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In the figure shown , the coefficient of static friction between B and the wall is (2)/(3) and the coefficient of kinetic friction between B and the wall is (1)/(3). Other contacts are smooth. Find the minimum force F required to lift B, up. Now if the force applied on A is slightly increased than the calculated value of minimum force, then find the acceleration of B. mass of A is 2m and the mass of B is m. Take tantheta=(3)/(4) |
Answer» Solution :  The FBDs of A and B are For A to be in equilibrium: `F=Nsintheta` .(1) For B to just lift off, `Nsintheta=mg+mu_SN^(`)` .(2) For horizontal equilibrium of B, `N^(`)=Nsintheta` .(3) From (2) and (3) `N(costheta-mu_Ssintheta)=mg` or `N((4)/(5)-(2)/(3)xx(3)/(5))mg` or `N=(5)/(2)mg`.(4) From equation (1) `F=Nxx(3)/(5)` `because` `F=(3)/(2)mg` (ii) The acceleration of the BLOCK A be a and B be b `F-Nsintheta=2ma` .(1) `Ncostheta-mg-mu_SN^(`)=MB` .(2) `N^(`)=Nsintheta` .(3) From constraint `=` `asintheta=bcostheta` Solving (1),(2),(3) and (4) we get `impliesb=(3g)/(7)` |
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