Newton's Law solved problem
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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