Two equal masses(1 and 2) are connected by an ideal rope across a pulley. This system is on a larger block A; 1 is on top of A, while 2 is hanging off the side, and the pulley rests on the corner of A. The question is what is the minimal acceleration of A such that masses 1 and 2 remain stationary. No variables given, please HELP!
To find the minimal acceleration of block A such that masses 1 and 2 remain stationary, we need to consider the forces acting on the system.
Let's assume that block A has a mass of M, and masses 1 and 2 have masses of m. We need to find the minimal acceleration, so we can start by assuming that block A is moving upwards with an acceleration of a.
Considering the forces acting on mass 1:
1. The weight of mass 1 acts downward with a force of m * g, where g is the acceleration due to gravity.
2. The tension in the rope acts upward with a force of T (assuming that the rope does not stretch or have any significant mass).
Considering the forces acting on mass 2:
1. The weight of mass 2 acts downward with a force of m * g.
2. The normal force (contact force) between mass 2 and block A acts towards the left with a force of m * g (to balance the weight of mass 2).
3. The tension in the rope acts towards the right with a force of T.
Considering the forces acting on block A:
1. The weight of block A acts downward with a force of M * g.
2. The normal force (contact force) between block A and the ground acts upward with a force of M * g.
3. The tension in the rope acts downward with a force of T (to balance the tension forces acting on masses 1 and 2).
Since the system is in equilibrium (masses 1 and 2 remain stationary), the sums of forces in both the x and y directions must be zero.
In the x direction:
The tension force T in the rope is the only force with an x-component. Therefore, the sum of the x-components of the forces must be zero:
T = 0
In the y direction:
For mass 1: T - m * g = 0
For mass 2: m * g - m * g = 0 (normal forces cancel each other out)
Now, let's consider the forces acting on block A:
The sum of the y-components of the forces must be zero:
2 * m * g - M * g = 0
Now we can solve for the minimal acceleration of block A:
2 * m * g = M * g
2 * m = M
a = g
Therefore, the minimal acceleration of block A required to keep masses 1 and 2 stationary is equal to the acceleration due to gravity, g.