TEACHING NOTES (XIII) Chapter-5 Newton’s Law Motion (Complete in 3 Lecture) Newton’s first law and inertial frames Newton’s first law of motion, sometimes called the law of inertia, defines a special set of reference frames called inertial frames. This law can be stated as follow : If an object does not interact with other objects, it is possible to identify a reference frame in which the object has zero acceleration. Such a reference frame is called an inertial frame of reference. Any reference frame that moves with constant velocity relative to an inertial frame is itself an inertial frame. For our purposes we can consider the Earth as being such a frame. The Earth is not really an inertial frame because of its orbital motion around the Sun and its rotational motion about its own axis, both of which result in centripetal acceleration. However, these accelerations are small compared with g and can often be neglected. For this reason, we assume that the Earth is an inertial frame, as is any other frame attached to it. Another statement of Newton’s first law In the absence of external forces, when viewed from an inertial reference frame, an object at rest remains at rest and an object in motion continues in motion with a constant velocity (that is, with a constant speed in a straight line). If nothing acts to change the object’s motion, then its velocity does not change. From the first law, we conclude that any isolated object (one that does not interact with its environment) is either at rest or moving with constant velocity when viewed from an inertial frame. The tendency of an object to resist any attempt to change its velocity is called inertia. Newton’s second law Newton’s first law explains what happens to an object when no forces act on it. It either remains at rest or moves in a straight line with constant speed. Newton’s second law answers the question of what happens to an object that has a nonzero resultant force acting on it. When viewed from an inertial reference frame, the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. Thus, we can relate mass, acceleration, and force through the following mathematical statement of Newton’s second law, by choosing proper units so constant of proptionality is 1: F = ma (2) we have indicated that the acceleration is due to the net force F acting on an object. The net force on an object is the vector sum of all forces acting on the object. In solving a problem using Newton’s second law, it is important to determine the correct net force on an object. There may be many forces acting on an object, but there is only one acceleration Note that equation 5.2 is a vector expression and hence is equivalent to three component equations. Fx = max Fy = may Fz = maz
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Asking Question An object experiences a net force and exhibits an acceleration in response. Which of the following statements is/are always true? (a) The object moves in the direction of the force. (b) The acceleration is in the same direction as the velocity. (c) The acceleration is in the same direction as the force. (d) The velocity of the object increases. Force is the causes of changes in motion Force does not cause motion. We can have motion in the absence of force, as described in Newton’s first law. Force is the cause of change in motion as measured by acceleration. ma is not a force Equation 2 does not say that the product ma is a force. All forces on object are added vectorially to generate the net force on the left side of the equation. This net force is then equated to the product of the mass of the object and the acceleration that results from the net force. Do not include an “ma force” in your analysis of the forces on an object. Definition of the newton The SI unit of force is the newton, which is defined as the force that, when acting on an object of mass 1kg, produces an acceleration of 1m/s2. From this definition and Newton’s second law, we see that the newton can be expressed in terms of the following fundamental units of mass, length, and time 1N 1kg.m/s2 Newton’s third law If two objects interact, the force F12 exerted by object 1 on object 2 is equal in magnitude and opposite in direction to the force F21 exerted by object 2 on object 1. F12 = – F21 The third law, which is illustrated in figure , is equivalent to stating that forces always occur in pairs, or that a single isolated force cannot exist. The force that object 1 exerts on object 2 may be called the action force and the force of object 2 on object 1 the reaction force. In reality, either force can be labeled the action or reaction force. The action force is equal in magnitude to the reaction force and opposite in direction. In all cases, the action and reaction forces act on different objects and must be of the same type. For example, the force acting on a freely falling projectile is the gravitational force exerted by the Earth on the projectile Fg = FEp (E = Earth, p = projectile), and the magnitude of this force is mg. The reaction to this force is the gravitational force exerted by the projectile on the Earth FpE = – EEp. The reaction force FpE accelerates the projectile toward the Earth. However, because the Earth has such a large mass, its acceleration due to this reaction force is negligibly small.
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Caution This is such an important and often misunderstood concept that it will be repeated here. Newton’s third law action and reaction forces act on different objects. Two forces acting on the same object, even if they are equal in magnitude and opposite in direction, cannot be an action-reaction pair. Asking Question Which of the following is the reaction force to the gravitational force acting on your body as you sit in your desk chair? (a) The normal force exerted by the chair (b) The force you exert downward on the seat of the chair (c) Neither of these forces. Forces : Now we will learn about the nature and behaviour of major forces that we will encounter in mechanics. Weight : The weight of an object arises because of gravitational pull of earth. Using W for the magnitude of the weight, m for the mass of the object, and ME for the mass of the earth, it follows from Newton’s law of gravitation. W = G
MEm r2
Normal contact force : A contact force perpendicular to the contact surface that prevents two objects from passing through one another is called the normal contact force. Explain weightlessness here. Free body diagram : In this diagram the object of interest is isolated from its surroundings and the interactions between the object and the surroundings are represented in terms of forces. N Does not always equal mg In the situation shown in figure and in many others, we find that n = mg (the normal force has the same magnitude as the gravitational force.) However, this is not generally true. If an object is on an incline, if there are applied forces with vertical components, or if there is a vertical acceleration of the system, then n mg. Always apply Newton’s second law to find the relationship between n and mg.
The Tension force : Force are often applied by means of strings or ropes that are used to pull an object. The magnitude of the force acting between adjacent sections of rope is called tension. Page # [3]
We will assume that a light string connecting one object to another is massless, unless stated otherwise. Tension force in a rope remains same from one end to the other even when the rope passes around smooth objects such as a smooth pulley in fig. Also if a free pulley is massless than tension remains same even if there is friction between pulley and rope Also it is important to remember that ropes can change tension instantaneously. You may ask students to draw FBD for various situations as:-
m2
a)
C
T bc
B
Tab
A
F
m3 T2
F
F=6N
b) m1 T1
c)
A
B (a)
60°
P1
d)
m1 F
(e)
F1
M
Ideal Pulleys : A pulley can change the direction of the force exerted by a cord.
F (a)
(b)
Application Ex. (i) Suppose that blocks A and B have masses of 2 and 6kg, respectively, and are in contact on a smooth horizontal surface. If a horizontal force of 6N pushes them, calculate (a) the acceleration of the system and (b) the force that the 2kg block exerts on the other block. Sol. (a) Considering the blocks to move as unit, M = m0 + mb = 8 kg, F = Ma = 6N, a =0.75 m/s2. F=6N
A
B (a)
(b) If we now consider block B to be our system, the only force acting on it is the force due to block A, Fab. Then since the acceleration is the same as in part (a), we have Fab = Mba = 4.5 N Three blocks, of masses 2.0, 4.0 and 6.0 kg, arranged in the order lower, middle, and upper, respectively, are connected by strings on a frictionless inclined plane of 60°. A force 120 N is applied upward along the incline to the uppermost block, causing an upward movement of the blocks. The connecting cords are light. What is the acceleration of the blocks?
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Sol.
The situation is depicted in fig. with F = 120 N m3 F m1 = 2.0 kg m2 = 4.0 kg T and m3 = 6.0 kg m2 2 Applying Newton’s second law to each block, we have T m1 1 F – T2 – m3g sin 30° = m3a T2 – T1 – m2g sin 30° = m2a =30° T1 – m1g sin 30° = m1a Adding these equations, F – (m1 + m2 + m3)g sin 30° = (m1 + m2 + m3) a a = 5 m/s2
Ex.
A heavy block of mass M hangs in equilibrium at the end of a rope of mass m and length connected to a ceiling. Determine the tension in the rope at a distance x from the ceiling. //////////// m
l
M
Sol. Ex.
m (l – x)g + Mg l The device in fig. is called an Atwood’s machine. In terms of m1 and m2 with m2 > m1. What is the tension in the light cord that connects the two masses? Assume the pulley to be frictionless and massless.
T (x) =
T a m1
T m2 a
m1g m2g
Sol.
Ex.
Isolate the forces on each mass and write Newton’s second law T – m1g = m1a and m2g - T = m2a. Eliminating T gives a = (m2 – m1) g / (m1 + m2). From the above equations, T = 2m1m2g/(m1 + m2) Two bodies of masses m1 and m2 are connected by a light string going over a smooth light pulley at the end of an incline. The mass m1 lies on the incline and m2 hangs vertically. find accln of m1.
m1
m2
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[Sol.(a) Figure shows the situation with the forces on m1 and m2 shown. Take the body of mass m2 as the system. The forces acting on it are :
T T m1g
m2g
(i) m2g vertically downward (by the earth) (ii) T vertically upward (by the string) This gives m2g - T = m2a ........(i) Next, consider the body of mass m1 as the system. The forces acting on this system are (i) m1g vertically downward (by the earth) (ii) T along the string up the incline (by the string) (iii) N normal to the incline (by the incline) As the string and the pulley are all light and smooth, the tension in the string is uniform everywhere. Taking components parallel to the incline, T - m1g sin m1a ........(ii) Taking components along the normal to the incline, N = m1g cos Eliminating T from (i) and (ii), a= (b)
m 2g m1g sin (m1 m 2 )
Find force acting on the pulley due to rod which attaches it to incline. The pulley is in equilibrium and ropes passing over pulley must be applying forces on pulley. Forces applied by ropes and the rod on pulley should cancel. We have to remember that they are vectors. We construct FBD of pulley direction of force due to rod on the pulley is unknown we can not say that it is along the rod only. T Fx
T cos T sin Fy T
nd
Writing Newton’s 2 Law Fy – T sin – T = 0 Fy = T (1 + sin ) Fx – T cos = 0 Fx = T cos F Fx2 Fy2 F T 2 2 sin Ans
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Ex.
Sol.
Apparent weight in an Elevator A passenger weighing 600N rides in an elevator. What is the apparent weight of the passenger in each of the following situations? In each case, the magnitude of the elevator’s acceleration is 0.500 m/s2. (a) The passenger is on the first floor and has pushed the button for the 15th floor ; the elevator is beginning to move upward. (b) The elevator is slowing down as it nears the 15th floor. Lets understand this situation by making F.B.D. of the person inside the elevator. Please note that ma is not being shown in F.B.D. as ma is not the force it is effect of the net force acting on the person. Only two actual forces act on the person in elevator normal contact force by the scale’s surface and weight. By writing Newton’s 2nd law we can find the normal force from the known weight and the acceleration. W = 600 N ; magnitude of the acceleration is a = 0.500 m/s2. To find : W. We expect the apparent weight W = N to be greater then the true weight – the floor must push up with a force greater then W to cause an upward acceleration.
y
Vector sum of forces
N
N
mg
a
mg
Free-body diagram (a)
(b)
F = N + mg = ma F
F is upward so N > mg (c)
N – W = may Since W = mg, we can substitute m = W/g. N = W + ma = W +
W a =W g y
ay 1 g
0.500 m / s 2 = 600 N × 1 2 = 630 N 10 m / s By Newton’s third law same force will be acting on the scale. Hence scale will measure your weight larger than the actual weight.
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(b)
When the elevator approaches the 15th floor, it is slowing down while still moving upward ; its acceleration is downward (ay < 0) as in fig.2. y
Vector sum of forces
N
N
a
mg F = N + mg = ma
F
mg
Free-body diagram (a)
F is downward so N < mg
(b)
(c)
The normal force must be less than the weigtht to have an downward net force. 0.500 m / s 2 1 = 600 N × 2 = 570 N 10 m / s If cable of lift break what will be the reading shown by scale. Explain your result
a N = W 1 g asking ques qualitatively. Ex.
(a)
A man of mass M stands in a basket of mass m as shown in the figure. A rope is attached to the basket and passing over a pulley as shown. The man raises himself and the basket by pulling the rope downward. (a) With what minimum force should the man pull the rope so as to prevent himself from falling down. (b) If the man pulls the rope with a force F greater than the minimum force, then determine the acceleration of the (man + basket) system. (c) Determine the normal reaction between the man and the trolley in part b. Let the whole system moves upward with an acceleration a. Applying Newton's Second law, 2F – (M + m)g = (M + m)a When F = Fmin ; a = 0, thus
(b)
( M m )g 2 When F > Fmin, then acceleration of the system is
Sol.
Fmin =
a=
2F –g mM F
F F
F
a
a Mg N
(A)
(M + m)g (B)
(C)
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(c)
Considering the free body diagram of the man, we have from Newton's Second Law, F + N – Mg = Ma or
2F g F + N – Mg = m M m
or
F (m M) g N = (M – m) mM
Spring Force : Many springs follow hooke’s law for small extension and compression. That is , the extension or compression – the increase or decrease in length from the relaxed length - is proportional to the force applied to the ends of the spring. Hooke’s law for an ideal spring : F = k L In Eq., F is the magnitude of the force exerted on each end of the spring and L is the modulus of change in length of the spring from its relaxed length. The constant k is called the spring constant for a particular spring. The SI units of a spring constant are N/m. When we say an ideal spring, we mean a spring that obeys Hooke’s law and is also massless. Since we have assumed spring to be massles we know forces acting on both ends have to be equal and opposite, to have net force on spring to be zero. _____________________________________________________________________________ CAUTION If we look at F.B.D. of the spring we will note that force on spring must act from both ends. F
F
l0 + x l0 is natural length of spring
lets say Rahul and Sachin are pulling a spring from two ends as shown. Rahul moves x2 and Sachin moves x1. Rahul
x2
l0
Sachin
x1
The force acting on Rahul and Sachin is k (x1 + x2), Not kx2 on Rahul and kx1 on Sachin. Force due to spring is kx where x is defined as |l – l0|, where l is present length and l0 is natural length. ______________________________________________________________________________ Equivalent spring constant : When Springs are connected in Parallel then we can replace them by single spring of spring consant ke where ke = k1 + k2. For more than two spring k = k1 + k2 + k3 + ...... (b) When Springs are connected in sereies then we can replace them by single spring of spring consant ke where 1/ke = 1/k1 + 1/k2. As spring constants are not equal so extensions will not be equal, but total extension y can be written as sum of two extensions y = y1 + y2
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For more than two springs 1 1 1 k k1 k 2 + ..........
Objective : Change in tension in a string / spring force
Q.
The system shown in the figure is in equilibrium. Find the initial acceleration of A, B and C just after the spring-2 is cut.
[Sol.
3mg = KX3 ....(1) 2mg + KX3 = KX2 2mg + 3mg = KX2 5 mg = KX2 ....(2) KX1 = 6 mg ....(3) when spring 2 is cut spring force in other two strings remain unchanged.
KX1 – mg = ma3 a3 = 5g
KX3 + 2mg = 2ma2 a2 =
5g 2
acceleration of 3 m will be zero.
]
Objective : It is important to remember that ropes can change tension instantaneously while spring need to move to change tension, so in this example tension in spring is not changing instantaneously Q.
Find the acceleration of rod A and wedge B in the arrangement shown in fig. if the mass of rod equal that of the wedge and the friction between all contact surfaces is negligible. Take angle of wedge as 45o. [Ans: a =g/2 ] Page # [10]
[Sol. Perpendicular to the plane of contact displacement must be same. ds B 2
=
ds A 2
dsB = dsA Differentiating, aB = aA mg – N/2 = ma N/2 = ma -----------------mg = 2ma a = g/2 ]
Ex.
...(1) ...(2)
There is no friction at any contact. Wedge is free to move Find force acting on wedge due to block. Also find acceleration of wedge. 1
m
m2
Sol. Students may want to directly reach to conclusion that answer is m1g cos . Explain that it is being solved in refrence frame of wedge which may beacclerating. Horizontal component of normal contact force applied by block on wedge will accelerate the wedge. Thus refrence frame attached to wedge is non-inertial refrence frame. Acceration vector of block in ground frame is sum of acceleration of wedge and acceleration of block w.r.t. wedge (ab / w ) a b = ab / w + aw aw
ab
ab/w
Consider F.B.D of wedge. Take horizontal component of normal contact force and apply Newton’s 2nd Law N sin = m2 aw Consider F.B.D. of block and acceleration vector of block, Take horizontal and vertical component of forces and acceleration and apply Newton’s second law. ax = ab/w cos – aw ay = ab/w sin N sin = m1 (ab/w cos – aw) m1 g – N cos = m1 (ab/w sin ) N N
Mg N 1
aw
mg
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Solving we get m1m2 g cos N = (m m sin 2 ) 2 1 m1 g cos sin (m1 m2 ) g sin aw = (m m sin 2 ) , ab/w = (m sin 2 m ) 2 1 1 2
Pseudo force Motion in Accelerated Frames : Till now we have restricted ourselves to apply Newton’s laws of motion, only to describe observations that are made in an inertial frame of reference. In this part, we learn how Newton’s laws can be applied by an observer in a noninertial reference frame. For example,consider a block kept on smooth surface of a compartment of train. If the train acclerates, the block accelerates toward the back of the train.We may conclude based on Newton’s second law F= ma that a force is acting the block to cause it to accelerate, but the Newton’s second law is not applicable from this non- inertial frame. So we can not relate observed accleration with the Force acting on the block. If we still want to use Newton’s second law we need to apply a psuedo force, acting in backward direction, ie opposite to the aceleration of noninertial reference frame. This force explains the motion of block towards the back of car. The fictious force is equal to -ma, where a is the acceleration of the non inertial reference frame. Fictitious force appears to act on an object in the same way as a real force, but real forces are always interactions between two objects. On the other hand there is no second object for a fictitious force. Ex. Sol.
A small ball of mass m hangs by a cord from the ceiling of a compartment of a train that is accelerating to the right as shown in Figure( ). Analyze the situaions for two observers A & B. The observer A on the ground, is inertial Frame. He sees the compartment is accelerating and knows that the deviation of the cord provides the ball, required horizontal force. The noninertial observer on the compartment, can not see the car’s motion so that he is not aware of its acceleration. Because he does not know of this acceleration, he will say that Newton’s second law is not valid as the object has net horizontal force (the horizontal component of tension) but no horizontal acceleration.
a
Noninertial observer Inertial observer
T mg
(a)
Ffictitious
T mg
(b)
For the inertial observer, ball has a net force in the horizontal direction and is in equilibrium in the vertical direction. For the noninertial observer, we apply fictitious force towards left and consider it to be in equilibirium. According to the inertial observer A, the ball experience two forces, T exerted by the cord and the weight. Apply Newton’s second law in in horizontal and vertical direction we get Inertial observer Tsinmg = 0 Tcosma According to the noninertial observer B riding in the car (Fig. b), the ball is always at rest and so its acceleration is zero. The noninertial observer applies a fictitious force in the horizontal direction of magnitude ma towards left. This fictious force balances the horizontal component of and thus Page # [12]
the net force on the ball is zero. Apply Newton’s second law in horizontal and vertical direction we get Noninertial observer Tsinmg = 0 Tcosma = 0 These expressions are equivalent to Equations (1) and (2). The noninertial observer B obtains the same equations as the inertial observer. The physical explanation of the cord’s deflection, however, differs in the two frames of reference. { you can solve eg 7, 10 and 12 now using psuedo force}
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