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Quiz2_solutions_web

# Quiz2_solutions_web - Group Problem 2kg 5kg 3kg Two masses...

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Group Problem. 3kg 5kg 2kg Two masses on a horizontal surface are joined by light strings over a light frictionless pulley to a third hanging mass as shown in the diagram. The coefficients of friction between the surface and the masses are 0.3 (static) and 0.25 (kinetic). If the system is released from rest will the masses move? If so determine the acceleration of the masses. What are the tensions in the strings? Draw force and acceleration diagram Assuming the weights are moving and accelerating with a m/s 2 Resolving vertically for M 3 and applying second law M 3 g-T 3 =M 3 a Eq -1 Pulley is frictionless therefore T 2 =T 3 Resolving horizontally for M 2 M 3 M 2 M 1 y x T 3 T 2 T 1 T 1 F 2 F 1 N 1 M 1 g N 2 M 2 g M 3 g a a

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T 2 -T 1 -F 2 =M 2 a Eq-2 The friction force is given by F 2 = µ N Resolving vertically N=M 2 g Therefore F 2 = µ M 2 g Similarly resolving horizontally and vertically for M 1 T 1 -F 1 =M 1 a F 1 = µ M 1 g Eq-3 Substituting for T 1 ,T 2 (=T 3 ), F 1 and F 2 in equation 2 M 3 g-M 3 a-(M 1 a+ µ M 1 g)- µ M 2 g=M 2 a g(M 3 - µ M 2 - µ M 1 )=(M 1 +M 2 +M 3 )a a will be positive (i.e system will move) if M 3 > µ (M 2 +M 1 ) where µ is the coefficient of static friction i.e 3>0.3x7 >2.1 so it moves When it is sliding µ is the coefficient of sliding friction =0.25 a=(3-0.25x7)x9.8/10 =1.2 m/s 2 T 1 =M 1 (a+ µ g)=2(1.225+0.25x9.8) =7.3N T 2 =T 3 =M 3 (g-a)=25N Units, Newtons, OK. Significant figures 2, OK Is it reasonable?
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Quiz2_solutions_web - Group Problem 2kg 5kg 3kg Two masses...

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