На гладкой горизонтальной
поверхности находится клин с высотой h = 30 см и шириной
основания d = 40 см. На его гладкой наклонной плоскости находится маленькая шайба, соединённая с клином при помощи
невесомой нерастяжимой нити, перекинутой через два блока (см. рисунок). Блоки невесомые и вращаются без трения,
масса клина в n = 8 раз больше массы шайбы. С каким ускорением начнёт двигаться клин после отпускания? Ускорение
свободного падения считайте равным g ≈ 9,8 м/с2
. Движение
клина — поступательное.
To solve the problem, let’s start by analyzing the forces at play and setting up the equations of motion for both the puck (the small mass on the incline) and the wedge (the larger mass on the horizontal surface).
Let’s denote:
m as the mass of the puck,
M = n \cdot m as the mass of the wedge (since the mass of the wedge is n times the mass of the puck),
a as the acceleration of the puck down the incline,
A as the acceleration of the wedge on the horizontal surface,
\theta as the angle of the incline with respect to the horizontal (which can be calculated from the given height h and base d of the wedge),
T as the tension in the string.
The forces on the puck will be the gravitational force component down the incline, mg\sin(\theta) , and the tension T up the incline. The forces on the wedge will primarily be the horizontal component of the tension.
Since the string is inextensible, the acceleration of the puck down the incline will relate to the acceleration of the wedge on the horizontal surface. Specifically, if the puck moves down the incline by a distance s , the wedge will move by s/2 due to the two pulleys system. Therefore, the accelerations are related by a = 2A .
You can set up two equations based on Newton’s second law for the puck and the wedge, and solve them simultaneously to find the unknown accelerations. You should include the tension and the gravitational forces in your equations.
If you provide your equations, I can help you check them and guide you to the correct solution.
