Posted by **Chelsea** on Tuesday, June 21, 2011 at 7:53pm.

Block B weighs 716 N. The coefficient of static friction between block and table is 0.31. Assume that the cord between B and the knot is horizontal. Find the maximum weight of block A for which the system will be stationary.

The figure has block B on a table and to the right there is a knot in the string from which block A hangs. A 30 degree angle is made from the knot to a wall on the right.

I'm taking physics as a 5 week summer course and we just learned forces and motion today. However, I am really confused and none of the example problems in the notes are like this... I would appreciate it if someone could explain this to me.

- Physics -
**Lawrence **, Wednesday, February 1, 2012 at 2:55pm
Since we know that block B is on the verge of sliding, the maximum weight that can he held is 716N*0.31= 221.96 N. And since the knot is not moving, (no acceleration, that means the force of 221.96N is also the same for the rope from the knot to the wall. And with a little trigonometry, you can solve for force on A, which is 221.96* tan (30) = opposed side which is the force on A, so that means A must equal 128.15N

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