Video: Finding the Tension in the String Connecting a Body on a Table to a Vertically Hanging Body through a Pulley

Two bodies of masses 15 and 16.5 kilograms, were attached to the opposite ends of a light inextensible string which passed over a smooth pulley fixed to the edge of a smooth horizontal table. The body of larger mass was placed on the smooth table while the smaller one was hanging vertically below the pulley. Determine the tension in the string, given that the acceleration due to gravity 𝑔 = 9.8 m/sΒ².

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Video Transcript

Two bodies of masses 15 and 16.5 kilograms were attached to the opposite ends of a light inextensible string, which passed over a smooth pulley fixed to the edge of a smooth horizontal table. The body of larger mass was placed on the smooth table, while the smaller one was hanging vertically below the pulley. Determine the tension in the string given that the acceleration due to gravity 𝑔 is equal to 9.8 meters per square second.

In any question of this type, we begin by modeling the situation with a diagram. We are told that the masses of the two bodies are 15 and 16.5 kilograms. Their corresponding downward forces will therefore be equal to 15𝑔 and 16.5𝑔, respectively, where 𝑔 is gravity equal to 9.8 meters per square second. Both the pulley and table are smooth, so we do not need to consider friction in this question. The tension 𝑇 in the string will be constant throughout. As the string is inextensible, when the system is released, both bodies will move with the same acceleration. On release, we assume that the free hanging body accelerates downwards.

We can now use Newton’s second law, force equals mass times acceleration, to set up some equations. If we consider the body on the table, this will move in a horizontal direction. The only force acting in this direction is the tension 𝑇. Therefore, 𝑇 is equal to 16.5 multiplied by π‘Ž. The free hanging body B will accelerate in a downwards vertical direction. The force 15𝑔 is acting in this direction, whereas the tension is acting in the opposite direction. This means that the sum of the forces is equal to 15𝑔 minus 𝑇. This will be equal to 15π‘Ž. As we are taking 𝑔 equal to 9.8 and 15 multiplied by 9.8 is 147, this equation simplifies to 147 minus 𝑇 is equal to 15π‘Ž.

We now have two simultaneous equations that we can solve to calculate the acceleration π‘Ž and the tension in the string 𝑇. Whilst we’re only interested in calculating the tension in this question, it is easier to work out the value of π‘Ž first. This is because we already have 𝑇 as the subject of equation one. Substituting this into equation two gives us 147 minus 16.5π‘Ž is equal to 15π‘Ž. We can add 16.5π‘Ž to both sides so that 147 is equal to 31.5π‘Ž. Finally, dividing by 31.5 gives us π‘Ž is equal to fourteen-thirds or 14 over three. We can now substitute this value back in to equation one. We need to multiply fourteen-thirds by 16.5. This gives us an answer of 77. We can therefore conclude that the tension in the string is 77 newtons.

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