- #1
- 63
- 0
No.Please see picture.
Here is my method. Is it good?
T=m2g
m2g=m1a
a=m2g/m1
-n+Fpush=Ma. (n=normal force of m2 on M).
n=m2a=(m2)^2g/(m1a)
-n+Fpush=Ma.
Fpush=Ma+(m2)^2g/m
Is the answer of the book wrong :O ?
They merely used proportions.Perfect, I will come back to it!
What about this problem? I solved the force values:
I get -(5/8)M^2G/r^2 for the first attraction of the big mass and M^2G/(2r^2) for the attraction of the small mass.
Where did they get 5/16 and 1/4?
Ok perfect!
Coming back to the previous question:
Would the equation for M be:
-T+Fpush-n=Ma?
Why is there a T? :O Is the pulley not frictionless?