Full Question: A particle A of mass 2.5kg is at rest on a smooth inclined plane at an angle of 25 degrees, it is connected to a particle B of mass 1.5kg by a light inextensible string which lies along a line of greatest slope of the plane and passes over a fixed smooth pulley at the top of the plane. B hangs freely and the system is released from rest with the string taut. Find the acceleration of the system and the tension in the string. -------------------------------------------------------------- First, we shall look at the details question has given us to draw out an appropriate model. We are told the plane is smooth, meaning we don't have to be concerned with friction. We are also told A and B are connected by a light inextensible string, the fact it is light means we can ignore the any effect it's weight may have, the fact it is inextensible means provided the string is taut (which it is), we can treat the particles as a single body in calculations. Finally the question states the string lies along a line of greatest slope of the plane and passes over a fixed smooth pulley at the top, this means there is no added friction at the pulley, and all force acting down the slope due to A will act against the weight of B. This property is especially helpful as it will allow us to 'flatten' our model later. I find it can be useful to draw a diagram at this stage to help show all the details of the model, including the slope and any forces acting on the particles A and B. Since there is a smooth pulley, we can flatten out this diagram to contain just the relevant information: 2.5g(sin(25)) T 1.5g <----------- A <-------> B --> As you can see from the forces diagram above, 2 different forces act on both A and B. For A there is the component of weight acting down the slope, and also the tension in the string. For B there is it's weight and the tension of the string. Now we have a simple model, we can figure out what the question has asked for. We'll start with acceleration, since A and B are connected we can treat them as one particle. Tension acts in both directions on the string so is cancelled out, leaving a resultant force as follows: f = 1.5g - 2.5g(sin(25)) = 0.4435g We know f = ma so 0.4435g = 4a a = 0.4435g/4 = 1.06 ms^(-2) To calculate the tension T, we must consider the forces acting on just one particle. So if we consider B we have: T 1.5g <-- B --> Using f=ma we can write: 1.5g - T = 1.5a Fortunately we have already calculated a when considering the whole system, so we can calculate T. T = 1.5g - 1.5a This can be calculated as T = 13.1 N