A block of mass 3 kg is sitting on another block of 7 kg, and they are connected

Question

A block of mass 3 kg is sitting on another block of 7 kg, and they are connected by a string around a pulley. A force is applied to the bottom block. mu sTop =.6, mu kTop =.4, mu sBottom = .5, mu kBottom = .45 If you are pulling with a force of 60 Newtons, will it move? If not, what is fs between the bottom block and the table and between the blocks? What is the minimum force to make it move? What is the acceleration of the system if you pull with this force once it has moved? What is the force needed to make it move with constant speed once it has moved?

Explanation / Answer

assuming it will not move
max fr force = (m1+m2)g +(m1)g

=0.5(3+7)*9.8 + 0.6*3*9.8 = 66.64 N

more than applied force so will not move

first sliding will occur at bottom surface so

fs =  (m1+m2)g = 0.5*(3+10)*9.8 = 49N

fs` = 60-49 = 11N

min force to move is 66.64 + 0.4*3*9.8 = 78.4N

F - fs - fs` - T = 7a

78.4- 0.45(3+10)*9.8 - 0.4*3*9.8 - T = 22.54 - T = 7a

T - fs` = m1a

T - 0.4*3*9.8 = 3a

22.54 - 11.76 = 10a

a = 1.078 m/s^2

by the above rule we can say that min force is

0.4*(3+7)*9.8 + 2*0.45*3*9.8 = 65.66N

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