Equilibrium For Coplanar Forces Elliott Free Body Diagrams

Equilibrium For Coplanar Forces Elliott

Free Body Diagrams In mechanics, we model things as point masses, which makes things rather simpler. We do this as a free body diagram which shows all the forces acting on an object as acting on the centre of mass.

Balanced Forces in equilibrium mean that they are balanced. Coplanar forces act in the same plane. Two balanced forces are equal in magnitude but opposite in direction to the other.

Basic But Don’t Forget! If the resultant force is zero, the system is in equilibrium. If there is an overall resultant force, movement will result. Strictly speaking, acceleration will result.

Triangle of Forces Rule If 3 forces acting at a point can be represented in size or direction by the sides of a closed triangle, then the forces are in equilibrium, provided their directions can form a closed triangle. Notice how: • The forces form into a closed triangle. • The directions of the forces go round the triangle.

The forces F 1 and F 2 ma ke a closed triangle, but they are not going noseto-tail. These forces are NOT in equilibrium. F 1 an d F 2 make up a resultant FR.

Now let’s reverse the directions of F 1 and F 2, but keep the magnitudes the same: If we move F 1 we find that the three forces form a closed triangle and the arrows go noseto-tail.

Resolving the Vectors by Accurate Drawing

Resolving Vectors Through Trigonometry Any vector in any direction can be resolved into vertical and horizontal components at 90 degrees to each other.

For this situation we know that weight always acts vertically downwards. We can resolve the two other vectors into their horizontal and vertical components: T 1 resolves into T 1 cos θ 1 (horizontal) and T 1 sin θ 1 (vertical); T 2 resolves into T 2 cos θ 2 (horizontal) and T 2 sin θ 2 (vertical).

• We know that the three forces add up to zero, so we can say: • T 1 cos θ 1 + - T 2 cos θ 2 = 0. This means that the forces are equal and opposite. • T 1 cos θ 1 = T 2 cos θ 2 • T 1 sin θ 1 + T 2 sin θ 2 + - W = 0 ( Weight is acting downwards and downwards is, by convention, negative). • T 1 sin θ 1 + T 2 sin θ 2 = W

Common Error Be careful that you don't assume that W is split evenly between T 1 sin θ 1 and T 2 sin θ 2. This is only true when the weight is half way between the ends. Many students write T 1 sin q = mg which is wrong.

Forces Acting Symmetrically Consider a mass m that is suspended, so that it hangs freely, half-way between two points X and Y. Its weight will be mg and it will cause a tension, T 1 in the left hand part of the string and a tension T 2 in the right hand part of the string, as shown: This is a symmetrical situation. T 1 = T 2

The horizontal components add up to zero because they are equal in magnitude, and in opposite directions. The vertical components add up to mg, the vertical downwards force

Quick Note • At AS level, you are most likely to encounter symmetrical systems like this. If the system is NOT symmetrical, don't panic. • Remember: The weight (downwards force) can be split into two force vectors. These add up to the weight. • The horizontal force vectors add up to zero. • If you are completely stuck, use accurate drawing! Whatever you do, don't leave a blank!

Check Your Progress A ship is in a dock and is secured to bollards on the harbour walls by two thick ropes. It is facing into a very strong wind which is putting a force of 28 000 N on the ship. This is shown in the diagram below: (a) Calculate the tension T in the top rope. (b) Calculate the force at 90 degrees to the wind that is acting on the bottom bollard. Which direction is it acting in? (c) What is the overall force acting on the ship? Explain your answer. (d) The bottom bollard was not very well put in, and it gets ripped out of the harbour wall. Calculate the resultant force acting on the ship immediately after the bollard gives way, and state the direction relative to the wind. 0 o is parallel to the wind.

Answer