A block of 5 kg slides on a horizontal surface, connected by massless ropes and

Question

A block of 5 kg slides on a horizontal surface, connected by massless ropes and frictionless pulleys to blocks of 8 and 10 kg as shown. Suppose the block slides without friction. Find the acceleration (positive if to the right, negative if to the left). Find the tension T1 and T2 if there is no friction. Suppose the coefficient of kinetic friction is MR = 0.4. Find the acceleration and T1 and T2 if the system is moving to the right. With Mk = 0.4 again, find the acceleration and T1 and T2 if the system is moving to the left.

Explanation / Answer

so lets call to the right positive then the left block feels T1 - 8 g = 8 a the middle block feels T2 - T1 = 5a and the right block feels 10g - T2 = 10 a add all these equations together and notice Ts cancel 10g - 8g = 8a + 5a + 10a 2g =23 a a = 2/23 g = 2*9.81/23=0.853 b) plug this back into previous equations T1 -8g = 8a T1 = 8*9.81+8*.853=85.3 and 10*9.81-T2 = 10*.853 T2 = 10*9.81 - 10*.853=89.57 c) if moving to the right friction points to the left so middle equation becomes T2 - T1 - 0.4 *5*g= 5a now when you add the equations you get 10g -8g - 0.4*5*g= 23 a 10g -8g - 2g = 23 a a = 0 so T1 =8g = 78.48 T2 = 10g = 98.1 d) if moving to the left friction points to the right so middle equation becomes T2 - T1 + 0.4 *5*g= 5a so when you add equations 10 g - 8g + 2 g = 23 a 4 g = 23 a a= 4/23 g=1.706 so T1 = 8*9.81+8*1.706=92.13 T2 = 10*9.81 - 10*1.706=81.04

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