IIT JEE , AIEEE Forum - Physics , Mechanics

ashish17

please reply the following ques from friction chapter of hc verma.

Q. 20, 23(b), 24(b),28, 27.

reply soon.

paddy.dude

ques 27

in the ques a force is applied to the mass m on the top of the table such that it just slips.........this means that its limiting frictionnal force

let the applied force be F

so F= mew x m x g

now if we consider the table as the system ,
there will be a frictional force in the right direction.......and it will be equal to

mew x m x g due to newtons third law

now there is a frictional force in the left direction coz of friction between the floor and the table.........since the table is in equilibrium
this must be balanced by the friction on top of the table

so ans is mew x m x g

paddy.dude

ques 24 b

let the acc be a and tension be T

F - mew x m x g - T = m x a ---------- 1

and the block of mass M
friction is on the left due to newtons third law so

T - mew x m x g = M x a --------- 2

solving these 2 equations u will get the
value of a

ashish17

@ paddy.dude

In Q.24(b) i m also solving these 2 equations using F=2 mew m g and answer is coming 0.

please check this.

paddy.dude

yaar u missed one part of the question
the ques says that " what will be the acc if the force is twice that in the first part "

so F = 4 mew x m x g
now u will get the ans

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