 N

(e) 8 N

The limiting frictional force between the slab and the block is

*μ*_{s}*mg*where*μ*_{s}is the coefficient of static friction between the slab and the block.
When the force

*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.
[The inertial force must be infinitesimally greater than the frictional force for the slipping to occur].

Therefore, we have

*μ*_{s}*mg = ma*so that*a = μ*_{s}*g =*0.2×10 = 2 ms^{–2}
If the floor were smooth, the force to be applied on the slab to attain the slipping condition would have been

*a*(*M+m*).
But since the floor is rough, the applied force has to overcome the force of kinetic friction [

*μ*_{k}(M+*m*)*g*] between the floor and the slab.
Therefore,

*F*=*μ*_{k}(M+*m*)*g*+*a*(*M+m*)
Thus

*F*= (M+*m*)(*μ*_{k}*g + a*) = 10(1+2) = 30 N.
[The above question could be made a little more difficult if the coefficient of static friction (say, 0.25) between the floor and the slab also is given (to distract you). You should remember that once the slab slides along the floor, the friction called into play is kinetic friction and hence your answer will be unchanged].

-->
(5) A man of mass

*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
(a) 50 N

(b) 100 N

(c) 200 N

(d) 400N

(e) 600 N

If the man exerts too much pulling force his feet will slip and hence the maximum pulling force is determined by the coefficient of friction and the angle of pull. If T is the maximum tension produced in the rope, the maximum pulling force,

*F*=*T*.
The vertical component of

*T*which acts upwards reduces the normal force*N*and we have*N*=

*Mg – T*sin 30º = 80×10 –

*T*/2

The maximum frictional force =

*μ*_{s}*N*= 0.6×(800 –*T/*2).
The horizontal component of the maximum tension

*T*=*T*cos 30º =*T*×(√3)/2
The frictional force should balance the horizontal component of tension so as to prevent the man from slipping.

Therefore we have

0.6×(800 –

*T/*2) = T ×(√3)/2
Or, 960 – 0.6

*T*= 1.732*T*so that*T*= 960/2.332 = 400 N, approximately
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