Experiment 6
Rigid Body Equilibrium This experiment is designed to introduce the student to the concept of torque. Additional definition of such quantities such as center of mass, equ ilibrium, static equilibrium, rigid body and movement arm will also be introduced. Theory
A particle (a mathematical point) is in equilibrium if the net force acting on it is zero. According to ewton!s second law ("#ma) the acceleration of the particle will be zero if the net force is zero. $t should be noted, however, h owever, that zero acceleration only means that the velocity of the particle is constant. $f we impose that the velocity is also zero, the particle is said to be in static equilibrium. %otational motion need not be considered con sidered for a point particle. $n the real world, however, bodies of appreciable extent are the rule and the conditions cond itions of equilibrium must be reconsidered. "or example, it is possible that the sum of the force vectors acting on a body may be zero and still produce a rotational acceleration. $n figure & we see two forces, which are equal, and opposite but still cause a rotational acceleration because they do not act along the same line. 'uch a force pair is called a couple. onsider the following case the same force is applied at varying points on a door ("igure *). The time required for the door to swing through some specified angle is different for all four points of application of the force ". $n fact, the speed of the door after swinging through the specified angle is directly proportional to the distance from the hinge to the point at which the force is exerted, with the exception of the force at position four. The force being applied at p osition four produces no rotation of the door. +e +e have observed that the resulting rotation is dependent up on the force that is being applied, ap plied, the point at which the force is applied, ap plied, and the direction of the force relative to the axis of rotation. 'ince the points on a body at which the forces are exerted are important, it is important that the body is rigid. y rigid, we mean that the atoms and molecules ma-ing up the body do not change their positions relative to each other. $nternal forces between these particles hold the body together and are strong enough to do this no matter what external forces are put on the body. The The body is rigid and unbro-en and the internal forces will be neglected in the remainder of this discussion.
The tendency of a force to produce rotation is called torque (τ) or the moment of force. The torque is equal to the magnitude of o f the applied force (") multiplied by the distance d istance (r) between the point at which the force is applied and the axis of rotation, times the sine of the angle between F and r . $n this case, it is surely true that a picture is worth a thousand words. athematically, this sin(θ ). cumbersome statement simply means that for the situation depicted in "igure / , τ = rF sin
F ⊥
F
θ
F
r
r
axis of rotation
r
⊥
otice that sin (θ) may be associated with either " or r (the choice being one of convenience to you).
τ = rF sisin(θ ) = r ⊥ F = rF ⊥ where
r ⊥ = r sinsin(θ ) and F ⊥ = F sinsin(θ ) The distance r is referred to as the lever arm or o r moment arm of the force ". ⊥ Torque Torque is a vector and, thus, there is a direction associated with it. The direction of the torque is perpendicular to the plane formed by the line of action of the force and the moment arm. "or our purposes here, it will be sufficient sufficient to say that torques tending to produce a countercloc-wise rotation are positive and torques tending to produce a cloc-wise rotation are negative. An important example occurs when the net torque on a rigid body about some axis is due to the force of gravity acting on the body. bo dy. The The force of gravity acts on every atom of the body on both sides of the axis of revolution. "or an axis at point & in "igure 0 most of the mass is on the right side of o f the axis and the body will rotate in a cloc-wise manner. An An axis through point * will produce a similar result but the difference in the mass on the left and right sides of the axis is not as great as was the case
of axis point &. An axis at point 1 will have a large torque producing a countercloc-wise rotation. As the axis point moves toward the left the net torque is reduced. 'ince the left axes (&,*) produce a cloc-wise rotation and the right axes (0,1) produce a countercloc-wise rotation, there must be some point / about which there is no torque on the body due to the force of gravity. This balance point is called the center of mass or sometimes, the center of gravity. The The center of mass is that point at which the entire mass of the body seems to be concentrated.
learly then, the equilibrium conditions necessary for an e xtensible body are that the sum of the forces acting on the body must be zero and that the sum of the torques acting on the body about any axis must be zero. The words 2any axis3 in the above statement is very important because in many problems there will be an un-nown force acting on the force exerted by an un-nown weight (e.g. the force exerted by a fulcrum). y 4udiciously choosing an axis to compute the torques acting on the body, you you can eliminate one of the un-nowns. 5xample A meter stic- with a fulcrum at the 16cm mar- has a mass of &66 grams hanging at the *6cm mar-. (ote The distance from the fulcrum to the &66 gram mass is 16cm minus *6cm which is /6cm.) +here must a *66 gram mass be positioned to balance the system7 +e +e will use grams as a force instead of dynes or o r ewtons. 8ne gram force is the force of g ravity acting on a one gram mass. The weight of the meter stic- and the force the fulcrum exerts on the meter stic- are un-nown. 9owever, if we compute the torque about an axis through the 16cm mar-, these forces will exert no torque about that axis since the moment arm is zero. τ counter − clockwise
=
( 30cm )(100 grams)
τ clockwise τ counter − clockwise
x
+
=
τ clockwise
3000 200
=
=
cm
= 3000
gram c
(− x )(200 grams)
gram cm 3000 gram =
−
( x )(200
mark 15 cm ( the 65 cm mar
Procedure :ou will need a Apparatus The equipment necessary for this experiment requires no explanation. :ou laboratory balance, a set of slotted masses, a meter stic-, a balancing stand, / meter stic- clamps and * slotted mass hangers. earlier, we will use the gram force as our unit un it of force rather than Experiment As was stated earlier, multiplying each mass by the acceleration due to gravity.
a) ;etermine ;etermine the mass mass of the meter meter sticstic- and the various various clamps. clamps. e sure sure to marmar- each clamp with a pencil in order to -eep trac- of them. %ecord the measurements on the data sheet.
b) %emove the hanger from the support clamp and put the support clamp on the meter stic- and place the meter stic- on the balancing stand. a-e certain the loc-ing screw on the support clamp is under the meter stic-. "ind and record the position at which the meter stic- is balanced. This is the location of the center of mass of the meter stic-. c)
(τ clockwise − τ counter − ) % difference =
clockwise
(τ clockwise + τ counter − )
× 100%
clockwise e)
