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

To find the maximum weight of block A for which the system will be stationary, we need to consider the forces acting on the system.

First, let's understand the forces acting on block B on the table. There are two forces acting on it: the weight force (mg) directed downwards, and the static friction force (fs) acting horizontally in the opposite direction.

The weight force (mg) can be calculated by multiplying the mass of block B (m) by the acceleration due to gravity (g):

mg = 716 N

Next, we need to calculate the maximum possible static friction force (fs_max) between block B and the table. The formula for static friction is:

fs_max = μs * N

where μs is the coefficient of static friction and N is the normal force acting on the object.

In this case, because the cord between B and the knot is horizontal, the normal force is equal to the weight force (mg) acting vertically:

N = mg = 716 N

Substituting the values into the formula, we can calculate fs_max:

fs_max = 0.31 * 716 = 221.96 N

Now, let's consider block A hanging from the knot. The only force acting on it is the weight force (mg) directed downwards.

To keep the system stationary, the static friction force (fs) acting on B must be equal to or greater than the weight force of A (mgA) hanging from the knot:

fs >= mgA

If the system is at its maximum limit, fs = fs_max. Therefore:

fs_max >= mgA

Substituting the values, we can solve for mgA:

221.96 N >= mgA

To find the maximum weight of block A, divide both sides of the inequality by g (acceleration due to gravity):

221.96 N / g >= mA

On Earth, the approximate value of acceleration due to gravity, g, is 9.8 m/s^2. Substituting this value, we have:

22.63 kg >= mA

Therefore, the maximum weight of block A is approximately 22.63 kg for the system to be stationary.