Mechanics Tutorial 3 - Moments and Equilibrium

 Contents

Mass and Weight

The mass of an object represents the amount of material in it.  It is measured in kilograms, and is a scalar.  The mass results in inertia, the degree in which a body opposes the change in velocity.  Mass does not change.

Weight is a force measured in newtons.  It is the force that is the result of gravity acting on a force.  Like all forces, weight is a vector, which always acts vertically downwards.  Weight depends on the acceleration due to gravity.

Weight (N) = Mass (kg) × acceleration due to gravity (m s-2)

W = mg

The acceleration due to gravity on Earth is 9.81 m s-2.  It is also expressed as force per unit mass, 9.81 N kg-1.  On the Moon the acceleration due to gravity is 1.6 m s-2.

Turning Effects of a Force

If we have hinged or pivoted body, any force applied changes the rotation of that body about the pivot. The turning effect is called a moment. The equation is:

Moment = force × perpendicular distance

In Physics Code:

G = Fs

This strange looking symbol, G, which looks like a gallows, is “Gamma”, a Greek capital letter ‘G’.  Units are newton metres (N m).  An alternative physics code for moment is t, ("tau", a Greek lower case letter 't').

Moments have a direction.  As they are turning effects, we can talk of clockwise and anti-clockwise moments. By convention, clockwise is positive. Question 1 The spanner in the picture is 30 cm long and the nut in question has to be tightened to a torque (moment) of 85 N m.  What force must the fitter apply? Consider a trap door held by a piece of string, BC.  P and Q are forces.  The trap door is hinged about point O. The perpendicular distance of the line of action of force Q is the length of the line OC.

Moment of P about O = P × OA

Moment of Q about O = Q × OC

Note that the Moment of Q about O is [Q ×OC]. This is because OC is the perpendicular distance of the force Q from the hinge O.

If the trap door remains in equilibrium:

Anticlockwise moment (Q × OC) = clockwise moment (P ×OA)

This is the Principle of Moments.

Since OC = OB sin q we can say that  Q × (OB sin q) =  P × OA

 Question 2 The trap door in the diagram has a mass of 12 kg.  The centre of mass is at A. The weight is force P.  What is the value of force P?  The angle q is 25 degrees.  The trap door is 100 cm wide.  What is the value of force Q? (Use g = 9.8 m s-2) Use the quantities in the diagram above to calculate the angle q.

The principle of moments is an important rule that says:
The sum of clockwise moments about a pivot… (1 mark)
• …is equal to the sum of anti-clockwise moments… (1 mark)
• …for a system in equilibrium. (1 mark)

When asked to state the principle of moments, you must include these three points.  A pivot is sometimes called a fulcrum.

Consider a beam of negligible mass, on which three forces in equilibrium are acting.  The forces are arranged as in the diagram:

• F1 is x m from the pivot;

• F2 is y m from the pivot;

• F3 is z m from the pivot. Using the principle of moments we can write:

Clockwise moments = anticlockwise moments

Therefore: Centre of Mass

We treat objects as point masses referring to a single point called the centre of mass.

In regular objects like a cube or a sphere, the centre of mass is in the middle.  In some objects the centre of mass is outside the object. The centre of mass (or centre of gravity) is the point through which the entire weight is said to act.  Objects with a very low centre of gravity tend to be very stable.  Some objects are so stable that they never fall over.  Objects with a high centre of gravity are unstable.

Many questions involve the balancing of see-saw around pivots.  Let us look at some situations: This is the simplest case.  The pivot is in the middle of a uniform bar.  It means that the object is totally regular and the centre of mass is in the middle.  Therefore we can ignore the mass of the bar.  If the bar is balanced, we can say that:

anticlockwise moments = clockwise moments

Ax = By

Let us look at a case where we move the pivot P to one side.  The centre of mass stays where it is, and is NOT above the pivot.  We now have to take it into account.  Mass is not a force, but weight is.  So we need the weight, W.

W = mg In this case, the line of action of the weight, W, is z metres from the pivot P.  Applying the principle of moments we can say:

Ax = By + Wz

 Question 4 Some children are playing on a see-saw as shown in the diagram.  The see-saw is a plank of wood 3.0 m long, with a pivot exactly in the middle.  The centre of mass is directly above the pivot, so we can ignore it. (a) What are  the weights of the children? (b) Child B is sitting 0.4 m from the pivot.  Where should child C sit so that the see-saw remains level?      (c) Child C misses its footing and falls off the end.  What will happen to the others?   Use g = 9.8 m s-2 Couples

If two forces act about a hinge in opposite directions, there is an obvious turning effect called a couple. The resulting linear force from a couple is zero. The couple is given by the simple formula:

G = 2 Fs

This strange looking symbol, G, is “gamma”, a Greek capital letter ‘G’.  Couples are measured in Newton metres (Nm).  The code t (tau) is also used.

The turning effect is often called the torque.  It is a common measurement made on motors and engines, alongside the power.  Racing engines may be quite powerful but not have a large amount of torque.  This is why it would not make sense for a racing car to be hitched to a caravan, any more so than trying to win a Formula 1 race in a 4 x 4.

Centre of Mass and Stability

A stable object does not tip over.  Objects that have a high stability do not tip over easily.  Their centre of mass is low down, near the base.  This candlestick has a low centre of mass so that it does not tip over so easily.

The candlestick is in stable equilibrium.  If you push the top, it will drop back to where it was. Now suppose we put it upside down.  This time the centre of mass is high up.  It is in unstable equilibrium and if you pushed the candlestick, it would tip over easily. Now we put it on its side: If you push it, it will roll, but will not tip over.  It is in neutral equilibrium.

This wine rack uses the idea of moments to form a stable object. Question 5 Look at the picture above. (a) Where is the centre of mass in the bottle? (b) Where is the pivot? (c) Where does the line of action from the centre of mass of the bottle act? (d) Use the principle of moments to explain how this system balances. Finding the centre of mass for irregular objects

We can find the centre of mass of an irregular object quite easily.  If we let it hang freely, the centre of mass is directly below where we hang it from.  We can find this line by hanging a plumb-line from where we hang the object.  In old text books, an object like this is called an irregular lamina.  (The word lamina is a Latin word for leaf.) We draw a line vertically downwards.

If we then hang the object from a couple of other points and draw the lines that go vertically downwards, the centre of mass is where the lines meet.

When the object is hanging freely, the centre of mass is vertically below the hanging point.  The vertical arrow is called the line of action of the weight.

We will see the importance of stability in the next tutorial.

Some Examples using Moments

Sack Trolley
Now let’s look at a sack trolley being pushed along a floor. The mass of the trolley and the load is
m. The trolley is at an angle to the vertical. The distance from the centre of mass and the pivot is s, while the total length of the trolley is l. For the sake of simplicity, the centre of mass is being taken as both the centre of mass of the trolley and the load.  We will take moments about the pivot.

The weight results in the clockwise moment: The force F (the lifting force from the person pushing the trolley) gives the anticlockwise moment: A trolley with its load has a combined mass of 50.0 kg. The total length of the trolley is 1.10 m and the centre of mass is 30.5 cm from the wheels. The trolley is being held at an angle of 40o to the vertical. Calculate the force that is needed to hold it at that angle. (Use g = 9.81 N kg-1)

Consider a uniform ladder propped up against a wall between points A and B of mass m and length l propped up against a wall at an angle q. There are four forces:

• The weight, W, which always acts vertically downwards from the centre of mass.

• The normal force, N, which acts at 90o to the ground.

• The frictional force, F, which acts horizontally towards the wall and stops the ladder from sliding outwards (which is not desirable).

• The support force, S, from the wall which acts in a horizontally from the wall.

A ladder can be treated using moments.  Since it's uniform, the centre of mass is in the middle, i.e. half way along its length.  The ladder is stable because all four forces form a closed rectangle: We can resolve vertically and horizontally:

N = W and S = F

The weight of the ladder is mg.

The ladder is in contact with the ground at point A. Point A acts as the pivot. So there is a clockwise moment acting about A which is given by: The force, S gives out an anticlockwise moment.  Now we can work out the anti-clockwise moment about A: This is because the moment is the force S multiplied by the perpendicular distance from the pivot: By the principle of moments we can write: which tidies up to give and rearranges to give: Notice that the l terms cancel out.  Therefore the support force of the wall on the ladder, S, that acts away from the wall, is independent of the length.  The frictional force, F, has the same value as S, but acts towards the wall.

 A ladder of mass 20 kg and 3.0 m in length is propped up against a wall at an angle of 72o.  Calculate the support force on the wall. Which data item is irrelevant? g = 9.81 N kg-1.