Torque

1. definition

We know that when we open or close a door, it requires more force 
to open or close it if we pull or push close to the hinges. But 
the farther we are from the hinges, the easier it becomes. The 
same effect occurs if we use a wrench to loosen or tighten a bolt. 
We introduce then the concerpt of torque that takes into 
account a force and a distance (or radius)from 
the center of rotation or the axis of rotation to 
the  point of application of this force. It is the torque 
that tends to cause an object to rotate.

Linear acceleration of an object stems from 
applying exterior force(s) to this object. Angular 
acceleration is caused by applying a torque.


The expression of a torque is: 

τ = r x F = r F sin θ
θ is the angle between r and F.


 τ = r x F = r F sin θ
θ is the angle between r and F


r is the radius, that is the distance from the rotation 
axis to the point of application of the force F. The force F 
is the tangential force, hence the radius is called 
the lever arm.

A torque is a vector, perpendicular to r and F. Its direction 
is along the axis of rotation. By convention, the sign of a torque 
is positive for a counter-clockwise rotation and negative for 
clockwise rotation. The magnitude of τ is r F sin θ
θ is the angle between the vectors r and F. Its unit is the N.m.
We can recall also that if the line of action of a force passes 
though the axis, the torque aboout that axis is zero

2. Example:

The wheel has two radii a = 10 cm and b = 20 cm. 
The forces exerted on the edges have the following 
magnitudes: F1 = 3 N, F2 = 4 N, and F3 = 15 N. The angle θ 
is equal to 30o. What is the magnitude of the produced 
net torque τ?

The components of the net force are:
Fx = F1 - F3 cos θ = 3 - 15x(31/2 /2) = - 9.99
Fy = F2 + F3 sin θ = 4 + 15/2 = 11.5

The magnitude of the resultant force is:
F = [Fx2 + Fy2]1/2 = 15.23 N

Its orientation is:
θ = tg - (11.5/-9.99) = - 49o 

The clockwise equivalent torque has the magnitude of :
M = b F1  + b F2  -  a F3 cosθ -  0 F3 sin θ = + 10.10  N.cm. It is 
oriented along with the z-axis pointing towards us.

The force F = 15.23 N and the moment M 10.10 N.m can be reducted to 
a single force at the distance d = M/F = 10.10/15.23 = 0.66 cm

3. Static equilibrium

To be in static equilibrium, arigid body must be in both translational 
equilibrium  Σ Fext  = 0 and rotational equilibrium. 
The condition for the latter is  written as a vector equation:
Σ Τext  = 0, 
where  Σ Τext is the 
sum of the torques produced by the external forces acting on the 
rigid body. The rotational equilibrium balances the tendencies 
to rotate clockwise and counterclocwise about any axis.

static equilibrium: vector equations

The net external force is zero: Σ Fext  = 0 
The net external torque is zero: Σ Τext = 0

34. Example

What are the values of the forces F2 and F3 in the example 
above to set a static equilibrium ?

We have:
0 = F1 - F3 cos θ then:  F3 = F1/cos θ
0 = F2 + F3 sin θ then : F3 = - F2/sin θ
0 = b F1  + b F2  -  a F3 sin θ -  0 F3 cosθ 
Then: F3 = (b F1 + b F2 )/a sinθ 

F1/cos θ = - F2/sin θ or F2 = - F1 tan θ

Then: F2 = - 1.73 N

F3 = b(F1 + F2 )/a sinθ  = b(F1 - F1 tan θ )/a sinθ =  
b F1 (1 - tan θ )/a sinθ = 
20 x 3 x ( 1 - 3/1/2/3)/10 x (1/2) 
= 5.07 N