# Question Video: Finding the Tension in a String That Connects Two Bodies Hanging Freely and Passing through a Pulley after Projecting One of Them Downwards Mathematics

Two bodies of masses 270 and 𝑚 grams are connected by two ends of a string passing over a smooth pulley. The body of mass 𝑚 was projected downward at 105 cm/s, and 3 seconds later, it returned to its initial position. Find the value of 𝑚 and the tension 𝑇 in the string. Take 𝑔 = 9.8 m/s².

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### Video Transcript

Two bodies of masses 270 and 𝑚 grams are connected by two ends of a string passing over a smooth pulley. The body of mass 𝑚 was projected downward at 105 centimeter per second, and three seconds later, it returned to its initial position. Find the value of 𝑚 and the tension 𝑇 in the string. Take 𝑔 equal to 9.8 meters per second squared.

We begin by sketching a diagram to model the situation. We have two bodies of masses 270 grams and 𝑚 grams that will exert a downward force equal to their weight. This will be equal to the mass multiplied by gravity, and we are trying to calculate the value of 𝑚 together with the tension in the string 𝑇. We are told in the question that the body of mass 𝑚 is projected downward. However, as it returns to its initial position, the force acting on it must be upward.

We can find the acceleration of this body using the equations of motion or SUVAT equations. Taking the upward direction as positive, we know that the initial velocity 𝑢 is equal to negative 105 centimeters per second. As the acceleration of the body is uniform, the body is instantaneously at rest at the midpoint of the three-second time interval. This means that 𝑣 is equal to zero centimeters per second when 𝑡 is equal to 1.5 seconds. Rearranging the equation 𝑣 is equal to 𝑢 plus 𝑎𝑡, we have 𝑎 is equal to 𝑣 minus 𝑢 divided by 𝑡. And substituting in our values, we have 𝑎 is equal to zero minus negative 105 divided by 1.5. This is equal to 70. So, the acceleration of the body and hence the system is equal to 70 centimeters per second squared.

Adding this information to the diagram, we are now in a position to use Newton’s second law to calculate the values of 𝑚 and 𝑇. This states that the sum of the forces is equal to the mass multiplied by the acceleration. Considering the body of 𝑚 grams, the sum of our forces is equal to 𝑇 minus 𝑚𝑔. This is equal to 𝑚𝑎. And since the acceleration is equal to 70 centimeters per second squared, this is equal to 70𝑚. We are told to take 𝑔 equal to 9.8 meters per second squared. And since there are 100 centimeters in a meter, this is equivalent to 980 centimeters per second squared. Our equation becomes 𝑇 minus 980𝑚 is equal to 70𝑚. And adding 980𝑚 to both sides, we have 𝑇 is equal to 1050𝑚.

Let’s now consider the second body of mass 270 grams. Since this body is accelerating downwards, we have 𝑇 minus 270𝑔 is equal to negative 270𝑎. We can then substitute 𝑔 is equal to 980 and 𝑎 is equal to 70, giving us 𝑇 minus 264600 is equal to negative 18900. Adding 264600 to both sides gives us 𝑇 is equal to 245700. This has units of gram centimeters per second squared. We know that the standard unit of force is the newton, and this is equal to one kilogram meter per second squared. Since there are 1000 grams in a kilogram and 100 centimeters in a meter, we can divide our value of 𝑇 by 100000. The tension in the string is therefore equal to 2.457 newtons.

We are now in a position to calculate the mass 𝑚 by substituting our value of 𝑇 in gram centimeters per second squared into the equation 𝑇 is equal to 1050𝑚. Dividing through by 1050, we have 𝑚 is equal to 245700 divided by 1050, which is equal to 234. The mass of the second body is therefore equal to 234 grams. And our two answers are 𝑚 is equal to 234 grams and 𝑇 is equal to 2.457 newtons.