The mass of the upper block in the stack is 0.85 kg. The mass of the lower block in the stack is 1.67 kg. The coefficients of friction between the two blocks are: static 0.75, and kinetic 0.61. The child's mother, who likes to encourage his experiments, has oiled a small strip of the horizontal floor so that it is very slick; the coefficient of kinetic friction between the oiled section of floor and the lower block is only 0.12 and the coefficient of static friction is insignificantly different.
Before the pushing starts, here are some questions about the vertical forces acting on the two blocks.
b) What is the vertical component of the contact force on the upper block by the lower block? HINT: You must apply Newton's First Law.
Since at this point in time the boxes are stationary
The Attempt at a Solution
I know the Fg of upper box is 8.33N, I also know that along with the normal force on the upper block there is also a FB2B1 block 2 being the upper block.
I believe Fg,total= 24.696N, after many other attempts I tried to say Fg,total = NB2, therefore
ƩF = n + FB2B1 - Fg,B2
= 24.696N + FB2B1 - 8.33N = 0
this gives me a negative result for FB2B1 which I know is incorrect.
Here are my free body diagrams, I believe I may be missing something in my FBD which is why I cannot get the answer I am looking for.
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