0.2 Force, momentum and impulse  (Page 18/35)

Forces in equilibrium

At the beginning of this chapter it was mentioned that resultant forces cause objects to accelerate in a straight line. If an object is stationary or moving at constant velocity then either,

• no forces are acting on the object, or
• the forces acting on that object are exactly balanced.

In other words, for stationary objects or objects moving with constant velocity, the resultant force acting on the object is zero. Additionally, if there is a perpendicular moment of force, then the object will rotate. You will learn more about moments of force later in this chapter.

Therefore, in order for an object not to move or to be in equilibrium , the sum of the forces (resultant force) must be zero and the sum of the moments of force must be zero.

Equilibrium

An object in equilibrium has both the sum of the forces acting on it and the sum of the moments of the forces equal to zero.

If a resultant force acts on an object then that object can be brought into equilibrium by applying an additional force that exactly balances this resultant. Such a force is called the equilibrant and is equal in magnitude but opposite in direction to the original resultant force acting on the object.

Equilibrant

The equilibrant of any number of forces is the single force required to produce equilibrium, and is equal in magnitude but opposite in direction to the resultant force.

In the figure the resultant of $F_1$ and $F_2$ is shown. The equilibrant of $F_1$ and $F_2$ is then the vector opposite in direction to this resultant with the same magnitude (i.e. $F_3$ ).

• $F_1$ , $F_2$ and $F_3$ are in equilibrium
• $F_3$ is the equilibrant of $F_1$ and $F_2$
• $F_1$ and $F_2$ are kept in equilibrium by $F_3$

As an example of an object in equilibrium, consider an object held stationary by two ropes in the arrangement below:

Let us draw a free body diagram for the object. In the free body diagram the object is drawn as a dot and all forces acting on the object are drawn in the correct directions starting from that dot. In this case, three forces are acting on the object.

Each rope exerts a force on the object in the direction of the rope away from the object. These tension forces are represented by $T_1$ and $T_2$ . Since the object has mass, it is attracted towards the centre of the Earth. This weight is represented in the force diagram as $F_g$ .

Since the object is stationary, the resultant force acting on the object is zero. In other words the three force vectors drawn tail-to-head form a closed triangle:

A car engine of weight 2000 N is lifted by means of a chain and pulley system. The engine is initially suspended by the chain, hanging stationary. Then, the engine is pulled sideways by a mechanic, using a rope. The engine is held in such a position that the chain makes an angle of ${30}^{\circ }$ with the vertical. In the questions that follow, the masses of the chain and the rope can be ignored.

1. Draw a free body representing the forces acting on the engine in the initial situation.
2. Determine the tension in the chain initially.
3. Draw a free body diagram representing the forces acting on the engine in the final situation.
4. Determine the magnitude of the applied force and the tension in the chain in the final situations.

1. Initial free body diagram for the engine

   There are only two forces acting on the engine initially: the tension in the chain, $T_{chain}$ and the weight of the engine, $F_g$ .

   

2. Determine the tension in the chain

   The engine is initially stationary, which means that the resultant force on the engine is zero. There are also no moments of force. Thus the tension in the chain exactly balances the weight of the engine. The tension in the chain is:

   $$\begin{array}{ccc}\hfill T_{chain}& =& F_g\hfill \\ & =& 2000\mathrm{N}\hfill \end{array}$$
3. Final free body diagram for the engine

   There are three forces acting on the engine finally: The tension in the chain, the applied force and the weight of the engine.

   

4. Calculate the magnitude of the applied force and the tension in the chain in the final situation

   Since no method was specified let us calculate the magnitudes algebraically. Since the triangle formed by the three forces is a right-angle triangle this is easily done:

   $$\begin{array}{ccc}\hfill \frac{F_{applied}}{F_g}& =& tan{30}^{\circ }\hfill \\ \hfill F_{applied}& =& \left(2000\mathrm{N}\right)tan{30}^{\circ }\hfill \\ & =& 1155\mathrm{N}\hfill \end{array}$$

   and

   $$\begin{array}{ccc}\hfill \frac{T_{chain}}{F_g}& =& \frac{1}{cos{30}^{\circ }}\hfill \\ \hfill T_{chain}& =& \frac{2000\mathrm{N}}{cos{30}^{\circ }}\hfill \\ & =& 2309\mathrm{N}\hfill \end{array}$$