Torque

Forces and Rotation

In Linear Mechanics, The acceleration of an object was dependant only upon the Mass of the object and the Force applied to it. This gave rise to Newton's Second Law. When looking at Forces causing rotation, the picture is not a simple. Consider the two diagrams below :-

For a set point on a door, the larger the Force applied, the larger the turning effect, therefore the larger the Acceleration will be. This can be easily demonstrated and follows Newton's Second Law.

However, if the same sized Force is applied at different positions along the door, the further from the turning point, The larger the turning effect, therefore the larger the Acceleration will be.

The turning effect of a Force is known as the Torque ( T ) or moment, and is defined by :-

T = F x r

Where :-

T = Torque ( Nm )

F = Force Applied ( N )

r = Radius of Motion ( m )

Example 1 -

A Plumber tightens a nut as shown in the diagram above. If the Radius of motion is 20 cm and the Plumber applies a Force of 40 N, what is the magnitude of the Torque generated ?

T = F x r

T = 40 x 0.2

T = 8 Nm

The above formula applies only if the Force is applied at a right angle to the the radius of motion. If the Force is applied at any other angle the important distance is the perpendicular distance from the line of action to the axis of rotation. The diagram below shows how this is calculated :-

If the Force is applied at an angle ϕ to the axis of rotation, the then Torque can be found using the following formula :-

T = F r sinϕ

Where :-

T = Torque ( Nm )

F = Force Applied ( N )

r = Radius of Motion ( m )

ϕ = The Angle to the Axis of Rotation ( o )

Example 2 -

A Hammer is used to remove a Nail from a plank of wood as shown above. If the Hammer is 40 cm long and a Force of 15 N is applied at an angle of 85o , what is the size of the Torque applied to the Nail ?

T = F x r sin ϕ

T = 15 x 0.4 x sin 85

T = 6 Nm

Torque and Free Body Diagrams

As can be seen in the National 5 Course, Free Body Diagrams can be used to find the Unbalanced Force acting on an object by resolving all forces in each direction ( horizontally or vertically ) . The same can used to find the Unbalanced Torque acting on an object. In this case, all Torques act either clockwise or anticlockwise.

If the Torques acting clockwise are equal to the Torques acting anticlockwise, then the resultant Torque is equal to zero.

Example 3 -

A Castle Drawbridge is lowered using two sets of chains as shown in the image below.

Image result for drawbridge

The Drawbridge has a mass of 2500 kg and a length of 4m. When lowered, the chains make an angle of 60o to the horizontal. The Drawbridge is uniform and therefore its center of mass is at a distance of 2m from the axis of rotation. Calculate the Tension in each chain.

If the Drawbridge is stationary, then the resultant Torque must be zero. For the above example, the Torque due to the Weight of the Drawbridge must be equal and in opposite direction to the Torque due to the Tension in each Chain.

TorqueClockwise = TorqueAnticlockwise

2 ( T x 4 x sin 60 ) = 2500 x 9.81 x 2

6.93 T = 49050

T = 7078 N

Note - In the above example, the clockwise Torque has an extra factor of 2 due to there being two chains holding up the drawbridge .