The diagram below represents a balloon just before take-off. The balloon’s basket is attached to the ground by two fixing ropes.Calculate the tension in either of the fixing ropes.

The first and most important step is to highlight all the important information that is given in the question. Sometimes, it is very easy to miss out key numbers that you might require to get to the final answer. After having completed this, it is really helpful to sketch a small diagram on a paper to understand what forces are acting on the balloon. This gives you a really nice overview of the physical situation and will most of the time simplify the question a lot. Here the forces acting are: Gravity acting straight down, Tension along the two ropes and also an upward buoyancy/air force (which allows the balloon to be up in the air). Now: We know that there is no resultant force on the balloon (as it is stationary), so the forces acting upwards must equal the forces acting downwards.

The difference between the upward buoyancy force and gravity is: 2.15x10^3 - (1.95x10^2 x 9.81)=237N

This difference in forces must be equal to the tension (otherwise there would be a resultant force, and we said the overall force must be zero!) If you look on the diagram, it is then clear that both ropes have their own tension component, which makes up the 'total' tension and using Soh Cah Toa, and the angle of 50, we find that each tension must be equal to sin50 * x.

So, this 237N must be equal to 2Tsin(50). Equating this and solving for T gives you a tension of 150N.