Often in tasks in mechanics it is necessary to deal with blocks and cargoes, suspended on threads. Cargo pulls thread, under its action thread is affected by tension force. Just the same on the module, but force, opposite in the direction, affects from thread cargo according to the third law of Newton.
It is required to you
- Athwood's car, small weights
Instruction
1. For a start it is necessary to consider the simplest case when the cargo suspended is based upon thread. Cargo in the vertical direction is affected by Ftyazh gravity = down with mg where m is the mass of cargo, and g is acceleration of gravity (on Earth ~ 9.8 m / (с^2). As cargo is not mobile, and except gravity and force of a tension of thread other forces do not affect it, according to the second law of Newton of T = with Ftyazh = mg where T is thread tension force. If cargo at the same time moves evenly, that is without acceleration, then T is also equal to mg according to the first law of Newton.
2. Let now cargo with a mass of m move with acceleration of a down. Then under the second law of Newton Ftyazh-T = mg-T = ma. Thus, T = MGA. These two simplest cases, given above, it is also necessary to use in more difficult backs for determination of force of a tension of thread.
3. In tasks in mechanics the important assumption usually becomes that thread of a nerastyazhim and is weightless. It means that the mass of thread can be neglected, and force of a tension of thread is identical on all length. The simplest case of such task - the analysis of the movement of cargoes by Athwood's car. This car is the fixed block via which thread to which two cargoes are suspended by the mass of m1 and m2 is thrown. If the mass of cargoes are various, then the system comes to progress.
4. The equations for the left and right bodies by Athwood's car will register in a look: - m1*a1 = - m1*g+T1 and m2*a2 = - m2*g+T2. Considering properties of thread, T1 = T2. Having expressed thread T tension force from two equations, you receive: T = (2*m1*m2*g)/(m1+m2).