Slide 3 : P. K. BHARTI, IIT KHARAGPUR A sees the box and finds it is stationary
wrt himself. So, he draw FBD of box as:
Therefore A writes eqn as:
mg – N = 0.
which is wrong.
It is wrong, because A also has acceleration.
It means, A is a non-inertial frame of reference.
Therefore, Newton’s laws can not be correct in frame of A.
So, what to do to write Newton’s law correctly in non- inertial frame?
Well ! We apply an additional force of magnitude ma opposite to the direction of elevator’s acceleration.
Therefore, eqn in frame of A becomes
mg – N – ma = 0
? mg – N = ma which is same as eqn. (1) in previous slide.
Thus to write Newton’s law correctly in non-inertial frame we apply a force ma opposite to the acceleration of frame. This ma is called as pseudo force. 3 N mg ma
When & how to apply pseudo force : P. K. BHARTI, IIT KHARAGPUR When & how to apply pseudo force In general, when we draw FBD of a particle in a non-inertial frame, we have to draw pseudo force ma, opposite to the accln a of frame. Here m is the mass of the particle [not the mass of frame] whose FBD we are drawing in non-inertial frame.
Let us see some situation where we should apply pseudo force to make the things easier.
Example (a). Here block m is kept in
an elevator going with accln a upward.
FBD of block in the frame of ground:
As from ground (inertial frame),
an observer will find block m has
acceleration upward, so FBD will be
as shown in next slide. 4 m a www.vidyadrishti.com
When & how to apply pseudo force : P. K. BHARTI, IIT KHARAGPUR When & how to apply pseudo force FBD of block in the frame of ground :
Therefore, we have,
N – mg = ma …(1)
FBD of block in the frame of elevator:
As elevator is a non-inertial frame (because
elevator has acceleration a in upward direction),
we have to apply pseudo force ma opposite to a, i.e, in downward direction .
Therefore, equation becomes:
N – mg – ma = 0
? N – mg = ma which is same as (1)
. Till now you must be wondering, why to use pseudo
if it could have done in inertial frame. Well! u ll get
your answer in coming slides. 5 a N mg N mg ma,
pseudo www.vidyadrishti.com
When & how to apply pseudo force : P. K. BHARTI, IIT KHARAGPUR When & how to apply pseudo force Example (b). Here elevator is going with accln a1 and block sliding down with an accln a2 the incline wrt elevator.
Here, if u want to draw FBD of block wrt earth, u will have to find accln of block wrt to earth, which will be tedious procedure.
So, better we draw FBD of block in elevator frame
(non-inertial frame) as we know the accln of block
wrt elevator is a2 downward the incline.
Thus, we will have to apply pseudo ma1 opposite
to a1, ie, in downward direction.
FBD of block in the elevator frame:
. 6 m a1 m a2 N mg ma1 , pseudo www.vidyadrishti.com
When & how to apply pseudo force : P. K. BHARTI, IIT KHARAGPUR When & how to apply pseudo force Example (b) continued…
As the plane is inclined, therefore,
first we have to take a coordinate system.
We resolve ma1 .
We resolve mg.
Now, we ready to use Newton’s Laws
Along x-axis:
ma1sin? + mg sin? = m a2
Along y-axis:
ma1cos? + mg cos? - N = 0
Thus, you must have learned how and when to use pseudo force. 7 m a2 N mg mg cos? ? ? x y ma1 ma1sin? ma1cos? mg sin? www.vidyadrishti.com
PSEUDO FORCE : P. K. BHARTI, IIT KHARAGPUR PSEUDO FORCE When we are working from non-inertial frame (i.e., accelerating frame), we have to apply a force ma (pseudo force) opposite to that of accln a of the frame, to apply Newton’s law.
EXAMPLE 6. Cart is moving with accln a. Find angle ? in equilibrium. a ? 8 www.vidyadrishti.com
EXAMPLE 6 : P. K. BHARTI, IIT KHARAGPUR EXAMPLE 6 F.B.D. of bob wrt cart frame [non inertial frame].
For equilibrium,
SFx = O
? Tsin? – ma = 0
? Tsin ? = ma …(1)
& SFy = O
? Tcos? – mg = o
? Tcos ? = mg …(2)
Solving we get,
? = tan-1 (a/g) ? T mg ma
(Pseudo) NOTE: We have applied Pseudo force ma, opposite to that of a of cart. x y 9 www.vidyadrishti.com
Example 7 [Finding accln] : P. K. BHARTI, IIT KHARAGPUR Example 7 [Finding accln] METHOD 1:
Let the acceleration of mass 1 be ao
upwards. The acceleration of pulley 2
will also be ao downwards because
string connecting mass 1 and pulley 2
is constant in length.
Also the string connecting mass 3 and mass
4 is constant in length. This implies that
the decrease in the separation between
between 3 and 2 is equal to increase in
separation between 3 and 4. So, the
upward acceleration of 3 wrt 2 equals
downward acceleration of 4 wrt 2. Let
this accln be ar. 1 2 3 4 ao ao 10 www.vidyadrishti.com
Example 7 continued : P. K. BHARTI, IIT KHARAGPUR Example 7 continued 1 2 3 4 The accln of 3 wrt ground = ao-ar (downward).
The accln of 4 wrt ground = ao +ar (downward).
Let tension T1 in the upper string
and T2 in the lower string.
NOTE: wrt = with respect to
accln= acceleration ao ao ao-ar T1 T1 T2 T2 ao+ar 11 www.vidyadrishti.com
Example 7 continued : P. K. BHARTI, IIT KHARAGPUR Example 7 continued T1 - M1g = M1a
...(1) M1 M1g T1 ao FBD of mass M1 12 www.vidyadrishti.com
Example 7 continued : P. K. BHARTI, IIT KHARAGPUR Example 7 continued M3 M3g T2 ao-ar M3g – T2 =M3(ao-ar)
…(2) FBD of mass M2 13 www.vidyadrishti.com
Example 7 continued : P. K. BHARTI, IIT KHARAGPUR Example 7 continued M4 M4g T2 ao+ar M4g – T2 =M4(ao+ar)
…(3) FBD of mass M4 14 www.vidyadrishti.com
(As pulley is assumed massless, m=0) T2 +T2 - T1 = 0 x ao ?2T2 -T1 = 0 …(4) Solving these 4 equations, we can find all accelerations and tensions. : P. K. BHARTI, IIT KHARAGPUR (As pulley is assumed massless, m=0) T2 +T2 - T1 = 0 x ao ?2T2 -T1 = 0 …(4) Solving these 4 equations, we can find all accelerations and tensions. 2 T1 T2 T2 ao FBD of Pulley 2 15 www.vidyadrishti.com
Alternative Method for Example 7 [using pseudo force concept] : P. K. BHARTI, IIT KHARAGPUR Alternative Method for Example 7 [using pseudo force concept] We can draw FBD of masses m3 and m4 using pseudo force concept. In this method we draw FBD of masses m3 and m4 in pulley 2 frame, which is non-inertial frame.
As pulley 2 is accelerating with an acceleration a0 downward wrt ground, we have to apply pseudo force in the direction opposite to that of a0.
All other FBD will be same as was in first method. 16 www.vidyadrishti.com
Alternative Method for Example 7 [using pseudo force concept] : P. K. BHARTI, IIT KHARAGPUR Alternative Method for Example 7 [using pseudo force concept] M3ao +T2 -M3g =M3ar
? M3g – T2 =M3(ao-ar)
Which is same as equation (2)
in method 1. M3 M3g T2 M3ao
(pseudo) ar 17 FBD of mass M3 www.vidyadrishti.com
Alternative Method for Example 7 [using pseudo force concept] : P. K. BHARTI, IIT KHARAGPUR Alternative Method for Example 7 [using pseudo force concept] M4g -T2 -M4ao = M4ar
? M4g – T2 =M4(ao+ar)
Which is same as equation (3)
in method 1.
All other FBD will be same as in
earlier method M4 M4g T2 ar M4ao
(pseudo) 18 FBD of mass M4 www.vidyadrishti.com
Exercises [Time 1-2 hrs for beginners] : P. K. BHARTI, IIT KHARAGPUR Exercises [Time 1-2 hrs for beginners] After learning the concept of pseudo force, solve these problems:
From H. C. Verma [Ch: Newton’s Law] :
Solve Examples: 9, 10, 11.
Page no: 80, Qs. no. 16,
Page no: 82, Qs. no. 33.
From I. E. Irodov:
Page no: 23-25, Qs. no. 73, 81 & 82. 19 www.vidyadrishti.com
Constraint Equations : P. K. BHARTI, IIT KHARAGPUR Constraint Equations Constraints mean that the bodies are not free to move the way they want. The accelerations between them are dependent on each other.
The constraints equations gives the relation between accelerations or velocities of different masses. For the case of pulley string system no. of constraints equation = no. of strings. 20 www.vidyadrishti.com
How to write Constraint Equations : P. K. BHARTI, IIT KHARAGPUR How to write Constraint Equations There are so many methods to write constraint equation. But some of the steps may be as follows:
STEP 1: Find out the relationship between displacements of different bodies.
STEP 2: Differentiate this relationship of step 1 wrt time t , to get relationship between velocities of different bodies.
STEP 3: Differentiate this relationship of step 2 wrt time t , to get relationship between acceleration of different bodies. 21 www.vidyadrishti.com
Slide 22 : P. K. BHARTI, IIT KHARAGPUR Ex. 8. Let the displacement of vertical bar be y downward and that of wedge be x leftward.
Clearly, tan? = y/x ? y = x tan?
Differentiating wrt time t, we get, vy = vx tan?
Again differentiating wrt t, we get, ay = ax tan? ? ? x y 22 www.vidyadrishti.com
Constraint eqn in pulley string system : P. K. BHARTI, IIT KHARAGPUR Constraint eqn in pulley string system In this type of situation, we will use the advantage of constant length of string. Steps for writing constraint equations will be as follows:
STEP 1: Express length of the string in terms of different segments of the string. This expression will have some constant length part and some changing parts.
STEP 2: Differentiate and double differentiate expression of STEP 1 wrt time t. This will give the relationship between velocities and accelerations of different bodies.
You can use your common sense to find the direction of velocities and accelerations of different bodies.
Let us some examples. 23 www.vidyadrishti.com
Example 9 : P. K. BHARTI, IIT KHARAGPUR Example 9 One thing is clear here, P & B moves together. Therefore,
accln of B = accln of P. Also, if B moves
downward, then A moves rightward.
Now, String length AD = L (say)
L = AC +[length of circular arc of C = c1 (say)]
+ CP + [length of circular arc of P = c2(say)] + PD
? L = AC + c1 + CP + c2 + PD
? L = AC + c1 + CP + c2 + CP + CD
[because length PD = CP + CD]
? L = x + c1 + y + c2 + y + CD
? L = x + 2y + c1 +c2 + CD
Now, as A moves x changes and as P & B moves y changes, but length c1, c2 and CD remains constant. Thus, L, c1, c2 and CD are constant.
…CONTINUED A C P B D x y 24 www.vidyadrishti.com
Example 9 continued… : P. K. BHARTI, IIT KHARAGPUR Example 9 continued… L = x + 2y + c1 +c2 + CD
Now, when we differentiate both side of this expression wrt time t, we get
0 = vA + 2vB
[Because differentiation of x wrt t gives velocity of A and similarly derivative y wrt t gives velocity of P which is same as velocity of B .]
? vA = -2vB
? Magnitude of velocity of A = 2 x (magnitude of velocity of B).
Again , differentiating both sides of this eqn. wrt t we get
aA = - 2aB
?Magnitude of acceleration of A = 2 x (magnitude of acceleration of B).
[See slide no. 23 and compare result] 25 www.vidyadrishti.com
Slide 26 : P. K. BHARTI, IIT KHARAGPUR Ex 10. Let the length of the strings be l1 and l2 respectively
x1 +c1+x2=l1 ? a2 = - a1
& (x3 - x2)+c2+ (x4 - x2) = l2
? (a3 - a2)+(a4 - a2) = 0
? a3+a4-2a2 =0
? a3+a4+2a1 =0 26 1 2 3 4 x3 x1 x2 x4 www.vidyadrishti.com
Example 11 : P. K. BHARTI, IIT KHARAGPUR Example 11 Find the accelerations of block A and B of example 9 assuming masses of each A and B be m.
SOLUTION:
For this type of problems we have to first find the relationship between accelerations of A and B, which we have already found out in ex 9 itself using constraint eqn. It was,
Magnitude of acceleration of A = 2 x (magnitude of acceleration of B).
Let acceleration of B be a downward
It means acceleration of A will be 2a rightward.
Let the tension in the upward and downward strings be T1 and T2 respectively [see fig of ex 9].
Now, we draw FBD of A, B and pulley P. 27 www.vidyadrishti.com
Example 11 continued… : P. K. BHARTI, IIT KHARAGPUR Example 11 continued… Block A:
Along Horizontal: T1 = m (2a) ? T1 = 2ma …(1)
Along vertical: N – mg = 0. (2)
Block B:
mg - T2 = ma …(3)
Pulley P [mass m=0]: T1+T1=T2 ? 2T1 = T2 …(4)
Solving these eqns. we get: a = g/5.
Therefore, accln of block A = 2a = 2g/5 rightwards. [Ans]
And, accln of block B = a = g/5 downwards. [Ans] 28 A B 2a a N mg a mg T1 T1 T1 T2 T2 P www.vidyadrishti.com
Exercises: : P. K. BHARTI, IIT KHARAGPUR Exercises: After learning the concept of constraint eqns., solve these problems:
From H. C. Verma :
Solve Examples: 6 and 7 [page no. 73]
Page no: 81-82, Qs. no. 30, 32, 33 and 34.
From I. E. Irodov:
Page no: 23-24, Qs. no. 72, 75, 76. 29 www.vidyadrishti.com
Example 12 : P. K. BHARTI, IIT KHARAGPUR Example 12 At the moment t=0 the force F = kt is applied to a small body of mass m resting on a smooth horizontal plane [k=constant]. F always makes angle q with the horizontal. Find:
(a) The velocity of the body at moment of its breaking off the plane.
(b) The distance traversed by body up to this moment.
SOLUTION:
First of all read question twice.
Now, answer this question, what does breaking off the plane mean ?
Breaking off the surface means, it leaves the contact with surface.
As there is no contact, means there is not any normal force.
First we draw FBD of block before its leaves the contact with surface. 30 F q www.vidyadrishti.com
Example 12 continued… : P. K. BHARTI, IIT KHARAGPUR Example 12 continued… FBD OF BLOCK:
Now, we take our coordinate system.
Resolve F into its component.
Let the accln be a [ clearly it will be rightwards].
Along x- axis :
F cosq = ma
? kt cosq = ma [because F = kt, from question]
?a = kt cosq/m ? dv/dt = kt cosq/m [because a =dv/dt]
?? dv = ? (kt cosq/m) dt = (k cosq/m) ?t dt
[because, as body in question starts from rest, therefore, @ t= 0, u = 0
and, @ t= t, v=v] 31 F q N mg F cosq F sinq x y a 0 v 0 0 t t www.vidyadrishti.com
Example 12 continued… : P. K. BHARTI, IIT KHARAGPUR Example 12 continued… Integrating, we get
v= (k cosq/m) t2/2 …(1)
Along y axis:
N + F sinq –mg = 0 ? N + kt sinq –mg=0
At the moment block breaks off the surface, N=0
Therefore , eqn becomes: kt sinq – mg = 0
? t = mg/k sinq …(2)
(a). Putting this t in (1), we get the velocity at the moment of break off:
v = mg2 cosq /2asin2q 32 www.vidyadrishti.com
Example 12 continued… : P. K. BHARTI, IIT KHARAGPUR Example 12 continued… (b) Similarly, use v = dx/dt and integrate eqn. (1) to find x.
Then , put t from eqn (2).
Do it yourself.
Ans of this part: x = (m2g3cos q)/(6a2sin3q). 33 www.vidyadrishti.com
Final Exercises : P. K. BHARTI, IIT KHARAGPUR Final Exercises I. E. Irodov
Page no. 20-25
Qs. No. 59, 69,71.
Solve all remaining problems of H. C. Verma including objective types.
Please don’t take help of anyone. 34 www.vidyadrishti.com