Moment of a Force

 

The magnitude ( τ) of the moment of a force about a fixed point (O) is the product of the magnitude (F) of the force and the perpendicular distance (d) from the point O to the line of action of the force:

 

A turning moment twists the object and is also known as torque and has SI units Newton metres (N m), which is the same unit used for work (1 Joule = 1 N m) but is not the same thing!

Torque

is the cross product

Where r is the position vector of the point at which the force is applied.

so

This can also be displayed as

Where r_⊥represents the perpendicular distance from the point to the force.

 

Convention labels an anticlockwise moment with a superscript +

and a clockwise moment with a superscript -

or showing final force directions as

and

 

 

Examples

Calculate the magnitude of the moment applied by the 5 newton force about the point A.

 

 

 

Extending for the perpendicular,

 

 

For a see saw in equilibrium, the total moment about the pivot is zero.

Example

Given that the see saw is balanced, calculate force P.

 

 

Static Rigid Bodies

The total moment about any fixed point of the body is zero.

 

To solve :

1. Draw a force diagram.
2. Resolve each force in a chosen direction.
3. Equate the sum of the forces to zero.
4. Repeat for all force directions.
   Calculate the moment of each force about a chosen point .
5. Classify these results as anticlockwise or clockwise.
6. Equate the total moment to zero.

 

Example

A uniform beam, 3m long and of mass 20 Kg, is supported horizontally by two vertical ropes.

Each of the ropes is attached at a distance of 50 cm from the edge of the beam.

What weight in Newtons should be placed on one end of the beam

a) to remove the tension in one of the ropes?

b) to make the tension in rope B double that of rope A?

 

a)

Place a weight (W) on the far right hand edge of the beam:

 

The immediate affect of W is to turn the beam clockwise, which will loosen the left hand rope. So when no tension, T₁ = 0

Taking moments about B:

A weight of 40g N (approximately 392.4 N) is required.

b)

Again, place a weight (W) on the far right hand edge of the beam:

Taking moments about the right hand edge:

so

Resolving forces vertically,

A weight of 40g/7 N (approximately 56.06 N) is required.

 

Example

A uniform ladder of mass M kg and length l m rests against a smooth vertical wall. The base of the ladder rests on rough horizontal ground. Find the minimum angle ,θ, between the ladder and the horizontal for which the ladder can remain static. Assume that μ, the coefficient of static friction, is 1.

 

Solution

The center of gravity is the midpoint of the ladder.

S and R are the horizontal and vertical normal reaction of the ladder on the wall.

T is frictional force of the base of the ladder.

 

Taking moments about the bottom of the ladder :

 

 

 

Resolving forces vertically:

 

Resolving forces horizontally:

 

 

The inverse cotangent must lie between 0 and π

Using

 

 

Couples

 

Like forces are parallel and act in the same direction.

 

Unlike forces are parallel and act in opposite directions.

 

A couple is formed by two unlike forces of equal magnitude which are not acting along the same line.

Example

Show that the forces about the rectangle in the following diagram form a couple ; show that the moment of the couple is independent of the point about which the turning is taking place ; find the moment of the couple required to attain equilibrium.

 

Resolving forces parallel to BA gives

Resolving forces parallel to BC gives

The system has no translatory effect, so must either be in equilibrium or reduce to a couple.

 

Taking moments anticlockwise about O

Taking moments anticlockwise about A

 

Taking moments anticlockwise about B

 

Taking moments anticlockwise about C

 

Taking moments anticlockwise about D

 

The moment of the couple is 2 N m clockwise, so a couple of 2 N m anticlockwise must be applied to put the system into equilibrium.