Lesson Video: Applications of Newton’s Second Law: Two Masses Hanging from a Pulley Mathematics

In this video, we will learn how to solve problems on the motion of a system of two bodies suspended vertically by a string passing over a smooth pulley.

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Video Transcript

In this video, we will learn how to solve problems on the motion of a system of two bodies suspended vertically by a string passing over a smooth pulley. We will solve the problems using Newton’s second law of motion, calculating resultant forces, and using the equations of motion or SUVAT equations.

Newton’s second law states that the net force on an object is equal to its mass multiplied by its acceleration. This is written as 𝐹 equals 𝑚𝑎. In this video, we will measure the mass in kilograms or grams. The acceleration of the objects will be measured in meters per square second or centimeters per square second. In order to convert between these units, we recall that there are 1,000 grams in one kilogram. There are 100 centimeters in one meter. Therefore, 100 centimeters per square second is equal to one meter per square second. The forces that are acting on the object will be measured in newtons or dynes. When the mass is in kilograms and acceleration in meters per square second, the force will be in newtons. Likewise, multiplying a mass in grams by an acceleration in centimeters per square second will give us a force in dynes.

We will now consider how we can use Newton’s second law to solve problems involving a system of bodies and a smooth pulley. In the system modeled in the diagram, we have a light inextensible string passing over a smooth pulley. At each end of the string is attached an object of mass lowercase 𝑚 and capital 𝑀 kilograms, respectively. These will each have forces acting in the downward direction equal to the mass multiplied by gravity. As the string has no mass and the pulley has no friction, there will be equal tensions acting upward from the masses. When the system is released, the heavier body will accelerate downwards and the lighter body will accelerate upwards. As the string is inextensible, these accelerations will have the same magnitude.

We can now use Newton’s second law. The sum of the net forces will be equal to the mass multiplied by the acceleration. The left-hand body is moving upwards. If we assume that this is the positive direction for this body, the sum of the forces is equal to 𝑇 minus 𝑚𝑔. This is equal to the mass multiplied by the acceleration. We get the equation 𝑇 minus 𝑚𝑔 is equal to 𝑚𝑎. We can now repeat this process for the second body. This body is accelerating downwards, which we will assume is the positive direction. The sum of the net forces is equal to capital 𝑀𝑔 minus 𝑇. This is equal to capital 𝑀, the mass of the body, multiplied by the acceleration 𝑎. We now have a pair of simultaneous equations.

In the questions that follow, we will be able to solve these to calculate any unknowns. In our first question, we need to calculate the acceleration.

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 equal to 9.8 meters per square second.

We can begin by sketching the system as shown. The two bodies have masses of 12 kilograms and 18 kilograms, respectively. This means that they will have a downward force of 12𝑔 and 18𝑔, where gravity is 9.8 meters per square second. As we have a light string which passes over a smooth pulley, the vertical tensions will be equal. Once the system is released, the heavier body of mass 18 kilograms will accelerate downwards and the 12-kilogram body will accelerate upwards. As the string is inextensible, the magnitude of these accelerations will be equal.

We can now use Newton’s second law, 𝐹 equals 𝑚𝑎, to create two simultaneous equations. Body A is accelerating vertically upwards. Therefore, the sum of its forces is 𝑇 minus 12𝑔. We take the positive direction to be vertically upwards. This is equal to 12𝑎, the mass multiplied by the acceleration. As body B is accelerating downwards, we take this to be the positive direction. This means that the sum of the forces is equal to 18𝑔 minus 𝑇. This is equal to 18𝑎.

We now have two simultaneous equations, and we can eliminate 𝑇 in order to calculate the acceleration 𝑎 by adding equation one and equation two. Negative 12𝑔 plus 18𝑔 is equal to six 𝑔, and 12𝑎 plus 18𝑎 is 30𝑎. We can then divide both sides of this equation by 30 so that 𝑎 is equal to six 𝑔 divided by 30. Taking 𝑔 equal to 9.8, we can type this into the calculator, giving us an answer of 1.96. The acceleration of the system is equal to 1.96 meters per square second. We could substitute this value back into equation one or equation two to calculate the tension, but this is not required in this question.

In our next question, we need to calculate the mass of one of the bodies.

Two masses 𝑚 and 88 grams are attached to the ends of a light string passing over a smooth pulley. Determine the value of 𝑚, given that when the system was released, the other mass descended 11.76 meters in two seconds. Take the acceleration due to gravity 𝑔 equal to 9.8 meters per square second.

We begin by modeling the system as shown. We are told that the masses are equal to 𝑚 and 88 grams. We recall that there are 1,000 grams in one kilogram, and as the gravity is given in meters per square second, we need the mass to be in kilograms. 88 grams is equal to 0.088 kilograms. This means that the downward forces of the two bodies are 𝑚𝑔 and 0.088𝑔. We have a light string and a smooth pulley, which means that the vertical tensions will be equal. The entire system will move with the same acceleration, and we are told that the 88-gram body descends. This means that the body of mass 𝑚 will accelerate upwards.

We can now use Newton’s second law, 𝐹 equals 𝑚𝑎, to create equations for body A and body B. Body A is accelerating vertically upwards. Therefore, the sum of its forces is equal to 𝑇 minus 𝑚𝑔. This is equal to 𝑚𝑎. Body B is accelerating downwards. Therefore, the sum of its forces is 0.088𝑔 minus 𝑇. We assume the downward direction is positive. This is once again equal to the mass multiplied by the acceleration. At this stage, we have three unknowns: the tension, the mass of body A, and the acceleration of the system. We will need to use the information from the question that tells us that the body of 88 grams descends 11.76 meters in two seconds.

We can use the equations of uniform acceleration to calculate the acceleration. These are often referred to as the SUVAT equations, where 𝑠 is the displacement — in this case equal to 11.76 meters. 𝑢 is the initial velocity, 𝑣 is the final velocity, 𝑎 is the acceleration, and 𝑡 is the time, in this question, two seconds. We know that 𝑠 is equal to 𝑢𝑡 plus a half 𝑎𝑡 squared. Substituting in our values, we have 11.76 is equal to zero multiplied by two plus a half multiplied by 𝑎 multiplied by two squared. The right-hand side simplifies to two 𝑎, so this is equal to 11.76. Dividing both sides of this equation by two gives us 𝑎 is equal to 5.88. The acceleration of the system, once it is released, is 5.88 meters per square second.

Substituting this as well as 𝑔 equals 9.8 into our equation for body A gives us 𝑡 minus 9.8𝑚 is equal to 5.88𝑚. Adding 9.8𝑚 to both sides of this equation gives us 𝑡 is equal to 15.68𝑚. We will call this equation one. Substituting the value for the acceleration into the equation for body B gives us 0.8624 minus 𝑇 is equal to 0.51744. Rearranging this equation gives us a value of 𝑇 equal to 0.34496. We can now substitute the tension, which is equal to 0.34496 newtons, into equation one. This tension force is equal to 15.68𝑚. Dividing both sides by 15.68 gives us 𝑚 is equal to 0.022. This is the mass of the body in kilograms. By multiplying this answer by 1,000, we can convert to grams, which is the same units as the other mass. The mass 𝑚 is equal to 22 grams.

In our final question, we will calculate the force exerted on a pulley.

Two bodies of masses 𝑚 grams and 𝑚 plus 56 grams are connected to each other by a light string which passes over a smooth fixed pulley. The system was released from rest when the two bodies were at the same horizontal level. One second later, the vertical distance between them was 128 centimeters. Find the magnitude of the force exerted on the pulley while the bodies were in motion. Take the acceleration due to gravity 𝑔 equal to 9.8 meters per square second.

Let’s begin by modeling the situation as shown. The two bodies have masses 𝑚 grams and 𝑚 plus 56 grams. This means that they will each exert a downward force of mass multiplied by gravity. We notice that the gravity is given in meters per square second. As the masses are in grams and the distance in centimeters, we need to convert this to centimeters per square second. As there are 100 centimeters in one meter, the acceleration due to gravity is 980 centimeters per square second. We have a light string passing over a smooth pulley. Therefore, the vertical tensions will be equal. When the system is released, the magnitude of the acceleration of both bodies will be equal. The body of mass 𝑚𝑔 will accelerate upwards, and the body of mass 𝑚 plus 56𝑔 will accelerate downwards.

We are told that the bodies start at the same horizontal level, and one second later, the distance between them is 128 centimeters. This means that each body has moved a distance of 64 centimeters, one in the downward direction and one in the upward direction. We can use our SUVAT equations or equations of motion to calculate the acceleration. We know that the initial velocity was zero meters per second as the body started at rest, and we are dealing with a time of one second. We will use the equation 𝑠 is equal to 𝑢𝑡 plus a half 𝑎𝑡 squared. Substituting in our values gives us 64 is equal to zero multiplied by one plus a half multiplied by 𝑎 multiplied by one squared. The right-hand side simplifies to a half 𝑎. We can then multiply both sides of the equation by two, giving us 𝑎 is equal to 128. The acceleration of the system is equal to 128 centimeters per square second.

Our next step will be to use Newton’s second law, force equals mass multiplied by acceleration. As body A is accelerating upwards, the sum of the forces is equal to 𝑇 minus 𝑚𝑔. This is equal to 𝑚𝑎. We can substitute in our values for 𝑔 and 𝑎. Adding 980𝑚 to both sides of this equation gives us 𝑇 is equal to 1,108𝑚. As body B is accelerating downwards, we take this as the positive direction. This gives us the equation 𝑚 plus 56𝑔 minus 𝑇 is equal to 𝑚 plus 56𝑎. Substituting in our values for 𝑔 and 𝑎 and replacing 𝑇 with 1,108𝑚 gives us 980𝑚 plus 54,880 minus 1,108𝑚 is equal to 128𝑚 plus 7,168. Simplifying by collecting like terms gives us 256𝑚 is equal to 47,712.

We can then divide both sides by 256, giving us 𝑚 is equal to 186.375 grams. We can now use this value to calculate the tension 𝑇. 1,108 multiplied by 186.375 is equal to 206,503.5. As our masses were in grams and the acceleration in centimeters per square second, the tension will be measured in dynes. This isn’t the final answer, however, as we want the magnitude of the force exerted on the pulley. This is equal to two 𝑇 as there are two tension forces acting on the pulley. The magnitude of the force exerted on the pulley is therefore equal to 413,007 dynes.

We will now briefly summarize the key points from this video. When solving problems on the motion of a system of two bodies suspended vertically by a string, we use Newton’s second law, 𝐹 equals 𝑚𝑎, where 𝐹 is the net force, 𝑚 is the mass, and 𝑎 is the acceleration. We use simultaneous equations to calculate any unknowns. We also saw it was important to use the correct units, either kilograms, meters per square second, and newtons or grams, centimeters per square second, and dynes. We also used the equations of motion to help us solve more complicated problems. These are sometimes known as the SUVAT equations.

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