What role does the mass play in balancing the see-saw? (Hint: Would you need to balance the rigid body in a “weightless” environment?) EQUILIBRIUM OF A RIGID BODYObjective:In this experiment a balance is used as a rigid body to illustrate the application of the equationsof rotational equilibrium.The torque equation will be verified for a balanced system of two or more masses.The torque equation will also be applied to determine the magnitude of an unknown mass.Introduction:If a rigid body is in equilibrium, then the vector sum of the external forces acting on the bodyyields a zero resultant and the sum of the torques of the external forces about any arbitrary axis isalso equal to zero.Net Force = 0ΣF0Eq. 1Sum of forces on the left = Sum of forces on the rightSum of forces upwards = Sum of forces downwardsNet Torque = 0Στ0Еq. 2Sum of clockwise torques= Sum of counter-clockwise torquesIn this experiment force is defined as the mass, m times gravitational acceleration, g:F = mgEq. 3where m is in kg and g in m/s² therefore, force F is in newtons N.Torque is a measure of how effective a given force is at twisting or turning the object it isapplied to. Torque is defined as the force F times the moment arm or lever arm r of the forcewith respect to a selected pivot point x. In other words r is the distance from the pivot to thepoint where a force is applied. If the force is perpendicular to r, then:T = FrEq. 4 The unit of torque is Newton-meter, N-m. The sign for torque is defined as positive (+) whenrotating in counter-clockwise direction and negative (-) when rotating in a clockwise direction.For this experiment it is not only important to familiarize yourself with the equations, but alsowith sketching free-body diagrams (FBD). Using FBDS help solve the problems in questioneasier and faster.Example # 1: Below is an example where the torque equation is applied to prove the system isin equilibrium:Fulcrum2 1.75 1.5 1.25 1 0.75 0.5 0,25Meters0.25 0.5 0.75 1 1.25 1.5 1.75 2Metersm,*gM*gm,*gFig. 1. System in equilibrium and corresponding FBDSince Et= 0 and t = F:r we can develop the equation of equilibrium based on the FBD takinginto account the signs of the torques. r is the distance from the fulcrum or pivot point to thepoint where the force is applied. In the case above we have F1 applied at a moment arm rị thatwill cause a counterclockwise rotation with respect to the pivot point, therefore, the torque ispositive. F2 is acting at a moment arm r2 that will cause a clockwise rotation with respect to thepivot point, thus, making the torque negative.Therefore,Et = T, – T, = 0T1 = T2Rewriting in terms of F and r:(F,)(ri) = (F2)(r2)where F=mg therefore,(m1 9)(ri) = (m2g)(r2)Еq. 5In the system we have: m1= 10 kg, rı= 1.5 m; m2= 15 kg, r2= 1.0 m.Substituting the given values along with g=9.8 m/s² into Eq. 5:(10 kg * 9.8 m/s²)(1.5 m) = (15 kg * 9.8 m/s²)(1 m)147 N :т %3 147 N :тTherefore, the net torque:Et = 147 N ·m – 147 N ·m = 0The system is in equilibrium.

The condition for equillibrium according to newtonian mechanics is,Net force on the object must be zero, ∑F = 0 NNet torque on the object about any point on the object must be equal to zero, ∑T = 0 NmNow in balancing the see-saw, mass can get the see-saw out of balanceor even fet it into balanced state(equillibrium)But it is not mass alone that casn do this, there must bea non-zero acceleration due to gravity acting on the mass for, the mass to show its effect on see-saw( or any mechanical system so to speak).In an weightless environment, that is where gravity is neglegiblemass cannot exhert force on the see-saw.Therefore we can say WEIGHT is important quantity which can make or break the balance of the see-saw and not the mass.