Parallel Forces-merged

Saturday, February 19, 2022
9:48 PM

Parallel Forces

Resultant of two like parallel forces

If two like parallel forces P and Q act at points A and B then resultant R = P + Q acts at a
point C where P. AC =Q. BC

If two unlike parallel forces P and Q act at points A and B then resultant R=P-Q

Exercise

1) Find two like parallel forces acting at a distance of 2.5 m apart, which are equivalent
to a given force of 30 N, the line of action of one being at a
distance of 50 cm from the given force.
Solution
Let P and Q be two like parallel forces acting at points A and B and R = 30 N be their
resultant which acts at a point C Where

AC =50 cm =0.5 m
AB =2.5 m
CB =2 cm
Now

2) Replace a force of magnitude 50 kg. wt. by two unlike parallel distance of 2
m and the other at 8 m from the given force.
Solution
Let P and Q be two unlike parallel forces (P > Q) and R = 50 kg resultant which
acts at a point C where AC = 2 m, BC = 8 m AB= 6 cm

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acts at a point C where AC = 2 m, BC = 8 m AB= 6 cm

3) The uniform rod, 12 m long and weighing 17 N can turn freely about a poit
and the rod is in equilibrium. When a weight of 7 N is hung at one end ,how
far from the end is the point about which it can turn?
Solution
Let AB be a uniform rod of length 12 m whose wt. 17 N acts at its centre O. Let
the rod turns freely about a point C which is x m from the end A when a wt. of 7
N is hanged 17 N

4) Two like parallel forces P and Q act at points 18 m apart, if the force is 9 N
and acts at a distance 6 m from P, find Q.
Solution
Let two like parallel forces P and Q act A and B A where AB = 18 m. Their
resultant R = 9 N acts at a point C where AC = 6 m, BC = 12 m

5) The A uniform bar 4 m long and and weighting 3N passes a prop and is supported
in a Horizontal position by a force
Solution

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Solution

Let AB be a uniform bar of length 4 cm whose wt. 3N. acts at its centre O where AO
=OB=2 cm let it passes over a prop C. It is supported in a horizontal position by a
force IN acting vertically upwards at other end A. Let OC = x rn.
Now,

The distance of prop from the centre

6) Two men, one stronger than the other have to remove a block of mass weighing 270

N with a light plank of length 6m. If the stronger man is able to carry 180 N, how
must the block be placed so as to allow him that share of the weight?
Solution
Let P = 180 kg be the wt. of stronger man and Q be the wt. of weaker man. Let C be
the position of block of stone whose wt. is 270
270=180+Q

Q =90
If Ac=x then BC=6-x

7) The extremities of a straight bamboo poles rest in a horizontal position on two pegs A and B a
heavy load hand from a point ,If 2AC=5BC and presure at B is 75 more than A fond the weight
of the load
Solution

Let P N be the pressure at A then pressure at B = (P + 75)N. Let the wt. of load at C
is W N

2 AC=5 BC

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8) uniform beam, 7 m long is supported in a horizontal which are 4 m apart,
so that the beam projects 1 m beyond gel&, If the beam weighs 80N, find
the pressure on the two
Solution
Let AB be a uniform rod of length 7 m whose 80 wt. N acts at its centre C.
AC=BC=3.5
Let E and D be the two props such that BD=1 m
CD=2.5 and EC=1.5

Let P and Q be two pressure at two props

9) straight weightless rod 48 cm in length, rest in a horizontal position
between two pegs placed at a distance of 6 cm apart. one peg being at the
rod and a weight of 2 N is suspended from the other pressure on the pegs.
Solution

Let AB be a straight weightless rod of the ' length 48 cm resting and C where AC = 6
cm BC = 42 cm
Let P and Q be the pressures on pegs at A and C,
The system is in equilibrium when the weight of 2 N is suspended at B

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The system is in equilibrium when the weight of 2 N is suspended at B
Let Q be the resultant forces P and 2 N,
Now,

10) A man carries a bundle at the end of a stick which is place over shoulder , if
the distance between his hand and shoulder is changed , how does the pressure
on his shoulder change?

Solution
Let AB length of stick =l
C is the position of the shoulder
W be the weight of the bundle
R be the reaction of the shoulder
P be the pressure due to the hand
AC= x suppose
BC= l-x

From second and third

Reaction is inversely propositional to the distance between the hand and shoulder

11) Two parallel forces P and Q, act at a given points of a body, if Q is changed
show that the line of action of resultant is the same as it would be if the forces
are simply interchanged.
Solution
Let two parallel forces P and Q act At A and B

When Q is replaced by Their resultant is P+

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When Q is replaced by Their resultant is P+

from I and II
If

If P and Q are simply interchanged and their resultant act at D

BD=BC

12) Two like parallel forces P and Q acting at the end points A and B of a road AB
of l , If opposite forces each of magnitudes S and added to P and Q , then prove
that the line of action of the new resultant will be distance through a distance
Solution
Let two like parallel forces P and act at points A and B where AB = l, Let their

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Let two like parallel forces P and act at points A and B where AB = l, Let their
resultant R = P + Q acts at a point. C

When the forces S and -S are added to P and Q then their resultant
R=P+Q

BD=

Now
BD-BC =

13) Two like parallel forces P and Q (P > Q) acting at A and B. Show that!
interchange their positions, the point of application of the resultant is displaced
by
Solution
Let two like parallel forces P and act at points A and B where AB = l, Let their
resultant R = P + Q acts at a point. C

If P and Q are simply interchanged and their resultant act at D

AD- AC=
14) unlike parallel forces P and Q (P > Q) are acting at a distance d apart. If

each force is increased by S, show that the resultant will be unaltered in
magnitude but its point of application will be moved through a distance
solution
Let the resultant of force R= P-Q of two unlike force

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When each forces P and Q are increased by S then new resultant is R=P-Q act at C'

now
CC'=

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21.1

Monday, February 28, 2022
6:54 PM

Newton first law of motion
Newton second law of motion
Newton third law of motion
Important formula
Exercise 21,1

1) a. A body of mass 50 gram is hanging on a roof. Find the pull of the earth on it. (g =
9.8ms)
Solution
Mass= 50 grams

=0.05 kg
Acceleration (g)= 9.8 m/ s2
Pull of earth = mg

=

b. A car of mass 750kg is brought to rest by applying a force of 1500S. Find the
average retardation.
Solution
Mass= 750 grams

=0.75 kg
Force (F)=15000
Retardation =?
We have

c) Find the mass of a body, when a force of 50N is applied on it produces an acceleration of
5.5 m/s2

Solution
Force (F)=50
Acceleration (a)=5.5
Mass (m) =?
We have

d) A constant force of 10 N acting on an object reduces its velocity 15m/s to 5 m/s
in 2 seconds. Find the mass of an object.
solution
Force (F) = 10 N
initial velocity ( u) = 15 m/s
Final velocity (v) = 5 m/s
time (t) = 2 second
we have
Acceleration =

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Acceleration =

a. A player kicks of ball of 0.25 kg and produces a velocity of 6 m/s for 0.2
sec. How much was the force applied to the ball?
solution
Mass =0.25 k g
Velocity =6 m/s
time (t) = 0.2 second
we have
Acceleration =

b) A body of mass 50 kg falling from a certain height is brought to rét after
striking the ground with a speed of 5m/s. If the resistance forced the ground is
500 N, find the duration of contact

solution
Force =500 N
Mass =50 kg
Velocity =5 m/s
time (t) = ?
we have

Acceleration =
Again we have

2c) A man of 50 kg mass jumped from a certain height and came down the
ground with a velocity of 10 m/s If he is brought to rest in one tenth of a second,
find the force acting on the man
solution
Mass =50 kg
Velocity =10 m/s
time (t) =
we have

Acceleration =
Again we have

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3 a) Rain drops falling vertically on a flat roof at the rate of 0.2 kg/s come to
rest after hitting the roof. If the resistance force of the roof is 2 N, find the velocity

of rain.

Solution

Mass per second

Force = -2N
Final velocity = 0
Initial velocity = u
We have

F=ma

3b) Sand is allowed to fall vertically at a steady rate hits a horizontal floor
with a speed 0.05 m/s . If the force exerted on the floor is 0.005 N, find the
mass of sand falling per second.

Solution

Mass per second

Force =- 0.005
Final velocity = 0
Initial velocity = 0.05
We have

F=ma

4) How large force is required to bring a 1000 kg car moving with a to rest
(n) nt distance of 1000 m? (b) in 20 second?
, solution
Mass of car (m)= 1000 kg
Initial velocity (u) = 100
Final velocity =0
distance travelled =100 m
we have

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a= -5

b) Time =20 second

Force = m

5 a) a. A body of mass 7 kg changes its velocity 7 m/s to 21 hrs force of 5 kg wt.

Through what distance must the body travelled
solution
mass ()=7 kg
Initial velocity (u) = 7 m/s
Final velocity = 21 m/s
Force (F) = 5 kg weigt

=

distance =?
we know

a=7

s=28

5b) A train whose mass is 100 tonnes moving at the rate of 3 km per minute is
brought to rest in 10 seconds by the action of uniform force. Find how far the
train runs before coming to rest and calculate magnitude of the force.
Solution

force (F)= weight of 20 kg
=

Mass = 2 metric tones
= 2000

we have ,

196=
a=0.098
Initial velocity =0
Distance travelled = 39.2 m
time =?
we have

t=28.28

5 c) A mass of 10 kg falls 4 m from rest and is then brought to rest by penetrating
into some sand, find the average thrust of the sand on it. (g

Solution

Mass (m)=100 tones = 100000 kg

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Mass (m)=100 tones = 100000 kg m/s =50 m/s

Initial velocity (u)= 3 Km per minutes =

final velocity (V) =0 m/s

Distance travelled (s) = ?
We know,

Again

F=

6)

A mass of 10kg falls 4 m from rest and is then brought to rest by penetrating
into some sand, find the average thrust of the sand on it. (g

Solution

Mass (m) =10 kg

Initial velocity (u)=0
Distance travelled (s) = 4m
Acceleration (g)=10

Final velocity (v)=?
We know,

After penetrating into sand,

Initial velocity (u)=
Final velocity (v)=0

Let R be the thrust due to sand

R=

7) A man on lift moves (i) up (ii) down with an acceleration of 2 ms-2 in each
case calculate the reaction of the floor on a man of mass 50 kg on lift.
solution
Acceleration (a) = 2ms-2

Mass (m) = 50kg
Let R be the reaction of the floor on a mass and man in moving up,

R — mg =ma

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R — mg =ma
R =mg + ma

=50 9.8+50
=590

As the man is moving downwards

R= 390

8. a. A stone of mass 1 kg falls from the top of a vertical cliff. After (i)
falling for 3 seconds (ii) descending 300 cm, it reaches cliff and
penetrates 25 cms into sand. Find the resistance sand. (g = 10 m s -2)
Solution
Time of descend (t)=3

Second initial velocity (u)=0

Acceleration (g) =
Final velocity before penetrating sand,
After penetrating Initial velocity into sand

=

After penetrating in to the sand

Initial velocity =30
Final velocity (v) =0
Distance(s)=25 cm

=0.25 m
We have

a=1800

let R be resistance offered by sand
R -mg= ma

=
=1810 N
ii)

Distance trevelled (s) =300 cm =3m
Initial velocity (u)=0
Final velocity before penetrating (v) = ?

We have,

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=

After penetrating,
Initial velocity (u)=0
Final velocity (v)=0

Distance travelled (s) = 25 cms=0.25 m
Again let R be resistance offered by sand then,
R - mg = ma
R = mg + ma

8b) ' A bullet fired into a target loses half of its velocity after penetrating 6 cms.
How much further will it penetrate?
Solution
Let u be the initial velocity of the bullet.

Then by question

Final velocity (v)
Sl =6 cm

Again,

Again
Final velocity (V)=0

In second time distance penetrated by the bullet

8-6=2

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9) A mass m is acted on by a constant force P kg. wt and in t secs it mov5i
distance x metres from rest and acquires a velocity ofvm/s. Show that

Solution

Force (F)=P kg weight
=P g Newton

In time t , it travels x meters from rest

x=

x=
we have
F=ma
P g=m a

Putting in equation I
x=
Again

putting value of a

10)
An airship of total mass M kg is falling with acceleration f weight of the load
which must be thrown out in order that it NY the same acceleration fm/sec2.

Solution

Mass of Airship=M
When it is moving downward

kg mass is taken out so that it rises up with same accclcration fm/s2:

11) A bglloon is rising with acceleration f, Jirovc that the fraction of the weight or

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the bglloon which must be emptied out of the balloon in order to double the
acceleration is

solution

Let M be the mass of balloon and R be upward force due to air, As the balloon
rises up, reaction is in upward direction

i.e.

R= Mf+ Mg ...... (i)
let x kg be mass that is taken out in order to double the acceleration 2 f

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21.2

Tuesday, March 1, 2022
7:23 AM

Exercise 21.2

A force of 15N acts on a body for 10 seconds. Find the impulse of the force
on the body.
Solution
Force (F) = 15N
Time (T) = 10 seconds

Impulse = Force x times

=

15 x 10
. = 150 NS.
1b) body of mass 0.3kg initially at rest is subjected to a force of3N for 2
seconds. Calculate the change in momentum of the body.

Solution
Mass (m) = 0.3kg

Initial velocity (u)=0
Final velocity (v)=?

Force (F)=3 N
Time (T) = 2 second
we have
F=ma

a=10
again we have

v=20
change in momentum=

2) A bullet of mass 30g is fired from a rifle of mass 1500 gram with a

50 kmh-l. Find the velocity of the recoil of the rifle.
solution

Mass of bullet (m) 30 g = 0.03 kg

Mass of rifle (M) 1500 gm = 1.5 kg

Velocity of bullet (v) 50 km h-1

Velocity of rifle (v)=0

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Velocity of rifle (v)=0
We know,
From conservation of linear momentum,

3)

A 10g shot is fired from a 10kg gun with a velocity of SOO n/s. This
enters a block of wood of mass 900g. Find the velocity of the gun
common velocity of shot and block of wood.

solution

4) a shot is fired from a 10kg guan with a velocity of 800 m/s
this bullet enter a block of wood of mass 900 g . Find the
common velocity of shot and block of wood.
Solution

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Q 5)

A shot of 2 kg is discharged by a gun of mass 400kg with a velocity
of 800 m/sec. Find the constant force which would be required to stop
the recoil of the gun in i) 2 meter ii)

Solution

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Q 6 A gun of mass 1 metric tonne, fires a shot of mass 14kg and recoils
smooth inclined plane and rising to a height of 1.6m, find the initial
velocity of projectile
Solution
7) A gun of mass 25 kg resting on an inclined of 40 in 49 fires a shot
horizontally with a velocity of 500 m/s. Find the velocity of the recon
gun and the distance it moves up the inclined plane before coming to
rest
Solution

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8) A shell of mass m is fired horizontally from a gun of mass M with the
velocity just sufficient to carry the shell to a height h if fired vertically
upwards. Show that the velocity of recoil of gun is
Solution

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projectiles

Tuesday, March 1, 2022
1:38 PM

defination
formulas

1) A projectile is fired with a velocity of 320m/s at an angle of 300 to the
i the time to reach the greatest height.
ii. the horizontal range.
With the same velocity, what is the maximum possible range?
Solution

1) cricket ball is thrown to a maximum range of I remain in the air and to
what height will it attain?
solution

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3)

The highest point reached by a projectile is 20m above the horizontaL
If the initial velocity is 20T2 m/s, find the angle of projection. (g =
10m/s2).

4) Find the angle of projection when the range on a horizontal
plane is times the greatest height attained.
Solution

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5) If R is the horizontal range and T is the time of flight of a projectile,
that Inna e: 2R whcrc is the angle of projection

Solution

6) Find the velocity and the direction of projection of a shot which
passzu horizontal direction just over the top of a wall which is 200m
off and 1k high. (g = 9.8 m/s2).

Solution

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7) components of initial velocity of a projectile are U
If R is the horizontal range and Il the greatest height

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8) From a point on the ground, at a distance x from the foot of a
vertical wall, a ball is thrown at an angle of 450 which just clears
the top of the wall and afterwards strikes the ground at a distance y
on the other side. Prove that the
solution

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9) From a point on the ground at a distance of 6m from a vertical wall,
a thrown at an angle of 450 which just clears the wall and strikes the
ground q a distance of 3m on the other side. Find the height of the
wall.
Solution

10)

1f U is the horizontal component of initial velocity, R is the
the greatest height attained, prove that
Let u be the velocity of projection and abe the angle of projection

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11) A partical
solution

12) If
Solution

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13) A stone is thrown
Solution

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