Parallel Forces

Parallel Forces:

Two parallel forces are said to be like when these act in the same sense, i.e., when these act in the same direction, and unlike if these are opposite in sense, i.e., when these act in the opposite directions.

Resultant of Two Like Parallel Forces:

Let P and Q be two like parallel forces acting at points A and B of a rigid body.

Let these be represented by AX and BY respectively. At A and B, apply two equal and opposite forces each equal to F and represented by AD and BE respectively. These do not produce any effect upon the equilibrium of the body. The parallelograms AXLD and BYME are completed. The forces P at A represented by AX and F at A represented by AD are equivalent to a resultant AL. Let it be R₁ and let its point of application be transferred to point O.

Similarly, the forces Q at B represented by BY, and F at B represented by BE are equivalent to a resultant represented by BM. Let it be R₂ and let its point of application be transferred to point O. CO is drawn parallel to AX or BY to meet AB at C. Now the force R₁ at O is equivalent to two forces. One is P in the direction CO and parallel to AX and the other is F parallel to AD.

Similarly, the force R₂ is equivalent to two forces. One is Q in the direction CO and parallel to BY and the other is F parallel to BE.

But the forces F, F parallel to AD and BE balance each other being equal and opposite. Hence we are left with two forces P and Q in the direction CO and the resultant of two like parallel forces P and Q acting at A and B is equivalent to a single force (P + Q) acting at C in the direction CO.

To determine the position of point C since the triangles OAC and LAD are similar,

Again, since the triangles OCB and MEB are similar,

Hence P . AC = Q . CB or AC/CB = Q/P, i.e., point C divides the line AB internally in the inverse ratio of the forces.

Resultant of Two Unlike Parallel Forces:

Let two unlike parallel forces P and Q (P > Q) act at two points A and B of a rigid body along AD and BE respectively.

AB is drawn. At A and B, introduce two equal and opposite forces each equal to S acting in the directions BA and AB respectively. These two forces balance each other and have no effect on the equilibrium of the body.

Let AD and AL represent P and S at A, and BE and EM represent Q and S at B, in magnitude and direction.

Now the forces P and S at A have a resultant R₁ represented by AF, and the forces Q and S at B have a resultant R₂ represented by BG.

AF and GB are produced to meet at O and through O draw OC parallel to P and Q meeting BA produced at C, and XOY parallel to AB. The points of application of R₁ and R₂ are shifted from A and B to O. The force R₁ at O is equivalent to its original components P and S along CO and OX respectively. So, also R₂ at O is equivalent to its original components Q and S along OC and OY respectively. Therefore, P at A and Q at B are equivalent to P along CO, S along OX, Q along OC, and S along OY.

But of these S along OX and S along OY being equal and opposite, balance and the other two are equivalent to (P – Q) at O along CO, i.e., to (P – Q) at C in the direction of the force P (as P > Q).

Hence, the resultant R of the unlike parallel forces P at A and Q at B is a parallel force (P – Q) acting at C in the direction of the force P.

To determine the point C, where R acts, we have from similar triangles COA and LFA,

Again, from the similar triangles OCB and BEG,

Hence P . AC = Q . CB or AC/CB = Q/P, i.e., point C divides the line AB externally in the inverse ratio of the forces.

Note:

(1) If the two unlike parallel forces be equal, i.e., P = Q, the two triangles FAL and GBM become congruent. ∴ ∠FAL =∠GBM = ∠ABO.

In this case, the diagonals AF and BG become parallel and shall not intersect at a finite distance. Hence, the construction fails.

Accordingly, two equal and unlike parallel forces cannot be compounded into a resultant force. They are then said to form a couple. This is the case of failure.

(2) Whether the forces P and Q are like or unlike if they are equivalent to a resultant R, we have in either case P/Q = BC/AC.

If P > Q, then BC > AC, i.e., the resultant passes nearer the greater force. If the forces are like and P = Q, then the resultant of two equal and like parallel forces lie mid-way between the components.

(3) We have in either case, P/Q = BC/AC or P/Q = BC/AC ⟹ (P± Q)/(BC ± AC) = R/AB (taking the upper signs when forces are like, and the lowest signs when the forces are unlike).

(4) It follows from (3) that if three forces P, Q, and R are in equilibrium, R is equal and opposite to the resultant of P and Q. Hence, each force is proportional to the distance between the other two.

Example 2: ABC is a triangle and O is its orthocentre. A force R acts along AO. R is resolved into two forces P and Q parallel to it and acting at B and C respectively. Find the magnitude of P and Q. Solution: AD is perpendicular to BC and O, the orthocentre lies on AD. Now BD/AB = cos B ∴ BD = AB cos B Similarly, CD = AC cos C ∴ BD/CD = (AB cos B)/(AC cos C) = (c cos B)/(b cos C) ……….(i) But a/sin A = b/sin B = c/sin C = k (say) ∴ a = k sin A, b= k sin B, c = k sin C ……….(ii) From (i) and (ii), we get BD/CD = (k sin C cos B)/(k sin B cos C) = tan C/tan B ……….(iii) Also, the line of action of the force R, i.e., AO meets BC in D, and P and Q are the components of R and parallel to it. ∴ R = P + Q ……….(iv) and P . BD = Q . CD or BD/CD = Q/P ……….(v) From (iii) and (v), we get Q/P = tan C/tan B or P/tan B = Q/tan C = (P + Q)/(tan B + tan C) = R/(tan B + tan C) [using (iv)] ∴ P = (R tan B)/(tan B + tan C) and Q = (R tan C)/(tan B + tan C)