Rigid Bodies: Equivalent System Of Forces
IntroductionIn general it is not feasible to consider bodies as point objects while considering the effect of forces. For tackling practical problems we need to take the dimension of bodies into account. A body which doesn’t deform by the application of force is termed as rigid bodies. Actual structures and machines, however, are never absolutely rigid and deform under the loads to which they are subjected
In the following unit we will be dealing with principle of transmissibility of forces , moment of a force about a point ,moment of a force about an axis ,moment of a couple ,addition of couple moments , resolution of a force into a force couple system ,resolution of a system of force into a force couple system , resolution of a system of forces into wrench . Letters in bold indicate vectors.
Principle of transmissibility, sliding vectorsIn the figure shown below, we are provided with two force systems. Let’s assume that a force F acts at on the rigid body. Let’s say that the force F’ has the same magnitude and direction and acts along the same line of action as that of F .Principle of transmissibility states that without altering the equations of equilibrium or motion and provided the above conditions are satisfied , the force F can be equivalently replaced by force F’. Therefore the force acting on the rigid body can be termed as a sliding vector, which is allowed to slide along its line of action.
Moment of a force MOMENT OF A FORCE ABOUT A POINT
Effect of a force not only depends on its magnitude and direction, but also depends on its point of application. Force is responsible for translational as well as rotational motion of a object. Let a force F be applied at a point A as shown in figure above. Position vector of A with respect to B is given by rAB. Therefore moment of the force F at A with respect to point B is given by : MB= rAB ×F The magnitude of MB is a measurement of rotating tendency of the object about B. Its SI unit is Newtonmetre (Nm). Points to note:
MB is a vector perpendicular to the plane containing rAB and F. Direction determined by right hand thumb rule. Rotation is anticlockwise if MB points outward to the plane and viceversa. MB= FrAB sinα, where α is the angle between force and position vector. In case of multiple forces acting at point B, moment of the resultant force is given by sum of individual moment of forces about pointB. MB= r × (F1 +F2+F3. . .) = r ×F1 + r× F2+r × F3. . . For an object with zero rotational motion the net moment of forces about every point should be zero.
MOMENT OF A FORCE ABOUT AN AXISEarlier we had seen that MO = rA × F . Now we define another useful term which is moment about about an axis. It is givem by : MP = î . (rA × F) where îis the unit vector in the direction of the axis OP. Moment about an axis measures the tendency of the force to rotate the body along the given axis. Points to note:
It is a scalar triple product which gives a scalar output. Result is independent of the point selected on the axis i.e. rBAcan be used intead of rA. To solve problems most convinient position vector should be selected in order to simplify calculations. For an object with zero rotational motion the net moment of forces about every axis should be zero.
Moment of a couple
 A couple system consist of two parallel forces equal in magnitude but opposite in direction. The above diagram shows the exact situation. Since the algebraic sum of force vectors is zero, the force will produce no translational effect on the body. But the body will tend to rotate in effect of the two forces.The derived result for moment due to couple turns out to be:MO = r ×FWhere F is the force vector and r is the vector joining points of application of two forces.We also get, MO=Fr sin α = Fd, where α is the angle between the two vectors , where ‘d’ is the perpendicular distance between two parallel force vectors(as shown in figure above). Points to note:
Since the results turns out to be independent of origin the same result would have been obtained if the origin was shifted. Therefore couple moment is a free vector which can be applied at any point. Rotation is anticlockwise if moment vector points outward to the plane of forces and vice versa. Since couple moments are vector quantities, it is derivable that individual couple moments acting upon a rigid body can be added up vectorialy. Two sets of couples producing same couple moment in the same direction are said to be equivalent couples. Couple moments are represented as vectors pointing outwards or inwards as the case maybe. Resolution of a force into a force couple system
In the above situation we have a forceFactingonthe rigid body at a point N. The principle of transmissibility does not allow us to shift forces parallel. To shift the force parallel we can add two forces at the point M equal in magnitude with the force Fbut opposite in direction directed parallel to initial force F. Hence we are in a situation where we can replace the two forces F and F by a singlecouple moment MO emerging out of the plane at the point M along with a shifted force vector. Thus, we can shift a force vector parallel to a point by adding a moment vector at a moment vector, created by the original force about the point. Proceeding backward in the above diagram we can state that a force couple system about a point can be equivalently replaced by a single force. For this we shift the force parallel until we have the moment of the force about the given point equal in magnitude and opposite in sense to the moment which is to be eliminated.
Equivalent Systems of forcesWe have just shown that a single force could be shifted parallel to a itself to a specific point by addition of a couple moment. This statement can be generalised for a system of forces acting on a system. Drawing parallel to the last topic we shift each individual force at ‘O’ and add moments created by the forces at ‘O’. Finally we add up the vectors. Resultant force at O:FR= F1+ F2+ F3 Resultant moment at O: MO =r1×F1 + r2×F2 + r3×F3 Thus, every complex system of forces can be thus reduced to a simple Couple – Force equivalent system. Unlike the previous case, we may not have resultant moment vector perpendicular to the resultant force vector. The equivalent force couple system is a characteristic of the system. Two systems are said to be equivalent if they reduce to the same force couple system at the same point.
Reduction of systems of forces to WrenchAny complex system of forces can be reduced to an equivalent force couple system as shown below:
MOis broken into two perpendicular components Ma and Mp, one along Frand other perpendicular to Fr as shown in figure. We shift Frparallel to a point Y such that: Mp= (rYX×R). Thus the additional moment added to the system on shifting of the force Fr cancels Mp. Since Ma is a free vector it can also be shifted to the point A. This particular forcecouple system is called a wrench because the resulting combination of push andr twist is the same as that would be caused by an actual wrench. The axis of force passing through Y is called the axis of a wrench. Pitch is defined as the ratio of: [Ma:Fr] Pitch is also given as = MO.Fr / Fr2 Axis = (pitch) * MO
Problem 1: Given force F =10N . OP=2m. Find :
b) Minimum force required and in what direction at P , to produce the same moment as in (a). c)where should be a force vertical of 25N be applied to obtain the same moment as in (a).
Problem 2:
In the following diagram we have three forces acting upon a thin rectangular plate. Compute the moment of various forces about of by the method of vectors.
Problem 3:
Compute the net couple moment of the following system :
Problem 4: Following figure indicates the arm of a factory robot. Structure has been simplified to bars as shown in figure. Find the reaction force and the moment at the support for equilibrium . Given weight components at X and Y to be 10N and 15N respectively .(Find out the equivalent force system at S and find the reaction components respectively for equilibrium).
Given S(0,4,0),x(1,1,5),y(3,2,2). Mz= 50 i 20 j + 40 k
Problem 5:Figure shows a force F ( 2i+5j3k) acting at the corner A of a cube of side s. Find the moment of the force about the axis (a)EC (b)FC
Problem 6:
Following figure consists of a symmetrical Tshaped wooden frame. Various dimensions and magnitude of forces are clearly indicated. Obtain the resultant of the system about the point O.
Problem 7: Reduce the following diagram to have a single force in the following system using the concept of couple moment and parallel shift of a force. Locate the final position of force F1. Also find out the final position of the force F1 if the couple pair were to be tilted at an angle of 60° with the vertical.
Given F1 = 30 N , F2 = 15 N , O is the midpoint of the bar .
Problem 8:The given diagram shows a pole BT (height 7m) supported by two cables AT and CT. Cable AT is supported by a block and CT supported by the ground. Tension in the cable AT is 50 N whereas 30N in the cable CT. All the dimensions mentioned. Find the moment of the forces about the bottom of the pole D.
Problem 9:Following figure indicate the forces acting on a cylindrical rod . Three forces indicated are parallel to three different coordinate axes. Find the resultant of the force system about O. Given magnitude of various quantities as : F1 = 10 N , F2 = 20 N , F3 = 15 N ,L1 = 50cm , L2= 30cm. Using this Also find the moment about z –axis.
Problem 10: Simplify the following setup to obtain a simple couple moment using the properties of couple moment.
Given F1 is parallel to x axis and F3 parallel to zaxis. F2 makes an acute angle of 30° with the x axis.  F1  = 10 N ,  F2 =20N ,  F3 =25N
Problem 11:
Following figure shows a T shaped bar clamped at one end at origin and supported by cables at other side. Find the moment of the tension in the cables about the origin.
Dimensions mentioned. T1 = 100 N , T2 = 125 N.
Problem 12:Given force F of magnitude of 10 dyne acting at the point D . All dimensions clerly mentioned .
Find : a) resultant of the system about O.
b) resultant of the system about B
c)horizontal force at C and a force at A.
Problem 12 :
Fig. below shows hexagonal plate acted upon by multiple forces. Force F1 acts along the line CB. Find the minimum magnitude and direction of force F2 so that the resultant moment of the system about the centre of the hexagon turns out to be zero.assume side of hexagonal to be a .Magnitude of F1 = 25 N . Problem 13: Find the equivalent of the given system at a) B b)C , also in part b) find the distance of a point (say M) from C , such that we can reduce the system into a vertical force at M point and a horizontal force at C.Problem 14:Find the minimum perpendicular distance between origin and the force F = (3i + 4j  5k) N. Also find the resultant of the system at point A. Problem 15: Find the resultant of the system about O and reduce the system into a wrench. Get the axis and the pitch of the wrench. Given F1 = 2i + 3j + 5k N , F2 = 10 k N , F3= 5k N. Problem 16:Reduce the following system into a wrench. specify the axis and the pitch . Given F1 = 10N , F2= (3j + 4k ) N ,F3= 5N , M=2Nm , AE = 10cm , AB =13cm. All coordinates in the diagram are in cm.Problem 17:Reduce the following system of forces acting on a cuboid into a wrench specify the pitch and axis .Given: F1 is aligned to FC. F1= 100N , F2 = 100N , F3 = 200N ,F4 = 150N, M = 5Nm Problem 18:In the following diagram reduce the couple moments and find out the equivalent of the system about O. Further reduce the system to a wrench. Specify the axis and the pitch of the given system.F1 = 5N, F2 = 10N, F3 = 20N, F4 = 7N
Problem 19:
A rectangular plate OCAB is acted upon by several forces as shown. Magnitude of forces F1, F2, F3 are 10N, 20N,30N respectively. Find the resultant of forces at O. Reduce the system into a single force.
Complete Tutorial with Problems, Figures and Solutions :
Rigid Bodies: Equivalent System Of Forces
