A body of mass five kilograms rests on a smooth plane inclined at an angle of 35 degrees to the horizontal. It is connected by a light inextensible string, passing over a smooth pulley fixed at the top of the plane to another body of mass 19 kilograms hanging freely vertically below the pulley. Given that the acceleration due to gravity 𝑔 equals 9.8 metres per second squared, determine the acceleration of the system.
In order to answer any question like this involving pulleys, we firstly draw a diagram with all the forces labelled. There’s a five-kilogram mass lying on the slope. This means that there will be a force of five 𝑔 acting vertically downwards. The body hanging freely has a mass of 19 kilograms. Therefore, this will have a force or weight acting vertically downwards of 19𝑔.
A tension force will act in the string towards the pulley. From the body hanging freely, it will act vertically upwards. And from the body on the slope, it will act parallel to the slope. As the string is inextensible, when the system is released, the acceleration will be the same for the whole system. We will call this 𝑎. As both the plane and the pulley are smooth, there will be no friction force. This means that the only other force we need to calculate is the component of the weight parallel to the plane. As the plane is inclined at an angle 35 degrees, this component will be five 𝑔 multiplied by sin 35.
Our next step is to use Newton’s second law to create two simultaneous equations using the formula 𝐹 equals 𝑚𝑎. When released, the freely hanging body will accelerate vertically downwards. This means that the positive force is 19𝑔. The tension force 𝑇 is negative. The mass of the object is 19 kilograms. And the acceleration is 𝑎. Therefore, 19𝑔 minus 𝑇 is equal to 19𝑎. We will call this equation one. When the system is released, the body on the plane will accelerate up the plane. This means that the tension force is positive. The five 𝑔 multiplied by sin 35 degrees force is negative. 𝑇 minus five 𝑔 multiplied by sin 35 is equal to five 𝑎. We will call this equation two.
As we’re trying to calculate the acceleration, we can eliminate 𝑇 from these equations by adding equation one to equation two. On the left-hand side, we end up with 19𝑔 minus five 𝑔 multiplied by sin 35. On the right-hand side, we have 24𝑎. We can then divide both sides of this equation by 24 to calculate the acceleration. 𝑎 is equal to 19𝑔 minus five 𝑔 multiplied by sin 35 divided by 24. Typing this into the calculator gives us 6.58728 and so on. We can round this to the nearest 100th or two decimal places. As seven is greater than five, we will round up.
The acceleration of the system to two decimal places is 6.59 metres per second squared.