τ = F r

- τ torque in Nm
- F force in N
- l lever arm (perpendicular distance from P tot the line of action of F)

If F tends to cause counterclockwise rotation around P then we choose τ > 0

If F tends to cause clockwise rotation around P then τ < 0

**Example 1 **

Torque of F_{1} with respect to P = F_{1 }l _{1}= – (3.0)( 0.02) =

– 0.06 Nm (if l _{1 }in cm then 6.0 Ncm )

Torque of F_{2}_{ }with respect to P = 0.06 Nm

**Example 2**

An uniform bar of 2.00 m with a weight of 30.0 N is supported in S. The bar can rotate about S in the vertical plane. The center of mass O is in the middle.

OS = 25 cm (see drawing)

The bar is in equilibrium.

**a**.Find the force on the end A of the bar (F_{A })

There are two conditions of equilibrium:

1.Σ τ = 0 with respect to every point

2.ΣF = 0 N

The sum of the torques must be zero

Σ τ = 0 with respect to S:

W OS – F_{A} AS = 0

(30.0)( 0.25) – (F_{A}) (0.75) = 0

F_{A} = 10 N

**b**.Find the force on the bar in S.

The sum of the forces on the bar must be also zero.

F_{S }= W + F_{A}

**Example 3 **

Given a uniform bar. Lenght = 1.00 m. The weight of the bar is 20.0 N

A force of 10.0 N acts perpendicular downward on the bar in A.

The bar is supported in S.

The bar is in equilibrium.

In B, at 40 cm from the center of mass, a force F acts upwards on the bar at an angle of 30 ^{o}.

** a.**Find F.

Resolve F into two directions, along the bar and perpendicular to the bar

Sin 30^{o} = F_{y}/F F_{y} = F sin 30^{o} = 0.5 F

cos 30^{o} = F_{x}/F F_{x} = F cos 30^{o }= 0.866 F

Sum of the torques is zero

Σ τ = 0 with respect to S

F_{A} AS – F_{y}_{ }BS = 0

(10)( 0.50) – (F_{y}_{ })( 40)= 0

F_{y }= 12.5 N

Now we can calculate F

F _{y} = 0.5 F

F = 12.5/0.5 = 25.0 N

**b.**Find magnitude and direction of the force that acts on the bar in S.

A second condition for equilibrium is

ΣF = 0 N

At first in the vertical direction: F_{SX }_{ }= 30.0 – 25.0 = 5.0 N (upward)

In horizontal direction : F_{SX} = F 0.866 F_{SX} = (25 )( 0.866) = 21.65 N (to the left)

Both components of F_{s }are known:

F_{S}_{ }= 22,2 N

( Pythagorean theorem F_{S}^{2}= F_{SX}^{2} + F_{SY}^{2})

Calculation of the direction :

tan α = 5.0/21.65 = 0.2309

α = 13 ^{o}