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Two small balls a and B with mass ma and MB are respectively connected to both ends of the spring, and the B end is fixed on a smooth inclined plane with an inclination of 30 ° by a thin line If the mass of the spring is not taken into account, the accelerations of balls a and B are 0 and [(MA + MB) / MB] · g / 2. B ball, I don't understand. How to analyze the force on the spring?
A's you understand, B's should also understand
At the beginning, a is at the bottom. When it is still together, the spring will deform, and its deformation force is equal to the component force of gravity on the inclined plane. When B is not sheared, it will be affected by the component force of gravity on the inclined plane and the spring force (actually from a), and then the tension of the rope is equal to the sum of these two forces. = MBG * sin30 + MAG * sin30 = (MB + MA) * g / 2
Now when the rope is cut, the pulling force from the rope disappears instantly. But the spring hanging under B has no time to deform, so it still keeps that shape. At that time, since it still keeps that shape at that time, it also shows that it is stretched. It has the natural "impulse force" to recover the original shape, so it has to pull B, plus the component force of B's own gravity on the inclined plane, So its total force is (MB + MA) * g / 2, and then divided by its mass, it is equal to (MB + MA) * g / 2MB
For a, even if you know it, don't say more. If you don't know, ask again
I'd like to say that for a, because the spring still has no time to deform, when cutting, a is still pulled upward by the spring along the inclined plane, which is equal to the gravity component of a downward along the inclined plane, so it is equal to o
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