posted by Chelsea on .
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.
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