IIT JEE , AIEEE Forum - Physics , Mechanics

cheeky_virus

Friction (Pg 98) :23, 28, 29.

Kinematics (Pg 54) :48

Circular Motion (Pg 110) : Ques No.12 - Explain why the tensions in the strings are showed in opposite direction.

Kindly Answer all these questions with necessary and appropriate free body diagrams and sketches.

karthik2007

If you are too lazy to type the question, then we are lazy to type the solutions as well

biki

i agree with karthik............

u r liable to be rebuked.....

but i m forgiving u

questions likha karo yaar.....!!!! it helps everyone

biki

for the circular motion one.....

read HCV chapter " Forces" so that u r not shocked when i say that tension is an electromagnetic force produced due to forces between electrons and protons of bodies involed.

If this is true, then the part considered in the problem has electrons, protons and all other stuff in all direction (both sides of the point C in the ring) which exert forces on point C as and when the point C tries to bulge out due to centrifugal force produced while rotation.

and what is the problem.........
in pulley sums also u consider tension in both direction along the string, dont you.......???

biki

28) [ friction ]

This one is a little tricky problem. To find the solution, not only you need to identify the forces acting on the blocks but also figure out the relative motion of the two as well.

As the string is inextensible , wheneve the lower end of the string (connected to M) displaces by x units to the right, the other end of the string (connected to block m) lowers down by 2x units. This is to be showed by 'constant length of the string' method as described well in HCV - 1 pg 73 , example 6 in ch. Newton's Laws of Motion

So accl^n of m (vertically downward) = 2 ( accl^n of block M in horizontal right direction )

Let accl^n of M = a
So accl^n of m = 2a.

Also, the block m is always in contact with the larger block M as far as motion in the horizontal direction is concerned. So accl^n of block m in horizontal direction is also a (same as that of M)

Now look at the figure.
The blue coloured are the forces on M and red ones act on m

Motion of m:

the forces are:
1) mg downwards
2) R ( contact force by M ) towards right
3)

₁R ( frictional force) upwards

4) T ( Tension) upward

For horizontal direction.

R = ma

In vertical direction

mg - T -

₁R = m(2a)

or T = mg - ma(2 +

₁)

Motion of M:

The forces on M are:

1) Mg downwards

2) R ( contact force by m ) towards left

3)

₁R ( reaction of frictional force on m ) downwards

4) T ( Tension due to string on the lower end ) towards right

5) N ( contact force by ground ) upwards

6)

₂N ( frictional force due to ground ) towards left

7) T ( Tension due to string on pulley attached to M ) towards right

8) T ( Tension due to string on pulley attached to M ) downwards

for vertical equilibrium

N = Mg + T +

₁R

or N = Mg + T +

₁ma {since R = ma}

In horizontal direction

2T - R -

₂N = Ma

Putting values of R, T and N.......

2T - ma -

₂(Mg + T +

₁ma) = Ma

or (2 -

₂)T - ma -

₂(Mg +

₁ma) = Ma

or (2 -

₂)[mg - ma(2 +

₁)] - ma -

₂(Mg +

₁ma) = Ma

or 2mg -

₂g(m + M) = 5ma + 2ma

₁ - 2ma

₂ + Ma

or a[M + m{5 + 2(

₁ -

₂)}] = g[2m -

₂(m + M)]

or a = g[2m -

₂(m + M)] / [M + m{5 + 2(

₁ -

₂)}]

waterdemon

Solution for (29) Friction:

The Force which may cause the tendency of motion or the
motion in the body is its own Weight & the applied horizontal
force of 15N.

We have Resultant of the forces as:
F_res =

(20)²+(15)²
F_res =

625
F_res = 25N.

In a Direction of:
@ = Tan^-1(15/20) = Tan^-1(3/4) = 37⁰ with the Vertical.

The Friction will always oppose the tendency of Relative
motion and will act in a direction opposite of Resultant F.
F = Force.

Now for the acceleration to be minimum:
Minimum Force required: (N=40=F applied against the wall)
F_min = F - uN
F_min = 25 - (0.5)40
F_min = 5N.

So the minimum Acceleration will be:
F_min=ma_min
5 = 2(a_min)
a_min = 2.5m/s².

Since @ = 37⁰ with the vertical so @ = 53⁰ with the
Horizontal.

Hope you find it useful.
Rate if useful.

Cheers!!!!!!!!!!!!! :)

waterdemon

Solution Friction (23)

1)When F=10N applied on 2kg Block on top:

So equation of motion for 2 kg block will be:
F - f = ma
10 - 2(0.2)(10) = 2(a_max)
6 = 2a_max
a_max = 3 m/s².

So the 2 kg block will move with a₁ = 3m/s².

Now since there is no Friction between Ground & 7kg block
So the frition force between 3 kg and 7 kg will drive them
together.

But since the force driving 3 kg block itself f = 4N which
is < f_3/7 = (0.3)(3)(10) = 9N.

Therefore the friction force between 2 kg and 3 kg will
drive the 3 kg and 7 kg blocks together.

So we have.
f = ma
4 = (3+7)(a)
4/10 = a
0.4 m/s² = a = a₂ = a₃.

2)When Force is applied on 3 kg Block:

Here 2 kg and 3 kg block act as system.
f = (2+3)(0.3)(10) = 5*3 = 15N.

Now this friction force of 15N > F applied = 10N so the
Force F = 10N will drive the Whole of System together.
F = (2+3+7)a
10/12 = a
a = 5/6 m/s².
a₁ = a₂ = a₃ = (5/6) m/s².

3)When Force is applied on 7 kg lower Block:
Since no Friction between ground adn 7kg block.so we have
F = (2+3+7)a
10 = 12(a)
5/6 = a

Therefore,
a₁ = a₂ = a₃ = (5/6) m/s².

Hope u find it useful.
Rate if useful.

Cheers!!!!!!!!!!!!

biki

(i m gonna use 'u' as the friction coefficient)
See, the 40 N force you have applied will act as the normal reaction for the body. So you get frictional force = u x 40 = 0.5 x 40 = 20 N

Thus the weight(downward) must nullify the frictional force(upward) if the body is to remain at rest. So you get weight = 20 N (which is again verified by the fact that mg = 2 x 10 = 20 N )

So on application of the 15N force as shown in figure, the net force acting on the body becomes

( 20² + 15² ) = 25 N

This 25 N force will be opposed by a frictional force = u x 40 = 20 N again as the 15 N force will have no effect on the normal reaction received by the body from the wall. This is because 15 N force makes an angle 90⁰ with the plane of the wall and has no component perpendicular to the wall.

Thus net force acting on the body becomes (25 - 20) N = 5 N.

This 5 N will make the body move.....

And this 5 N will act along the 25 N force, since frictional force is just opposite to the 25 N force.

Now let

be the angle made by the 5 N force with the 15 N force.

Thus (from figure) tan

= 20/15

So <theta> = tan^-1 (20/15)

= 53⁰

THE BODY MOVES BY MAKING AN ANGLE 53⁰ WITH THE 15 N FORCE......

waterdemon

Hey Biki,
I have already given the solution to (29) Sum of Friction.
Check 2 posts above :)

Cheers!!!!!!!!!!!!!

karthik2007

How do you draw such cool diagrams biki? Its surely not using paint... How do you type mu in paint?

biki

hit the 'A' button in the toolbox towards the left of the screen.

then do as shown in figure........

then type q for theta

w for omega

a,b,c for alpha, beta, gamma

m for mu

biki

click on the pic to download and then see clearly

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