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Mechanics
21 Dec 2007 22:25:49 IST
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Karthik M
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Joined: 1 May 2007
Posts: 2830
22 Dec 2007 11:10:29 IST
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If you are too lazy to type the question, then we are lazy to type the solutions as well
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22 Dec 2007 12:50:55 IST
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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.......???
in pulley sums also u consider tension in both direction along the string, dont you.......???
22 Dec 2007 12:59:27 IST
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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.
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
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
1) mg downwards
2) R ( contact force by M ) towards right
3)
1R ( frictional force) upwards4) T ( Tension) upward
For horizontal direction.
R = ma
In vertical direction
mg - T - 1R = m(2a)
1R = m(2a)or T = mg - ma(2 + 1)
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
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
2N ( frictional force due to ground ) towards left7) 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
1Ror N = Mg + T + 1ma {since R = ma}
1ma {since R = ma}In horizontal direction
2T - R - 2N = Ma
2N = MaPutting values of R, T and N.......
2T - ma - 2(Mg + T + 1ma) = Ma
2(Mg + T + 1ma) = Maor (2 - 2)T - ma - 2(Mg + 1ma) = Ma
2)T - ma - 2(Mg + 1ma) = Maor (2 - 2)[mg - ma(2 + 1)] - ma - 2(Mg + 1ma) = Ma
2)[mg - ma(2 + 1)] - ma - 2(Mg + 1ma) = Maor 2mg - 2g(m + M) = 5ma + 2ma1 - 2ma2 + Ma
2g(m + M) = 5ma + 2ma1 - 2ma2 + Maor a[M + m{5 + 2(1 - 2)}] = g[2m - 2(m + M)]
1 - 2)}] = g[2m - 2(m + M)]or a = g[2m - 2(m + M)] / [M + m{5 + 2(1 - 2)}]
2(m + M)] / [M + m{5 + 2(1 - 2)}]![[Thumb - 28.GIF]](http://www.goiit.com/upload/2007/12/21/7aca37bea85f5a4ee5106a9d5d712500_19520.GIF_thumb)
22 Dec 2007 13:10:05 IST
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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:
Fres =
(20)2+(15)2
Fres =
625
Fres = 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)
Fmin = F - uN
Fmin = 25 - (0.5)40
Fmin = 5N.
So the minimum Acceleration will be:
Fmin=mamin
5 = 2(amin)
amin = 2.5m/s2.
Since @ = 370 with the vertical so @ = 530 with the
Horizontal.
![[Thumb - HCV friction.JPG]](/../upload/2007/6/15/a1df69f08472b2f75c96b476b53060e7_19520.JPG_thumb)
Hope you find it useful.
Rate if useful.
Cheers!!!!!!!!!!!!! :)
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:
Fres =
(20)2+(15)2Fres =
625Fres = 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)
Fmin = F - uN
Fmin = 25 - (0.5)40
Fmin = 5N.
So the minimum Acceleration will be:
Fmin=mamin
5 = 2(amin)
amin = 2.5m/s2.
Since @ = 370 with the vertical so @ = 530 with the
Horizontal.
Hope you find it useful.
Rate if useful.
Cheers!!!!!!!!!!!!! :)
22 Dec 2007 13:40:17 IST
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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(amax)
6 = 2amax
amax = 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 < f3/7 = (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!!!!!!!!!!!!
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(amax)
6 = 2amax
amax = 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 < f3/7 = (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!!!!!!!!!!!!
22 Dec 2007 13:51:33 IST
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(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
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
( 202 + 152 ) = 25 NThis 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.
be the angle made by the 5 N force with the 15 N force.Thus (from figure) tan = 20/15
= 20/15So <theta> = tan -1 (20/15)
= 530
THE BODY MOVES BY MAKING AN ANGLE 530 WITH THE 15 N FORCE......
22 Dec 2007 21:23:10 IST
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@ WATER DEMON.......
while i was writing the sol^n and preparing the pic......u posted the answer.....
now while writing it was not possible to know that u hav already answered........whn i pressed submit......i saw that y hav already answered this......
:D:D:D:D:D:D:D
while i was writing the sol^n and preparing the pic......u posted the answer.....
now while writing it was not possible to know that u hav already answered........whn i pressed submit......i saw that y hav already answered this......
:D:D:D:D:D:D:D
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