Video: Studying the Motion of Two Freely Hanging Bodies Connected by a String through a Pulley

Two masses of 5π‘š and 2π‘š kilograms are connected to each other by a light inextensible string passing over a smooth pulley. Given that the system was released from rest and that the force exerted on the pulley during the motion was 238 N, determine the value of π‘š. Take 𝑔 = 9.8 m/sΒ².

05:41

Video Transcript

Two masses of five π‘š and two π‘š kilograms are connected to each other by a light in extensible string passing over a smooth pulley. Given that the system was released from rest and that the force exerted on the pulley during the motion was 238 newtons, determine the value of π‘š. Take 𝑔 equal to 9.8 metres per second squared.

Let’s begin by looking at three key words and what they tell us about this problem. As the string is light, we can ignore the weight of the string. As the string is in extensible, the accelerations are equal. This means that both bodies in the system will move with the same acceleration. We also know that the pulley is smooth, so there will be no friction. We can now draw a diagram with all the forces involved in this problem.

The two bodies have masses five π‘š and two π‘š. This means that they will have downward forces of five π‘šπ‘” and two π‘šπ‘”, respectively, where 𝑔 is gravity β€” in this case 9.8 metres per second squared. In the string, there will be vertical tension forces moving upwards. Once the system is released from rest, the five π‘š body will accelerate vertically downwards. The two π‘š body will accelerate vertically upwards with the same acceleration.

We will now be able to use Newton’s second law 𝐹 equals π‘šπ‘Ž or force equals mass times acceleration to set up two simultaneous equations. As the body on the left-hand side is moving downwards, the five π‘šπ‘” force is positive and the tension force 𝑇 is negative. Five π‘šπ‘” minus 𝑇 is equal to five π‘š multiplied by π‘Ž. We will call this equation one. As the body on the right-hand side is moving upwards, the tension force is positive and the two π‘šπ‘” force is negative. 𝑇 minus two π‘šπ‘” is equal to two π‘š multiplied by π‘Ž. We will call this equation two.

We’re also told in the question that the total force exerted on the pulley is 238 newtons. This means that the two tensions must sum to 238. As the tensions are equal, two 𝑇 is equal to 238. We can divide both sides of this equation by two. 238 divided by two is equal to 119. Therefore, the tension on each side of the pulley is 119 newtons. We can then substitute this value into equations one and two. We can calculate the acceleration of the system by adding equations one and two.

On the left-hand side, five π‘šπ‘” minus two π‘šπ‘” is equal to three π‘šπ‘”. Negative 119 plus 119 is equal to zero. On the right-hand side, five π‘šπ‘Ž plus two π‘šπ‘Ž is equal to seven π‘šπ‘Ž. We can cancel an π‘š on both sides of the equation. This gives us three 𝑔 is equal to seven π‘Ž. Dividing both sides by seven gives us a value of π‘Ž of three 𝑔 divided by seven. We’re told to take 𝑔 equal to 9.8. Three multiplied by 9.8 divided by seven is equal to 4.2. Therefore, the acceleration of the system is 4.2 metres per second squared.

We can now substitute this value for π‘Ž back into equation one or equation two to calculate π‘š. In this case, we will substitute π‘Ž equals 4.2 into equation two. Two multiplied by 9.8 is equal to 19.6 and two multiplied by 4.2 is equal to 8.4. Our equation becomes 119 minus 19.6π‘š equals 8.4π‘š. Adding 19.6π‘š to both sides of the equation gives us 119 is equal to 28π‘š. We can then divide both sides by 28. 119 divided by 28 is equal to 4.25.

The value of π‘š is, therefore, 4.25 kilograms. We could use this value to work out both masses. Five multiplied by 4.25 is equal to 21.25 and two multiplied by 4.25 is 8.5. The two masses are 21.25 kilograms and 8.5 kilograms.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.