APT ACADEMIC SOLUTIONS : APT ACADEMIC SOLUTIONS
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Slide 3 : www.aptacads.com Particle Dynamics - 1 Session
Session Objective : www.aptacads.com Session Objective 1. Free body diagram
Working With Newton’s Laws : Working With Newton’s Laws Strictly speaking Newton’s laws refer to a particle but in practical situations, we deal with extended bodies.
The laws as stated may be applied to the extended bodies provided all parts of the body - move with same velocity and acceleration.
Such a motion is called translatory motion.
Approach to Problem Solving : www.aptacads.com Choose the body:
A body to which Newton’s laws are applied may be a
particle,
a block,
a combination of blocks connected by a string etc. Approach to Problem Solving The only restriction is that all parts of the body have the same acceleration. Step 1
Step II : www.aptacads.com Step II Choose the frame of reference
Choose the frame in which the acceleration of the body is perceived
Consider a box standing in elevator moving up with an acceleration A upwards.
If the frame of reference is ground, the acceleration of box is A upwards.
If the frame of reference is elevator, the acceleration of box is zero.
Step III : www.aptacads.com Identify the forces:
Only forces that act on the body are to be considered.
There may be many forces in a given problem but one must pick up only those that act on the body.
Only external forces exerted on the body should be considered.
We do not consider internal forces that one part of the body exerts on the other part. Step III
Step IV : www.aptacads.com Make Free body diagram :
It is stripped down diagram in which only one body is considered at a time.
We remove all bodies in contact or influencing the chosen body and replace the removed bodies with the respective forces they exert on the chosen body. Step IV
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Question : www.aptacads.com Make free body diagrams of ball A and ball B connected by three strings as shown below Question
Solution : www.aptacads.com Solution A B A B F.B.D. Of ball A F.B.D. Of ball B T1 T3 T2 m1g T1 m2g
Question : www.aptacads.com Make free body diagrams of block B and wedge W as shown in the figure . All the surfaces are smooth. Question
Solution : www.aptacads.com Solution B W B W T N1 mg N1 N2 N3 Mg F.B.D. of box B F.B.D. of wedge
Question : www.aptacads.com Make free body of system of two blocks and a pulley as shown in the figure below..Masses of two blocks and pulley are ma, mb and mp respectively. Question
Solution : www.aptacads.com A B P A B P A B P T1 T1 T2 T3 T2 mag mbg mpg F.B.D of block A F.B.D of block B F.B.D of pulley P Solution
Question : www.aptacads.com A box of mass m is being dragged towards right by applying a force F along a rough horizontal force F. Make free body diagram of the box. Question
Solution : www.aptacads.com mg f N F Solution
Step- V : www.aptacads.com Choose Axis:
It is convenient to choose X-axis along the direction in which body is likely to have acceleration.
A direction perpendicular to this may be chosen as the Y-axis. Note: In rectilinear motion direction of acceleration is either in the direction of motion or opposite to it Step- V
Step VI : www.aptacads.com Write down separate equation along X-axis and separate equation along Y-axis.
By Newton’s second law of motion Step VI If X-axis is along direction of motion as stated in step V ,then
Example - 1 : www.aptacads.com Example - 1 A block of mass m = 10kg is placed on a table.
(a) Identify the action-reaction pair.
Find the normal reaction on the block ?
(c) If the table is given a vertical acceleration of 4.2m/s2, find the value of normal reaction ?
Solution : www.aptacads.com Solution (a) There are two action-reaction pairs
Example - 2 : www.aptacads.com Example - 2
Solution : www.aptacads.com Solution Applying Newton’s second law to the two blocks separately: Block m1: T – m1g = m1a ….(1) Block m2: m2g – T = m2a …..(2) Adding (1) and (2), m2g – m1g = (m1 + m2) a
Example - 3 : www.aptacads.com Example - 3 Two blocks of masses m1 and m2 are placed in contact with each other on a smooth horizontal surface as shown in the figure. A constant horizontal force F of 30 N is applied to block m1.
(a) What is the acceleration of the blocks?
(b) What is the magnitude of normal reaction between the two blocks?
Solution : www.aptacads.com Solution Block m1:F – N = m1a ... (1) Block m2:N = m2a ... (2) Adding equations (1) and (2),
F = (m1 + m2) a
Example - 4 : www.aptacads.com Example - 4 A block of mass m is placed on a smooth inclined surface which is inclined at an angle q to the horizontal (see figure). Find the acceleration of the block after it is released and the magnitude of normal reaction N exerted on block by the surface.
Solution : www.aptacads.com Solution From Newton’s second law: Solving eqn (1) and eqn (2) a = g sin q N = mg cos q Note :
Acceleration a of the block is independent of the mass of the block.
For q=90o; a = g and N = 0. This condition corresponds to free fall. Choose X-axis along the direction of acceleration and Y-axis perpendicular to it
Exercise - 5 : www.aptacads.com Exercise - 5 Figure shows a block of mass m = 15 kg hanging from three strings. What is the tension in the strings?
Solution : www.aptacads.com Solution Equation from F.B.D of m
Tc – mg = 0
\ Tc = 15 × 9.8 = 147.0 N Equations from F.B.D of knot
– (TA cos 60°) + TB cos 45° = 0 ...(i)
TA sin 60° + TB sin 45° –TC = 0 ... (ii) Solving equations (i) and (ii) TB = 107.6 N and TA = 152.19 N
Exercise - 6 : www.aptacads.com Exercise - 6 Figure shows a block of mass m1 on a smooth horizontal surface pulled by a massless string which is attached to a block of mass m2 hanging over a pulley as shown. Neglecting the mass of the pulley, find
(a) the acceleration of the system and
(b) tension in the string.
(c ) force exerted by the hinge on the pulley?
Solution : www.aptacads.com Solution Equation from F.B.D of mass m1 : T = m1a ... (i) Equation from F.B.D of mass m2 : m2g – T = m2a ... (ii) From (i) and (ii),
Exercise - 7 : www.aptacads.com Exercise - 7 Figure shows a man of mass m= 80 kg standing on a light weighing machine which is kept in a box of mass M = 40 kg. The box is hanging from a pulley fixed to the ceiling through a light cord, the other end of which is held by the man himself.
If the man manages to keep the box at rest, what is the weight shown by the machine?
(b) What force should he apply on the cord to get his correct weight on the machine?
Solution : www.aptacads.com Solution The weight shown by the machine will be equal to force N with which man presses on the machine. (a) N + T – mg = 0 (From F.B.D of man) ... (i)
T – N – Mg = 0 (From F.B.D of Box) ... (ii) Weight shown by machine
Solution : www.aptacads.com Solution (b) In this case, we want N = mg. T + N – mg = ma ... (iii) T – N — Mg = Ma ... (iv) Putting N = mg and solving (iii) and (iv) yields T = 2352 N
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Example : www.aptacads.com Example A river is flowing at a speed of 2 m/s. There is a boat in the river which can move in still water at 10 m/s.
Find the time taken by the boat to travel 100 m downstream when its engine is off?
Find the time taken by the boat to travel 100 m upstream when its engine is on?
If the engine is off, find the speed of the boat as observed by a stationary observer on the river bank?
When the engine is on and the boat is moving in a direction opposite to that of the flow of the river, find its velocity as observed by a stationary observer on the banks?
Solution : www.aptacads.com Solution When the engine is off, the boat flows with the river. Hence,
Velocity as observed from ground = 2 m/s
Time taken to travel 100m downstream = 100/2 = 50 s.
With the engine on, when it travels upstream,
speed = 10 – 2 = 8 m/s.
Time taken to travel 100 m upstream = 100/8 = 12.5 s.
Example : www.aptacads.com Example A boat can travel at a speed of 10 m/s in still water. Find the orientation of the boat such that it reaches point B, directly opposite to B across the river.
Solution : www.aptacads.com Solution There are three entities –
B – Boat; R – River; G - Ground
Example : www.aptacads.com Example A man who can swim at a speed of vm relative to “still” water decides to cross a river with water flowing at a speed vr (vr < vm) such that he “orients” his swimming at an angle ‘q’ to the perpendicular across the river as shown in the figure. Find the time taken by him to across the river as a function of ‘q’, given that the width of the river is ‘d’ and calculate the minimum possible time in which he can “across” the river.
Solution : www.aptacads.com Solution Three entities: man, river, ground. Seen from the ground, the man swims at an angle f w.r.t. the y-axis. If the man takes time t to cross the river, then t is minimum when cos q is maximum (= 1). Hence,
Example : www.aptacads.com Example
Solution : www.aptacads.com Solution There are three entities –
M – Man; R – Rain; G - Ground Since the man finds the raindrops striking vertically down,
Example (Important) : www.aptacads.com Example (Important)
Solution : www.aptacads.com Solution Position of ball w.r.t. trolley:
Solution : www.aptacads.com Since the ball eventually hits the trolley, hence Since the initial position vector of A is equidistant from the x and y axes, Solution
Solution : www.aptacads.com Solution
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