Question Video: Finding the Force Exerted on a Pulley by Two Attached Bodies Hanging Freely During Their Motion Mathematics

Video Transcript

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.