Torque of a force with respect to a point P

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 F1 with respect to P = F1 l 1=  –  (3.0)( 0.02) =

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

Torque of F2 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 (FA )

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 – FA  AS = 0

(30.0)( 0.25) –  (FA) (0.75) = 0

FA = 10 N

b.Find the force on the bar in S.

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

FS = W + FA

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 30o = Fy/F          Fy = F sin 30o = 0.5  F

cos 30o = Fx/F          Fx = F cos 30o = 0.866 F

Sum of the  torques is zero

Σ τ = 0    with respect to S

FA  AS – Fy  BS = 0

(10)( 0.50) – (Fy )( 40)= 0

Fy = 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:  FSX  = 30.0 – 25.0 = 5.0 N  (upward)

In horizontal direction   : FSX = F 0.866    FSX = (25 )( 0.866) = 21.65 N (to the left)

Both components of Fs are known:

 FS = 22,2 N   

 ( Pythagorean theorem   FS2= FSX2 + FSY2)

Calculation of the direction :

tan α = 5.0/21.65 =  0.2309

α = 13 o

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