Newton's Law solved problem

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Newton’s Law solved problem P a g e | 1 © www.vidyadrishti.com An education portal for future IITians by current IITians This problem has been asked by a registered student of Vidya Drishti. (www.vidyadrishti.com) Question: Find the mass M of the hanging block in figure which will prevent the smaller block from slipping over the triangular block. All surfaces are frictionless and the strings and the pulleys are light. Solution: Mass m has zero relative acceleration with respect to triangular block M’ because it is not slipping over triangular block. Therefore triangular block M' and smaller block m moves with same horizontal acceleration a towards right with respect to grand frame. As both the string and pulley are light, tension in every part of the string will be same i.e. T. (Why tension T is shown in both directions here? Go for Vidya Drishti Physics package at http://www.vidyadrishti.com/phypackage.php) Clearly, hanging mass M also moves with acceleration a (downwards). M’ M m M’ M m θ θ a a T T Newton’s Law solved problem P a g e | 2 © www.vidyadrishti.com An education portal for future IITians by current IITians F. B. D. of hanging block M w. r. t. ground: Mass M moves with acceleration (downwards). Forces are: Weight Mg (downward) Tension T (upward) Newton's second law along vertical direction becomes Mg – T = Ma ... (i) F. B. D. of triangular block M’ w. r. t. ground: Block M’ moves with acceleration a (rightwards). Forces are: Weight M’g (downward) Tension T (rightward) Normal N from ground (upward) Normal N’ from smaller block m (perpendicular to slope as shown) We will use Newton's second law along horizontal as well as in vertical direction. Along horizontal: T – N’ sinθ = M’a ... (ii) Along vertical: N – M’g – N’ cos θ = 0 ... (iii) T T M’g a M’ N’ N M Mg a T M’g a N’ N θ θ Newton’s Law solved problem P a g e | 3 © www.vidyadrishti.com An education portal for future IITians by current IITians F. B. D. of smaller block m w. r. t. ground: Block m moves with acceleration a (rightwards) w.r.t. ground. Forces are: Weight mg (downward) Normal N’ from triangular block M’ (perpendicular to slope as shown) We will use Newton's second law along horizontal as well as in vertical direction. Along horizontal: N’ sinθ = ma ... (iv) Along vertical: N’ cos θ – mg = 0 ... (v) From equation (v), we get N’, which is ' cos N mgθ = Putting this value of N’ in equation (iv), we get acceleration a, which is 'sintan a N m a g θ θ =  = Putting values of N’ and a in equation (ii), we get T, which is T – N’ sinθ = M’a ( ) ' 'sin ' tan sin cos ' tan T M a N M g mg T m M g θ θ θ θ θ  = + = +      = + Putting values of a and T in equation (i), we get M, which is Mg – T = Ma mmg a N’ mg a θ θ N’ Newton’s Law solved problem P a g e | 4 © www.vidyadrishti.com An education portal for future IITians by current IITians ( ) ( ) ( ) ' tan ' tan tan cot 1 ta ' t 1 n coT m M g m M g M g a g g g M m M θ θ θ θ θ θ + +  = = = − − −  = + − Vidya Drishti Presents Physics Package for IIT-JEE, AIEEE & Boards (Prepared by present IITians) Physics package is made by current IIT Kharagpur students Our aim is to provide crystal clear concepts and to boost problem solving skills of a student through animations and tricks Physics package is in the form of PowerPoint presentation to illustrate the concepts through animations A student who has read and understood the concepts found himself miles ahead of other competitors Our package develops the concepts so well, no external tutor is required for mastering concepts Package will be sent to a student throughout the year after subscription For more details visit http://www.vidyadrishti.com/phypackage.php Call us at 09333377572 or mail to vidyadrishti@gmail.com or mail to physics@vidyadrishti.com

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Newton's Law solved problem

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TSB .
By: TSB .
741 days 14 hours 50 minutes ago

good one

m.kesavamani
By: m.kesavamani
741 days 12 hours 33 minutes ago


Of negative gravity of the earth and high gravity in space-
A revisit to Newton’s law of Universal gravitation.
M. Kesavamani, M. V. R. Krishna Rao, R. M. C. Prasad,
V. Vidyasagar, C. Ramachandran,
214, Star Shelter Apartments, Saidabad Colony,
Hyderabad- 500 059, A. P., India.
E. mail: manikesava@rediffmail.com
ABSTRACT
Considerable confusion and debate exists in the literature on the subject of gravity right from the geophysical definition of the gravity field to standardization of gravity reduction and corrections as well as geodetic versus geophysical perspectives of the “gravity anomaly”. A common belief is that the gravity field decreases with height just because the distance factor is in the denominator of the equation of the Newton’s law of Universal gravitation. Contrary to this idea, the physicists deal with high gravity fields in space and the geophysicists with negative gravity field for mass distribution on the surface of the earth. The physicists and the geophysicists use the scalar gravity to compute the energy and gravity field for theoretical models respectively. We analyze and present here our perception of the Newton’s law of Universal gravitation and show that gravity increases with height.
The Newton’s law of Universal gravitation states that the mutual force of attraction between two point masses M and m is proportional to the product of masses and “inversely proportional” to the square of the distance Z between them. Then, k as the constant of proportionality, the gravitational force
[F = k m M/ Z2]…. (A)
But, in practice in a more general form, the gravitational attraction (F) exerted by M of the earth on m, is [F =- G m M/(R) 2 R] --- (B)
Here, R is a unit vector in the direction of increase of coordinate of radius of earth R, which directs away from the centre of reference at the mass M. The negative sign in the equation indicates that the force F acts in the opposite direction, towards the attracting mass M of the earth and G, the constant of universal gravitation. Thus, there are two equations, one with minus sign and the other with positive sign, for the Newton’ law of universal gravitation to compute the gravity field.
A cursory glance at the equations reveals that the gravity decreases as the distance increases because the distance factor is in the denominator. However, the negative sign in equation (B) results in a negative gravitational force away from mass M, as if the smaller mass m attracts mass M. It suggests that a similar large mirror image mass as that of M should exist at the same distance in the opposite direction (invisible mass above the surface of the earth?). However, the force F, masses M and m cannot be negative. Therefore, only the distance factor R can be negative. However, what we deal in practice is negative gravity field of the earth, which decreases with height. Now let us examine what this negative gravity field actually means.
Analyzing the Newton’s law of universal gravitation for a general case when M=m, the forces of attraction will be in opposite direction and the effect of the mutual attraction of the two masses is maximum at the mid point of the two masses. The effect of each mass on the other reduces as the distance increases away from the mid point of the masses. The forces vary inversely with the square of the distance from the mid point. When M is larger than m, the force of attraction is always towards the mass M because it attracts mass m. The mass M by virtue of its size exerts its influence over a large distance as compared to the mass m. Hence, the distance at which the maximum force occurs shifts away from the large mass towards the small mass. We can better visualize if we consider another mass Q smaller than that of M and m located between the two the mass then, Q experiences force of attraction of mass M or mass m depending upon its location and size. It is at this point of intersection, the acceleration due to gravity changes its direction towards the attracting masses. The mutual attraction between two bodies results only due the interaction of the excess and deficit masses of the two bodies respectively.
However, here, we consider two unequal masses and discuss about attraction of mass M on m. It is unidirectional in the case of two unequal masses. Therefore, it follows that the direction of force and the distance measurement are positive in the same direction towards the attracting mass M. We can measure the distance between the two masses from any direction, as it is a square term in the denominator. Therefore, it is apparent that the difference between the two equations is only the reckoning of the distance factor. In addition, gravity field is a vector quantity and the direction of force and measurement are in the same direction towards the attracting mass. This is implicit in the Newton’s law of universal gravitation, but requires a more correct mathematical expression for the words “inversely proportional”. Therefore, we emphasize that the gravity increases as the height increases. The equation without minus sign actually indicates the gravity field vector. Differentiating the gravity force with distance vector results in the change in signature in opposite direction.
Therefore, we show that this negative gravity with height is actually a mirror image of the gravity field of the earth obtained from the Newton’s law without the minus sign and is proportional to vertical gradient of gravity field vector as a change in sign as that observed in equation with minus sign. Consequently, the vertical gradient of gravity, which indicates the density of the body, decreases and the gravity increases with height. Therefore in space at the intersection of planets and galaxies high gravity fields occur. In addition, gravity decreases to zero towards the centre of the earth. The magnetic field also requires similar understanding.
INTRODUCTION: There has been considerable confusion and debate on the subjects right from the geophysical definition of the gravity field (Hualin Zeng and Tianfeng Wan, 2004) to standardization of gravity reduction and corrections (LaFehr, 1991; William Hinze et.,al, 2005) as well as geodetic versus geophysical perspectives of the “gravity anomaly ” (Hackney and Featherstone, 2003). Of some concern is that the practical realization of gravity anomaly remains open to question despite being the subject of many investigators.
That the gravity field decreases with height is so much ingrained in the minds of the scholars and students of science, it would be difficult to erase and convince with contrary ideas. In physics the field of a force is often more important than the absolute magnitude of the force. In geophysical applications, we are concerned with accelerations rather than forces. Classical theoretical gravity interpretation and modern Bouguer gravity anomaly interpretation differ and use the Newton’s law of Universal gravitation equation with positive and negative signs respectively. The Physicists deal with high gravity black holes in space and the Geophysicists with negative gravity field for mass distribution on the surface of the earth. In spite of, extensive theoretical work in gravity interpretation, the concept of gravity anomaly, is still an enigma. If gravity anomaly reveals the density, it should be proportional to vertical gradient of gravity. On the other hand, if it represents the gravity field, it cannot indicate the density. We analyze and present here our perception of the Newton’s law of Universal gravitation and show that gravity increases with height.
Our perception of Newton’s law of universal gravitation: The Newton’s law of Universal gravitation states that the mutual force of attraction between two point masses M and m is proportional to the product of masses and inversely proportional to the square of the distance Z between them. Then, k as the constant of proportionality, the gravitational force
[F = k Mm/ Z2]…. (A)
However, in practice in a more general form, the gravitational attraction (F) exerted by M of the earth on m (William Lowrie, 1997), presumably for bodies away from earth, is [F =- Gm M/(R) 2 R] --- (1)
where, R is a unit vector in the direction of increase of coordinate of radius of earth R, which directs away from the centre of reference at the mass M. The negative sign in the equation indicates that the force F acts in the opposite direction, towards the attracting mass M of the earth and G, the constant of universal gravitation. Also in geophysical applications, we are concerned in accelerations rather than forces.
Converting the force into acceleration, the force (William Lowrie, 1997) Is
F= mass (m) X acceleration (a)
So, [(F=m*a) or (F= m*g)]…. (2)
Here g is the acceleration due to gravity. The negative sign in equation (1) and distance measurement as that from M results in a negative gravitational force away from mass M, as if the smaller mass m attracts mass M. It suggests that a similar large mirror image mass as that of M should exist at the same distance in the opposite direction (invisible mass above the surface of the earth?). . However, what we deal in practice is negative gravity field of the earth, which decreases with height. Now let us examine what this negative gravity field actually means by converting the force into acceleration. Now, let us consider a general case. When M=m, the forces of attraction will be in opposite direction and the effect of the mutual attraction of the two masses is maximum at the mid point of the two masses. The effect of each mass on the other reduces as the distance increases away from the mid point of the masses. The forces vary inversely with the square of the distance from the mid point. When M is larger than m, the force of attraction is always towards the mass M because it attracts mass m. The mass M by virtue of its size exerts its influence over a large distance as compared to the mass m. Hence, the distance at which the maximum force occurs shifts away from the large mass towards the small mass. Another mass Q smaller than M and m, placed between the two the mass, Q experiences force of attraction of mass M or mass m depending upon its location and size.
However, here, we consider two unequal masses and discuss about attraction of mass M on m. Therefore, it is unidirectional in the cases of two unequal masses. Therefore, it follows that the direction of force and the distance measurement are positive in the same direction towards the attracting mass M. We can measure the distance between the two masses from any direction, as it is a square term in the denominator. However, it is the object of mass m, which is affected due to the attraction of mass M a vector and hence important for the direction of measurement of distance and force. Therefore, both are in the same direction towards mass M. Here g is the acceleration due to gravity of the earth M. If we replace m by unit mass, from equations (1) and (2) we have
[g=- G M/R2 ]… (3) which indicates that the gravity field decreases with height.
The Newton’s second law of motion states that the rate of change of momentum is proportional to the force acting upon it and takes place in the direction of the force. It is here the problem lies, which somehow escaped the attention of the doyens of physicists. Following the notation and the negative sign used in equation (1) The acceleration caused by attraction of mass M on mass m should be [(-F) =ma].
So, [(-F) =mg]. ----- (4)
From equations (1) and (4) we have,
[g= G M/(R)2]--- (5) which results in the gravitational acceleration of the earth and the distance are in the same direction towards mass M, as that obtained in equation (A). As force of attraction, acting on the unit mass is always towards the attracting mass and R is a square term in the denominator the gravity field remains the same irrespective of the measurement of distance as height or depth. Comparing equations (3) and (5), we observe only a change in sign that is a mirror image. Now, let us analyze, what actually the negative gravity field indicates.
Then what does the negative gravity field indicate?
Now differentiating equation (5) with respect to R, we get
[dg/dR=dg/dz= (-2GM)/(R)3]......... (6) that gives the vertical gradient of gravity in the direction of force of attraction of mass M, if z represents depth.
Now differentiating equation (3), we get
[dg/dR= 2GM/(R)3]... (7) that gives the vertical gradient of gravity in the opposite direction of attraction or in the direction of distance measurement. As the distance factor is reckoned from the centre of earth in equation (3) R should be replaced (–R) in equation (7), which gives the same equation as that of (6) that gives the vertical gradient of gravity in the direction of the force of attraction. The change in the sign is evident when the vertical gradient of gravity is calculated. Therefore, equations (3) and (5) are mirror images and indicate the vertical gradient of gravity and the gravity field respectively. The vertical gradient of gravity is proportional to twice the product of masses and inversely proportional to the cube of the distance from the attracting mass. The Vertical gradient of gravity, which is a mirror reflection of ‘g’, integrated with respect to height gives the acceleration due to gravity.
Therefore, the negative sign in equation (1) symbolizes a figurative expression that the forces are in opposite direction. Consequently, the vertical gradient of gravity decreases and the gravity field increases with height.
Potential energy and work: Assuming that F is constant through the short distance of the fall of an object, the work expended is (-F) h. This is the potential energy of an object in space before its fall. If the constant force F moves through a small distance dr in the same direction as the force the work done is [dW= F dr]
Therefore, the change in the potential energy (dE) is
[dE=dW= F dr]…. (8)
The gravitational potential is the potential energy of a unit mass in a field of gravitational attraction. If the potential is U, the potential energy of a mass m in a gravitational field is equal to mU. Thus, a change in potential energy [dE=mU].
Equation (8) becomes m dU= F.dr and by use of equation (4)
[m dU=m g dr].
Thus [g= (dU/dr) R]. … (9)
Equating equations (8) and (9) [dU/dr = - G M/ R2], the solution of which is [U= G M/R]… (10)
CONCLUSION: The negative sign introduced in the Newton’s law of Universal gravitation, to indicate the opposite forces, should be related to the distance vector. The negative gravity with height is actually a mirror image of the gravity field of the earth obtained from the Newton’s law without the minus sign and is proportional to vertical gradient of gravity field vector as a change in sign as that observed in equation with minus sign. Consequently, the vertical gradient of gravity, which indicates the density of the body, decreases and the gravity increases with height. Therefore in space at the intersection of planets and galaxies high gravity fields occur. In addition, gravity decreases to zero towards the centre of the earth. The magnetic field also requires similar understanding.
The absolute gravity measurements by Pendulums or by free fall method in one form or other indicate the distance related to height. Therefore, the measured value based on distance reckoned as height is proportional to the vertical gradient of gravity. This is the reason the hitherto calculated absolute value of the gravity is large at the poles of the earth as compared to that at equator. In addition, it leads to the conclusion that the object of attraction has zero potential energy in space. Besides, the gravity anomalies become negative and are proportional to vertical gradient of gravity (Kesavamani, 2001, 2002, Kesavamani et.al. 2005, 2006, 2007) These negative gravity anomalies indicate the density of the body or the negative density contrast there by, indicating the Himalayan heights as negative and ocean depths as positive gravity anomalies.
Therefore, maximum gravity occurs at the mutual attraction of masses in space and decreases towards the centre of masses. We can therefore have two data sets indicating the absolute gravity as well as its vertical gradient of gravity at different station elevations by change in sign in the Newton’s law of Universal gravitation. The magnetic field also needs similar attention.
REFERENCES:
Hualin Zeng and Tianfeng Wan, 2004. Clarification of the geophysical definition of a gravity field. Geophysics, Vol. no 5(Sept-Oct 2004), pp 1252-1254.
Kesavamani, M. 2001 Bouguer anomalies over the continents and oceans-Notes Jour. Geol. Soc. Ind., 58 466-467(2001)
Kesavamani, M.2002 Bouguer anomalies over the continents and oceans Correspondence, Jour. Geol. Soc. Ind. 60, (2002)
Kesavamani, M., Ramachandran, C., Krishna Rao, M.V.R., Prasad, R.M.C., Venkateswarlu, M. 2005 The Concept of gravity and magnetic anomalies. Notes. Jour. Geol. Soc. Ind. 66 511(2005)
Kesavamani, M., Ramachandran, C., Prasad, R.M.C., Krishna Rao, M.V.R., Reduction of gravity and magnetic data.2005 Notes. Jour. Geol. Soc. Ind. 66, 644. (2005).
Kesavamani, M.,Ramachandran,C,.,Krishna Rao, M. V. R., Prasad, R. M.C.,
Ram Mohan. P. K. The fallacy of Bouguer anomaly in Geophysical exploration and the new concept of the theory of gravity anomaly. Abs.46.Third International Seminar and Exhibition on Exploration Geophysics, AEG, (2006), Hyderabad, India.
M.Kesavamani, V.Vidyasagar, M.V.R.Krishna Rao, R.M.C Prasad, P.K.Rammohan. 2007. Does the gravity increase with the height?. June 2007. GSI SR News Vol.24(1

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