Question

Через неподвижный блок переброшена легкая нить, на одном конце которой висит груз массой 3m, а на другом − подвижный блок (рис.). Через подвижный блок переброшены на нити гири массами 2m и m = 1 кг. а) Каково будет натяжение нитей и сила давления на оси блоков при движении грузов? б) При каком значении массы среднего груза он будет находиться в равновесии? Массой блоков пренебречь.

Answer 1

Problem Analysis

We have a system consisting of two blocks connected by a light thread. One block is fixed and the other is movable. A weight of mass 3m is hanging from one end of the thread, and on the other end, there are two weights of masses 2m and 1 kg. We need to determine the tension in the thread and the force exerted on the axis of the blocks when the weights are in motion. Additionally, we need to find the mass of the middle weight for the system to be in equilibrium. We can solve this problem using Newton's laws of motion and the principles of equilibrium.

Solution

a) To find the tension in the thread and the force exerted on the axis of the blocks when the weights are in motion, we can consider the forces acting on each block separately.

Let's denote the tension in the thread as T, the force exerted on the axis as F, and the acceleration of the system as a.

For the movable block: - The weight of mass 3m exerts a downward force of 3mg. - The tension in the thread exerts an upward force of T. - The force exerted on the axis exerts a downward force of F.

Using Newton's second law, we can write the equation of motion for the movable block as: 3mg - T - F = 3ma For the fixed block: - The tension in the thread exerts a downward force of T. - The force exerted on the axis exerts an upward force of F.

Using Newton's second law, we can write the equation of motion for the fixed block as: T - F = 0 [[2]]

To solve these equations, we need one more equation. We can use the fact that the acceleration of the system is the same for both blocks: a = (2m + 1 kg) / (3m) [[3]]

Now we can solve the system of equations [[2]], and [[3]] to find the values of T and F.

b) To find the mass of the middle weight for the system to be in equilibrium, we need to consider the condition for static equilibrium. In static equilibrium, the net force and net torque acting on an object are zero.

Since the system is in equilibrium, the acceleration of the system is zero (a = 0). Using equation [[3]], we can write: (2m + 1 kg) / (3m) = 0

Solving this equation will give us the value of m for which the system is in equilibrium.

Conclusion

a) To find the tension in the thread and the force exerted on the axis of the blocks when the weights are in motion, we need to solve the system of equations [[2]], and [[3]]. b) To find the mass of the middle weight for the system to be in equilibrium, we need to solve the equation (2m + 1 kg) / (3m) = 0.

Please note that the specific values of m, g, and other parameters were not provided in the question, so we cannot provide numerical answers.