- #1
mrjoe2
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Homework Statement
the 1.0kg block is tied to the wall with a rope. it sits on top of the 2.9kg block. the lower block si pulled to the right with a tension force of 20N. the coefficient of kinetic friction at both the lower and upper surfaces of the 2kg block is mk=.40. what is the tension in the rope holding the 1kg block to the wall?
Homework Equations
Newtons third law, perhaps the second law. ffk=fn *mewk fnet=ma fg=mg
The Attempt at a Solution
when i tried thisi said that the backward friction in the 2kg block was equal to the forward fricttion in the 1kg block (action reaction pair. this also makes sense i think because the 1kg is tending backwards "slipping" as the 2.0kg object accelerates forward). i noticed that the fg of the 2kg box was actually 3*9.8 because it has the 1kg box on top of it. this would also be the Fn of the 2kg box. so ff = 11.76 subbing in fN and mewk. so there is a forward force on box 1kg. then we also have to take into account another action reaction pair. that is, the forward force 20N on box 2kg has a reaction pair on the 1kg block of 20N back. then we have yet another force acting on the 1kg, and that is its own Ffk which is Fk=9.8*1 *.4 = 3.92 Newtons forward. so the tension in the rope would be T = 3.92 + 11.76 - 20 = 4.32 N (back). There are no solutions for these questions and i just want to see if i did it right. i have a feeling its perfection but like in the title i infered that i wasnt sure about the action reaction forces i assumed (through Newtons second law)