### Video Transcript

A body of mass 16 kilograms rests on a smooth plane inclined at 65 degrees to the horizontal. It is connected, by a light inextensible string passing over a smooth pulley fixed to the top of the plane, to another body of the same mass hanging freely vertically below the pulley. Given that the acceleration due to gravity is 9.8 metres per second squared, determine the tension in the string.

When dealing with any problem involving pulleys, it is worth drawing a diagram first and marking on all the forces. We’re told that the mass of both of the bodies is 16 kilograms. This means that the force going vertically downwards will be 16𝑔, 16 multiplied by 9.8. There will be two equal tension forces acting towards the pulley. From the free hanging body, it will act vertically upwards. And from the body on the plane, the force will be parallel to the plane.

As the string is inextensible, when the system is released, the acceleration of the whole system will be the same. We will call this 𝑎. We are also told in the question that the angle of the plane is 65 degrees. As both the plane and the pulley are smooth, we do not need to consider friction in this problem. The only extra force we need to add to our diagram is the component of the weight parallel to the plane. Using right-angle trigonometry, we can see that this is equal to 16𝑔 multiplied by sin 65.

Our next step is to use Newton’s second law to create two simultaneous equations. This states that force is equal to mass multiplied by acceleration, or 𝐹 equals 𝑚𝑎. After the system is released from rest, the body on the slope will accelerate up the slope. This means that the tension force is in the positive direction. And the 16𝑔 sin [45] force is in the negative direction. This will be equal to the mass 16 multiplied by the acceleration 𝑎. We will call this equation one.

The free hanging body will accelerate downwards. This means that the force moving in the positive direction is 16𝑔. And the tension force is moving in the negative direction. Once again, this is equal to 16𝑎. We will call this equation two. As we’re trying to calculate the tension, we need to eliminate the acceleration from these two equations. As both equations are equal to 16𝑎, we know that 𝑇 minus 16𝑔 multiplied by sin 45 is equal to 16𝑔 minus 𝑇.

Rearranging this equation by adding 𝑇 and 16𝑔 sin 45 to both sides gives us two 𝑇 is equal to 16𝑔 plus 16𝑔 multiplied by sin 45. We can then divide both sides of this equation by two to calculate the tension. Typing this into our calculator gives us 149.4545 and so on. We can round this to the nearest 100th or two decimal places, giving us an answer of 149.45 newtons. If we did need to calculate the acceleration, we could then substitute this value back in to equation one or equation two.