InterviewSolution
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InterviewSolution
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Define Angle of friction and Angle of repose .Show that angle of friction is equal to angle of repose for a rough inclined plane. A block of mass 4 kg resting on a rough horizontal force of 30 N is applied on .If g=10m//s^(2). Find the total contactforce exerted by the plane on the block . |
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Answer» Solution :Angle of friction : The angle made by the resultant of the Normal reaction and the limiting with Normal reaction is called angle of friction `(phi)` Angle of repose : Let a body of mass m is placed on a rough inclined plane . Let the angle with the horizontal `'theta'` is gradually increased then for a particular angle of INCLINATION (say`alpha` ) the body will just slide down without acceleration . This angle`theta=alpha` is called angle of repose . At this stage the forces acting on the body are in EQUILIBRIUM. Equation for angle of repose : Force acting on the body in vertically downward direction - W - MG . By resolving this force into two compnents . (1)Force acting along the inclined plane in downward direction - mg `sintheta` . This component is resposible for downward motion . (2) The component mg `costheta` , which is balaneced by normal reaction .  If the body slides down without acceleration resultant force the body is zero , then mg`sin theta` = Frictional force `(f_(k))` mg `costheta` = Normal reaction (N.R) But coefficient friction `mu_(k)=(f_(k))/(N.R)=(mgsintheta)/(mgcostheta)=tantheta` `thereforemu_(k)=tanalpha` Hence tangent of angle of angle of repose`(tan theta)` is EQUAL to COEFFIECIENT of friction `(f_(k))`between the bodies . (b ) When the block rests on the horizontal surface , it is in equilibrium under the action of four forces . They are (i)Normal reaction (N) (ii) Weight of the block (mg) (iii) Horizontal force (30N) Limiting frictional force`(f_(L))`  If the applied horizontal force is equal to the limiting frictional force , then only the block will be ready to move on the rough horizontal surface i. e , `f_(L)` = horizontal force applied . `therefore`Total contact force = 30 N . |
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