Question 1
A body of Mass m moves along the X-axis such that a time t its position is given by following expression $x=at^{3/2} -bt + c$ . Where a, b and c are constant
(a) Calculate the acceleration of the body
(b) What is the force acting on it
(c) What is the force at t=1 sec
(d) What are the dimension of a Solution
Given
$x=at^{3/2} -bt + c$ ---(1)
(a)We know that
$ a=\frac{d^2x}{dt^2}$
So differentiating the equation (1) twice w.r.t to time
$a=\frac{3at^{-1/2}}{4}$
(b) Now force is given by
$F=ma$
So $F=\frac{3mat^{-1/2}}{4}$
(c) At t=1
$F=\frac {3ma}{4} \ N$
(d) From equation 1, it is clear that
$at^{3/2}$ has dimension L
So dimension of a will be $LT^{-3/2}$
Question 2
An 8 Kg object is subjected to three forces
(a) Find the acceleration of the object.
(b) If the object starts from rest from origin, what will be the location after 4 sec
(c) What is the magnitude of resultant force and its direction?
(d) What fourth force F4 should be applied on the body to make the net resultant force zero Solution
Net force on the object Fnet = F1+ F2+ F3 Fnet = (20i + 30j) +(22i -10j) +(6i+4j) = 48i + 24j
Now , Acceleration is given by a = Fnet/m = 6i+3j
(b) Now equation of motion
$\mathrm{s\ }=\mathrm{\ u}t+\frac{1}{2}at^2$
Here, u =0 and t= 4 sec
so s = (1/2) (6i+3j) *16 = 48i+24j
(c) Fnet =48i + 24j
Magnitude=$\left(48^2+24^2\right)^{1/2} =53.65$ N
Direction =${tan}^{-1}{2}$
(d) Fnet =48i + 24j
Now Fnet + F4 =0 F4=- Fnet = -(48i + 24j) N
Question 3
Consider a three body system shown in figure below
(a) Find the acceleration of the each object
(b) Find the contact force between all the objects Solution
Let a be the acceleration of the system
Let $f_1$ be the contact force between A and B
Let $f_2$ be the contact force between B and C
Let's draw free body diagram for each object
For object A
$F- f_1 = m_1a$ ---(1)
For object B
$f_1 - f_2 =m_2 a$ ---(2)
For object C
$f_2 =m_3 a$ ---(3)
Adding 1 ,2 and 3 we get
$a=\frac{F}{m_1+m_2+m_3}$
Substituting the value in 1 and 3 ,we get
$f_1 =\frac{(m_2+m_3)F}{m_1+m_2+m_3}$
$f_2 = \frac{m_3F}{m_1+m_2+m_3}$
Question 4
Three Block of mass m1, m2 Â and m3 are connected as shown in the figure below.
All the surfaces are frictionless and strings and pulley are light. Find the acceleration of all the masses Solution
Let take $a_1$ as the acceleration of block $m_1$ towards left. This will also be the acceleration of Pulley B because the strings length is constant.
Similarly, as the strings length is constant over pulley B, magnitude of Acceleration of block $m_2$ will be same as acceleration of block $m_3$ with respect to pulley B. Let assume that acceleration to be $a_2$.
Then Acceleration of Block $m_2$ wrt to ground will be = $a_1 - a_2$ (downward)
Then Acceleration of Block $m_3$ wrt to ground will be = $a_1 + a_2$ (downward).
Let take $T_1$ be the tension in upper string and $T_2$ be the tension in the lower strings
Let draw the free body diagram for each items
(a) Pulley B
(b) Block $m_1$
(c) Block $m_2$
(d) Block $m_3$
Since the pulley is light ,
We should have
$T_1 =2 T_2$
For block $m_1$
$T_1=m_1a_1$ ---1
For block $m_2$
$m_2g -T_2=m_2(a_1 - a_2)$
Or
$m_2g - \frac {T_1}{2}=m_2(a_1 - a_2)$ --2
Force block $m_3$
$m_3g -T_2=m_3(a_1 +a_2)$
or
$m_3g - \frac {T_1}{2}=m_3(a_1 +a_2)$ -3
Solving equation 1,2 and 3 we get
$a_1=\frac{g}{1+\frac{m_1}{4}(\frac{1}{m_2}+\frac{1}{m_3})}$
$T_1=\frac{m_1g}{1+\frac{m_1}{4}(\frac{1}{m_2}+\frac{1}{m_3})}$
Question 5
A small block of mass m1= 3 kg is placed at rest on a larger block of mass m2=5 kg .The situation is given in the figure below
The coefficient of friction between the two block is µ=.3
And the horizontal surface is smooth.A constant Force F is applied on the block
Find out the following
(a) Find the value of limiting friction between the two blocks
(b) What is the maximum acceleration by which the upper block can move
(c) What is the maximum value of F at which both the block move together
(d) if F=20 N,what is the acceleration of each block and what frictional force is acting between the block
(e) What is the normal contact force between the blocks
(f) what is the normal contact force between the larger block and the smooth surface
(g) if F=40 N,what is the acceleration of each block and what frictional force is acting between the block
(h) If the force is applied to the upper block,what will be the minimum force required so that there is relative motion between the block
Given g=10 m/sec2 Solution
(a) The limiting friction is given by
$=\mu N$
Where N is the normal contact force between the block
Now $N=m_1g=3 \times 10=30$N
So Limiting friction($f_{max}) =.3 \times 30=9 \ N$
(b) The only force responsible for acceleration on the upper block in the friction force
So
Maximum Acceleration=Limiting Friction/m
$a_{max}=\frac {9}{3}=3 \ m/sec^2$
(c) Maximum acceleration of the upper block=$3 \ m/s^2$
So both the block will move together if the acceleration of the lower -block does not exceed $3 \ m/s^2$
So maximum force=$(m_1+m_2) a_{max} = 24 \ N$
(d) For F=20 N ,Since it is less the maximum Force
So both the both the block will move together
$\text{Acceleration}=\frac {20}{8}=2.5 \ m/sec^2$
Friction force acting between the block=$m_1a=3 \times 2.5=7.5 \ N$
(d)Normal contact force between the block
$N=m_1g=3 \times 10=30 \ N$
(e) Normal Contact force between the larger block and surface
$N=(m_1+m_2)g=80 \ N$
(f) For F =41 N,since it is greater then maximum force,both the block will have different acceleration
For upper block
$F_{max}=m_1a_1$
Or $9=3a_1$ or $a_1=3 \ m/s^2$
For larger block
$F- F_{max}=m_2a_2$
$41-9=8a_2$
Or $a_2=4 \ m/sec^2$
(h) If the force is applied on the upper block
The only force responsible for motion of the Lower block is the frictional force
So Maximum acceleration of the lower block= Limiting Friction/m =9/5=$1.8 \ m/sec^2$
Maximum acceleration of the lower block=$1.8 \ m/sec^2$
So both the block will move together if the acceleration of the lower -block does not exceed $1.8 \ m/sec^2$
So maximum force=$(m_1+m_2) a_{max} = 14.4 \ N$
Question 6
A object of mass M is standing in a stationary lift. What pressure force N exerted by the object on the floor of the lift
(a)If the lift is stationary
(b) if the lift is moving upward with acceleration a
(c) if the lift is moving downward with acceleration a
(d) if the lift is falling freely
(e) if the lift is moving upward with constant velocity
(f) if the lift is moving downward with constant velocity Solution
The forces acting on the object
Weight of the body acting downward (Mg)
N pressure force exerted by the lift floor on the object in upward direction( this is the same pressure force exerted by the object on the lift floor)
(a) if the lift is stationary
$Mg - N =0$
So $N =mg$
(b) if the lift is moving upward with acceleration a
$N-Mg=Ma$
Or $N=M(a+g)$
(c) if the lift is moving downward with acceleration a
$Mg-N=Ma$
Or $N=M(g-a)$
(d) if the lift is falling freely
$N-Mg=Mg$
Or $N=0$
(e) if the lift is moving upward with constant velocity
$a=0$
so $N=Mg$
(f) if the lift is moving downward with constant velocity
$a=0$
so $N=Mg$
Question 7
A piece of uniform strings hangs vertically so that its free end just touches the horizontal surface of the table.The upper end of the strings is now released. Show that at any instant during the falling of string, the total force on the surface of the table is three times the weight of the part of the string lying on the surface Solution
Let y be the length of string lying on the surface of the table at any instant during the fall.If additional length dx of the strings falls on the surface in time dt.The velocity of this part when it strikes the surface of the table is given by
$v^2=u^2 + 2gx$
Now as u=0
So $v=\sqrt{2gx}$
Now the total force on the surface of the table
=rate of change of Momentum of the length dx+ weight of the length of the string lying on the surface of the table
$F=\frac{d}{dt}(mdxv) + mxg$
$ =mv\frac{dx}{dt}+mxg$
$=mv^2 + mxg$
$=2mgx + mxg$
$=3mgx$
$ =3Mg$
As M=mx
Question 8
Three blocks A, B,C are such as
M1=2 kg , M2=3 kg , M3=6 kg
They are connected as shown in the below figure.
The coefficient of friction between the block M2 and table is 0.2
Find out the following
(a) Draw all the forces acting on the system
(b) The acceleration of the system
(c) Frictional force between the block M2 and table
(d) Tension in the cord on the left and tension in the cord on the right
Given g=10 m/sec2
The pulley are light and friction less Solution
Let $T_1$ be the tension in the left cord
$T_2$ be the tension in the right cord.
The system will move from left to right as $M_3$ has the larger mass
Let a be the acceleration of the system
For Block A
$T_1 – 20=2a$ ----(1)
For block B
$T_2 – f – T_1 = 3a$
Now Frictional force $f= =.2 \times 3 \times 10=6 \ N$
As the body is in vertical equilibrium so N=mg
So
$T_2 – 6 – T_1 = 3a$ --(2)
For block 3
$60 –T_2=6a$ ---(3)
Adding 1 ,2 and 3 we get
$60 -6-20=11a$
Or $a=\frac {34}{11}=3.09 \ m/s^2$
Now from equation 1
$T_1=20+ 2 \times 3.09=26.18 \ N $
From equation 3,we get
$T_2=60-6 \times 3.09=41.46 \ m/s^2$
Question 9
A boxcar is moving such that  Initial velocity v=0
And Acceleration= (4 m/sec2 ) i. Two objects A and B of mass 2 kg is kept in the boxcar. Take g=10m/sec2
Find out following
(a) If the object A slid along the frictionless floor with the velocity v= (10m/s) i. Find out the equation of the motion of object from the frame of reference of Boxcar. Also what time would it take to reach it original position relative to box car
(b) The object B slid along the rough floor with the velocity v= (10m/s) i. Find out the equation of the motion of object from the frame of reference of Boxcar. Coefficient of sliding friction =.3, Coefficient of Static friction >0.5 Solution
(a) From the frame of reference of Box Car,A pseudo force has to be introduced to apply the Newton second law of motion
So Pesudo force=-8i N
So from Newton second law,
Acceleration would be
-8i=2a
Or a = (-4) i
Now we know from kinematics v=u+at
So v=10i -4t i
So
$\frac{dx}{dt}=10-4t$
Or
x=10t-2t2 +c
At t=0 x=c
Time at which it will return to its original place
c=10t-2t2 +c
or t=5 sec
(b) From the frame of reference of Box Car,A pseudo force has to be introduced to apply the Newton second law of motion
So Pesudo force=-8i N
Also since the object is moving,a frictional force will also act to prevent the relative motion .
Friction force=$\mu mg=.3 \times 2 \times 10=6 \N$
So from Newton law
-8i – 6i=2a
Or a=-7i
The object will move till the velocity becomes zero.At that point .Static friction will come into picture .Since the value of static friction=.5*2*10=10N is greater then pseudo force.The object will stay there only
Now we know from kinematics v=u+at
So v=10i -7t i
So
$\frac{dx}{dt}=10-7t$
Or
x=10t-3.5t2 +c
Question 10
A bead of mass M is fitted onto a rod with a length of 2s and can move it without any friction. At t=0,the bead is in the middle of the rod. The rod moves translationally in a horizontal plane with an acceleration a in a direction forming an angle α with the rod.
(a) Find the acceleration of the bead relative to ground
(b) Find the acceleration of the bead relative to rod
(c) The reaction force exerted by the rod on the bead
(d) The time when the bead will leave the rod Solution
Let’s first check out the force acting on the bead
The Reaction force of the rod on the bead which is at right angle to the rod. There is no other force acting. Let the reaction force is N. Let w be the acceleration of the bead wrt to ground, then
N=Mw
The relative acceleration of the bead with respect to rod will be given by wr = w – a
The vectors are shown below in figure
From the triangle ,it is clear that
wr=acosα
w=asinα
So N=Masinα
Now the time period which the bead moves along the rod
$s=\frac{1}{2}w_rt^2$
Or $t=\sqrt{\frac{2s}{acos{\alpha}}}$
Question 11
A mass less rod of length L with a small load (A) of mass m at the end is hinged at point X. And it is strictly vertically position touching a body B of mass M. A light jerk sets the system in motion. Let its separate out at angle α from the horizontal
Let i and j are the unit vectors across the X and Y axis
(a) what will be the velocity vectors of A and B at separation
(b) What will be centripetal acceleration of object A at time of separation?
(c) Find out the ratio of M/m in terms angle of separation
(d) If they separate out at angle α=30, what will the ratio of M/m
(e) What will be the velocity of A and B if α=30
(f) What will be value of tangential acceleration at separation? Solution
As long as the object A touched the body B, the velocity of the body B is equal to the horizontal components of the velocity of the load A.
Similarly acceleration of the body B is equal to the horizontal components of the acceleration of the load A.
Let a be the total acceleration of the load A. Now Since the load will be moving along circular path, this can be written as
$\mathbf{a}=\mathbf{a_t}+ \mathbf{a_c}$
Where $\mathbf{a_t}$ = Tangential acceleration of the load
$\mathbf{a_c}$=centripetal acceleration of the load
Now $a_c=\frac {v^2}{L}$ where v is the velocity of the load and it is directed towards center
The horizontal components of the acceleration can be written as
$ a_h=a_tsin{\alpha}-\frac{v^2}{L}cos{\alpha}$
Now body B also has the same acceleration, we can write
$N=Ma_h$
$N=Ma_tsin{\alpha}-M\frac{v^2}{L}cos{\alpha}$
At the moment of separation N=0
So
$a_t sin{\alpha}=\frac {v^2 cos {\alpha}}{L}$ ----(1)
Now the tangential acceleration at the time of separation will be because of force of gravity
So
$a_t=gcos{\alpha}$ ---(2)
Solving (1) and (2)
We get
$v=\sqrt{gLsin{\alpha}}$
or
$\mathbf{v}=(sin{\alpha}\sqrt{gLsin{\alpha}})\mathbf{i}-(cos{\alpha}\sqrt{gLsin{\alpha}})\mathbf{j}$
And velocity of the body B
$u=vsin \alpha$
or
$u=sin{\alpha}\sqrt{gLsin{\alpha}}$
Or
$\mathbf{u}=(sin{\alpha}\sqrt{gLsin{\alpha}})\mathbf{i}$
Centripetal acceleration of the body A =$\frac {v^2}{L}= g sin{\alpha}$
Now According to law of conservation of energy, we have
$mgL=mgLsin{\alpha}+\frac{1}{2}mv^2+\frac{1}{2}Mu^2$
$mgL=mgLsin{\alpha}+\frac{1}{2}mgLsin{\alpha}+\frac{1}{2}MgL{sin}^3{\alpha}$
Or
$\frac{M}{m}=\frac{2-3sin{\alpha}}{{sin}^3{\alpha}}$
For $\alpha=30$
M/m=4
Now velocities of A and B
$\mathbf{v}=(sin{\alpha}\sqrt{gLsin{\alpha}})\mathbf{i}-(cos{\alpha}\sqrt{gLsin{\alpha}})\mathbf{j}$
$\mathbf{v}=\sqrt{\frac{gL}{8}}\mathbf{i}- \frac{3gL}{8}\mathbf{j}$
$\mathbf{u}=\sqrt{\frac{gL}{8}}\mathbf{i}$
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