Torque: We have already studied in school physics that a turning effect is produced when a nut is tightened with a spanner (Fig. 2.13). the turning effect increases when you push harder on the spanner. It also depends on the length of the spanner: the longer the handle of the spanner, the greater is the turning effect of an applied force.
The turning effect of a force is called its moment or torque and its magnitude is defined as the product of force F and the perpendicular distance from its line of action to the pivot which is the point O around which the body (spanner) rotates. This distance OP is called moment arm I. Thus the magnitude of torque represented by τ is
τ = IF ……………….. (2.26)
Fig. 2.13: the nut is easy to turn with a spanner.
It is easier still if the spanner has a long handle.
when the line of action of the applied force passes through the pivot point, the value of moment arm I = 0, so in this case torque is zero.
We now consider the torque due to a force F acting on a rigid body. Let the force F acts on rigid body at point P whose position vector relative to pivot O is r. the force F can be resolved into two rectangular components, F sin θ perpendicular to r and F cosθ along the direction of r (Fig. 2.14 a). the torque due to F cos θ about pivot O is zero as its line of action passes through point O. therefore, the magnitude of torque due to F is equal to the torque due to F sin θ only about O. it is given by
τ = (F sinθ) r = r F sin θ ………….. (2.27)
alternatively the moment arm I is equal to the magnitude of the component of R perpendicular to the line of action of F as illustrated in Fig. 2.14 (b). thus
τ = (r sinθ) F = r F sin θ ……………… (2.28)
where θ is the angle between r and F
from Eq. 2.27 and Eq. 2.28 it can be seen that the torque can be defined by the vector product of position vector r and the force F, so
τ = r x F
or τ = (r F sin θ) n̂ ……………… (2.29)
where (rF sin θ) is the magnitude of the torque. The direction of the torque represented by n̂ is perpendicular to the plane containing r and F given by right hand rule for the vector product of two vectors.
The SI unit for torque is newton metre (N m).
Just as force determines the linear acceleration produced in a body, the torque acting on a body determines its angular acceleration. Torque is the analogous of force for rotational motion. If the body is at rest or rotating with uniform angular velocity, the angular acceleration will be zero. In this case the torque acting on the body will be zero.
Point to ponder
Example 2.6: the line of action of a force F passes through a point P of a body whose position vector in metre is î-2ĵ+k̂. if F = 2î -3 ĵ + 4k̂ (in Newton), Determine the torque about the point A whose position vector (in metre) is 2 Î + ĵ + k̂
Solution:
The position vector of point 
The position vector of point
relative to O.
The position vector of P relative to A is
AP = (î – 2 ĵ + k̂) – (2 î + ĵ + k̂) = -î -3 ĵ
The torque about A = r x F
= (-Î -3 ĵ) x (2 Î – 3 ĵ + 4k̂)
= -12 Î +4 ĵ +9k̂ N m
Can you do?
Equilibrium of Torques
Equilibrium of Torques – Second condition of equilibrium: let two equal and opposite forces act on a rigid body as shown in Fig. 2.17. although the first condition of equilibrium is satisfied, yet it may rotate having clockwise turning effect. As discussed earlier, for angular acceleration to be zero, the net torque acting on the body should be zero. Thus for a body in equilibrium, the vector sum of all the torques acting on it about any arbitrary axis should be zero. This is known as second condition of equilibrium. Mathematically it is written as
By convention, the counter clockwise torques are taken as positive and clockwise torques as negative. An axis is chosen for calculating the torques. The position of the axis is quite arbitrary. Axis can be chosen anywhere which is convenient in applying the torque equation. A most helpful point of rotation is the one through which lines of action of several forces pass.
We are now in a position to state the complete requirements for a body to be in equilibrium, which are
When 1st condition is satisfied, there is no linear acceleration and body will be in translational equilibrium. When 2nd condition is satisfied, there is no angular acceleration and body will be in rotational equilibrium.
For a body to be in complete equilibrium, both conditions should be satisfied, i.e., both linear acceleration and angular acceleration should be zero.
If a body is at rest, it is said to be in static equilibrium but if the body is moving with uniform velocity, it is said to be in dynamic equilibrium.
We will restrict the applications of above mentioned conditions of equilibrium to situations in which all the forces lie in a common plane. Such forces are said to be coplanar. We will also assume that these forces lie in the xy-plane.
If there are more than one object in equilibrium in a given problem, one object is selected at a time to apply the conditions of equilibrium.
Can You Do?
With your nose touching the end of the door, put your feet astride the door and try to rise up on your toes.
Example 2.8: a uniform beam of 200N si supported horizontally as shown. If the breaking tension of the rope is 400N, how far can the man of weight 400 N walk from point A on the beam as shown in Fig. 2.18?
Solution: let breaking point be at a distance d from the pivot A. the force diagram of the situation is given in fig. 2.19. by applying 2nd condition of equilibrium about point A
Στ = 0
400 N x 6 m – 400 N x d – 200 N x 3 m = 0
Or 400 N x d = 2400 Nm – 600 Nm = 1800 Nm
D = 4.5 m
2.9: A boy weighing 300 N is standing at the edge of a uniform diving board 4.0m in length. The weight of the board is 200 N. (fig. 2.20 a). find the forces exerted by pedestals on the board.div align=”center”>
>Solution: we isolate the diving board which is in equilibrium under the action of forces shown in the force diagram (fig. 2.20 b). note that the weight 200 N of the uniform diving board is shown to act at point C, the centre of gravity which is taken as the mid-point of the board, R1 and R2 are the reaction forces exerted by the pedestals on the board. A little consideration will show that R1 is in the wrong direction, because the board must be actually pressed down in order to keep it in equilibrium. We shall see that this assumption will be automatically corrected by calculations.
Let us now apply conditions of equilibrium
Σ Fx = 0 (No x-directed forces)
Σ Fy = 0 R1 + R2 – 300 – 200 = 0
R1 + R2 = 500 N …(i)
Στ = 0 (pivot at point D)
– R1 x AD – 300 N x DB – 200 N x DC = 0
– R1 x 1m – 300 N x 3 m – 200 N x 1m = 0
Substituting the value of R1 in Eq. (i). we have
– 1100 + R2 = 500
The negative sign of R1 shows that it is directed downward. Thus the result has corrected the mistake of our initial assumption.
Equilibrium of Forces: We have studied in school physics that if a body, under the action of a number of forces, is at rest or moving with uniform velocity, it is said to be in equilibrium.
First Condition of Equilibrium
A body at rest or moving with uniform velocity has zero acceleration. From Newton’s Law of motion the vector sum. Of all forces acting on it must be zero.
This is known as the first condition of equilibrium. Using the mathematical symbol Σ F for the sum of all forces we can write
Σ F = 0 ……………….. (2.30)
In case of coplanar forces, this condition is expressed usually terms of x and y components of the forces. We have studied that x-component of the resultant force F equals the sum of x-directed or x-components of all the forces acting on the body. Hence
ΣFx = 0 …………. (2.32)
It may be noted that if the rightward forces are taken as positive then leftward forces are taken as negative. Similarly if upward forces are taken as positive then downward forces are taken as negative.
Example 2.7: a load is suspended by two cords as shown in fig. 2.15. determine the maximum load that can be suspended at P, if maximum breaking tension of the cord used is 50N.
Solution: for using conditions of equilibrium, all the forces acting at point P are shown by a force diagram as illustrated in Fig. 2.16 where w is assumed to be the maximum weight which can be suspended. The inclined forces can now be easily resolved along x and y directions.
Applying ΣFx = 0
Putting the values
50 N + 0.866 + 26.6 N x 0.34 = w
Or w = 52 N
Interesting application
A concurrent force system in equilibrium. The tension applied can be adjusted as designed.