Video Transcript
Two bodies of mass 590 grams and 𝑚 grams are attached to the ends of a light inextensible string passing over a smooth pulley fixed to the edge of a smooth horizontal table. The first body rests on the table, and the other hangs freely vertically below the pulley. If the tension in the string is 90,860 dynes, determine the acceleration of the system.
Let’s first notice that we’ve been given mass of the bodies in grams and tension in the string in dynes. Now, a dyne is a unit of force that acts on a massive one gram and increases its velocity by one centimeter per second every second along the direction it acts. And so, for consistency, we’re going to convert 𝑔 into centimeters per square second. Well, there are 100 centimeters in a meter, so we multiply by 100. We find 𝑔 is 980 centimeters per square second. The next thing that we need to do when we’re dealing with pulley problems is to draw a sketch.
We have two bodies; let’s call them A and B. A sits on a smooth horizontal table. And B hangs vertically below the pulley from the piece of string. We really need to consider all the forces that act on each of our bodies. Well, we have the force acting downwards. It’s mass times acceleration. And the acceleration here is due to gravity. So it’s 590 times 𝑔. There is a reaction force acting in the opposite direction. And then on body A, we also have the tension in the string. The same amount of tension acts on particle B pulling it upwards. And since the string is light and inextensible, we know we don’t need to take into account its mass.
The force acting downwards on body B is mass times acceleration due to gravity. So this time, that’s 𝑚 times 𝑔. Since the table is smooth, there are no other forces. If the table wasn’t smooth, we would need to consider friction. Finally, we know that when we release the system from rest, it will move as shown. Let’s call the acceleration with what we’re trying to find, 𝑎 centimeters per square second. At this stage, we might want to consider the forces on both particles. But actually, we have enough information just to use particle 𝑎. We’re going to recall Newton’s second law of motion. That is, the sum of the forces 𝐹 on an object is equal to the mass of that object multiplied by the acceleration 𝑎 of that object.
The net force acting on particle A, in the horizontal direction at least, is 𝑇. Its mass is 590, and we’re looking for its acceleration; that’s 𝑎. Now, of course, we were told that the tension in the string is 90,860 dynes. So we replace 𝑇 with this number and we find that 90,860 equals 590𝑎. We solve for 𝑎 by dividing through by 590, such that 𝑎 is 90,860 divided by 590 which is equal to 154. And, of course, we’re working in centimeters per square second. So we’re finished. The acceleration in the system is 154 centimeters per square second.
