Parallelogram and Triangle law of forces

  

If two forces acting at a point are represented in magnitude and direction by the two adjacent sides of a parallelogram, then their resultant isrepresented in magnitude and direction by the diagonal passing through the point.

Parallelogram law of forces

If two forces acting at a point are represented in magnitude and direction by the two adjacent sides of a parallelogram, then their resultant isrepresented in magnitude and direction by the diagonal passing through the point.

Explanation

Consider two forces Vector P and Vector Q acting at a point O inclined at an angle θ as shown in Fig..

The forces Vector P and Vector Q are represented in magnitude and direction by the sides OA and OB of a parallelogram OACB as shown in Fig.

The resultant Vector R of the forces Vector P and Vector Q is the diagonal OC of the parallelogram. The magnitude of the resultant is

R = root[ P2 +Q2 + 2PQcos θ ]

The direction of the resultant is α = tan-1[ Qsin θ  /  P+Qcos θ ]

Triangle law of forces

The resultant of two forces acting at a point can also be found by using triangle law of forces.

If two forces acting at a point are represented in magnitude and direction by the two adjacent sides of a triangle taken in order, then the closing side of the triangle taken in the reversed order represents the resultant of the forces in magnitude and direction.

Forces Vector P and Vector Q act at an angle θ. In order to find the resultant of Vector P and Vector Q, one can apply the head to tail method, to construct the triangle.

In Fig., OA and AB represent Vector P and Vector Q in magnitude and direction. The closing side OB of the triangle taken in the reversed order represents the resultant Vector R of the forces Vector P and Vector Q. The magnitude and the direction of Vector R can be found by using sine and cosine laws of triangles.

The triangle law of forces can also be stated as, if a body is in equilibrium under the action of three forces acting at a point, then the three forces can be completely represented by the three sides of a triangle taken in order.

If Vector P , Vector Q and Vector R are the three forces acting at a point and they are represented by the three sides of a triangle then P/QA =Q/AB =R/ OB.

Experimental verification of triangle law, parallelogram law and Lami's theorem

  

Two smooth small pulleys are fixed, one each at the top corners of a drawing board kept vertically on a wall as shown in Fig.. The pulleys should move freely without any friction.

Experimental verification of triangle law, parallelogram law and Lami's theorem

Two smooth small pulleys are fixed, one each at the top corners of a drawing board kept vertically

on a wall as shown in Fig.. The pulleys should move freely without any friction. A light string is made to pass over both the pulleys. Two slotted weights P and Q (of the order of 50 g) are taken and are tied to the two free ends of the string. Another short string is tied to the centre of the first string at O. A third slotted weight R is attached to the free end of the short string. The weights P, Q and R are adjusted such that the system is at rest.

The point O is in equilibrium under the action of the three forces P, Q and R acting along the strings. Now, a sheet of white paper is held just behind the string without touching them. The common knot O and the directions of OA, OB and OD are marked to represent in magnitude, the three forces P, Q and R on any convenient scale (like 50 g = 1 cm).

The experiment is repeated for different values of P, Q and R and the values are tabulated.

To verify parallelogram law

To determine the resultant of two forces P and Q, a parallelogram OACB is completed, taking OA representing P, OB representing Q and the diagonal OC gives the resultant. The length of the diagonal OC and the angle COD are measured and tabulated (Table).

OC is the resultant R′ of P and Q. Since O is at rest, this resultant R′ must be equal to the third force R (equilibrant) which acts in the opposite direction. OC = OD. Also, both OC and OD are acting in the opposite direction. ∠COD must be equal to 180o.

If OC = OD and ∠COD = 180o, one can say that parallelogram law of force is verified experimentally.

Verification of parallelogram

To verify Triangle Law

According to triangle law of forces, the resultant of P (= OA = BC) and Q (OB) is represented in magnitude and direction by OC which is taken in the reverse direction.

Alternatively, to verify the triangle law of forces, the ratios , P/OA, Q/OB and R/O C,  are calculated and are tabulated (Table). It will be found out that, all the three ratios are equal, which proves the triangle law of forces experimentally

                                                                                      

To verify Lami's theorem

To verify Lami's theorem, the angles between the three forces, P, Q and R (i.e) ∠BOD = α, ∠AOD = β and ∠AOB = γ are measured using protractor and tabulated (Table 2.4). The ratios P/sin α, Q sin /β and R/sin γ

are calculated and it is found that all the three ratios are equal and this verifies the Lami's theorem.

                             Verification of Lami's theorem                           

                                                                                                                                                             

Conditions of equilibrium of a rigid body acted upon by a system of concurrent forces in plane

        If an object is in equilibrium under the action of three forces, the resultant of two forces must be equal and opposite to the third force. Thus, the line of action of the third force must pass through the point of intersection of the lines of action of the other two forces. In other words, the system of three coplanar forces in equilibrium, must obey parallelogram law, triangle law of forces and Lami's theorem. This condition ensures the absence of translational motion in the system.

            The algebraic sum of the moments about any point must be equal to zero. Σ M = 0 (i.e) the sum of clockwise moments about any point must be equal to the sum of anticlockwise moments about the same point. This condition ensures, the absence of rotational motion.