Question Video: Finding the Acceleration of Two Hanging Bodies Connected by a String through a Pulley

Two bodies of masses 12 kg and 18 kg are attached to the ends of a light inextensible string which passes over a smooth pulley. Determine the acceleration of the system. Take 𝑔 = 9.8 m/sΒ².

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Video Transcript

Two bodies of masses 12 kilograms and 18 kilograms are attached to the ends of a light inextensible string which passes over a smooth pulley. Determine the acceleration of the system. Take 𝑔 to equal 9.8 metres per second squared.

We’ll call the first mass listed 12 kilograms π‘š sub one. And the second mass 18 kilograms we’ll call π‘š sub two. We want to determine the acceleration of the system of these two masses connected by an effectively massless string over a frictionless pulley. We’ll call that acceleration π‘Ž. And on route to solving it, assume that the acceleration due to gravity is exactly 9.8 metres per second squared.

We can begin our solution by drawing a diagram of this scenario. In our system, we have two masses π‘š one and π‘š two connected by a weightless string strung over a pulley that runs without friction. Considering the system as a whole, we want to solve for its acceleration. To help us solve for the acceleration, we can recall Newton’s second law, which tells us that the net force on a system is equal to the mass of that system multiplied by its acceleration.

As we prepare to apply the second law to our scenario, let’s decide that motion when the pulley turns clockwise is motion in the positive direction. With that definition, we can write that the weight force due to mass two π‘š two times 𝑔 minus the weight force due to mass one π‘š one times 𝑔 is equal to the mass of this system multiplied by its acceleration.

Since the two masses are connected by a string, they move together and the mass of our system is simply their sum. If we rearrange this equation to solve for π‘Ž by dividing by the sum of our two masses, we find that π‘Ž is equal to 𝑔 times π‘š two minus π‘š one over π‘š one plus π‘š two.

We’ve been given the values for π‘š one and π‘š two as well as 𝑔 in the problem statement. So we’re ready to plug in and solve for π‘Ž. When we enter these values on our calculator, we find that π‘Ž is 1.96 metres per second squared. That’s the acceleration of this system of two masses.

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