# Equilibrium. To determine the mass of unknown objects by utilizing the known force requirements of an equilibrium

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1 Equilibrium Object To determine the mass of unknown objects by utilizing the known force requirements of an equilibrium situation. 2 Apparatus orce table, masses, mass pans, metal loop, pulleys, strings, two unknown masses, balance, force probe, computer with software. Theory There are two conditions on an object that must be met for it to be in equilibrium. They are: = 0 and τ = 0 () Since = m a and τ = I α, this implies that an object in equilibrium has neither linear nor angular acceleration, and thus constant linear and angular velocities. If these velocities are zero, they will remain so for as long as the net force and net torque remain zero. This is called static equilibrium. If either the linear velocity or angular velocity is non-zero yet the net force and torque are zero, the situation is referred to as dynamic equilibrium.. Part If only a point mass is considered, it will have no dimensions and thus can have no torques associated with it. The condition for equilibrium then reduces to = 0 (2) In general, since, the force, is a vector, it will have both x and y components given by x = cos(θ) and y = sin(θ) () where θ is the angle the force vector makes with the positive x axis. In this particular experiment the forces are produced by a string connected horizontally to a metal ring and then extended over a pulley and connected to a mass at the other end. The mass hanging from the end of the string will also be in equilibrium so equation 2 must also hold for it as well. There are only two forces acting on this mass, the tension force due to the connected string and the gravitational force down on the mass, W = mg. If the net force is zero, these two forces must be in opposite directions and have equal magnitudes. As a result, the resulting tension in the string connected to the hanging mass must be equal to the gravitational force on that mass T = W = mg (4) Assuming negligible friction in the pulley, the tension in the horizontal section of the string will be the same as the tension in the vertical section so the tension in the string connected to the ring will be equal in magnitude to mg.

2 In most cases there is more than one force acting on a given object (can an object with only one force acting on it be in equilibrium?), and the condition for equilibrium, Equation 2, can be written out as = = 0 (5) Each of these vectors may be written in component form using equation. Equation 5 can then be written in component form: x + 2x + x +... = 0 and y + 2y + y +... = 0 (6) where the component of every force acting on the object should be included. In this lab, we will y θ θ x θ 2 2 be considering the horizontal forces acting on the metal ring. Each string exerts a tension force on the metal ring, which we know is equal to the gravitational force on the mass suspended from that string. Combining this knowledge with equation 6 and defining the positive x axis as θ = 0 (see the figure above), we see that Tx = m g cos(θ ) + m 2 g cos(θ 2 ) + m g cos(θ ) +... = 0 Ty = m g sin(θ ) + m 2 g sin(θ 2 ) + m g sin(θ ) +... = 0 (7) Tx and Ty represent the component of the net force in the x and y directions, respectively. If the angle for each force is known and all the masses are known except for one, either of the relationships from equation 7 may be solved for the value of the unknown mass..2 Part 2 What if the object under consideration is not a point mass? In that case, there may be nonzero torques exerted on the object due to the different individual forces acting on it. However, if the object is in equilibrium, the conditions of equation must still hold true, and the different torques, when added together as vectors, must equal zero. Let us consider a long board of mass m and length L, supported, parallel to the ground, at one end by a hinge and at some distance x away from that hinge from underneath by some vertical force. This situation is pictured below in the figure. In accordance with equation, the net force acting on the board must be zero if the board is in equilibrium. The reaction forces on the board by the hinge in the vertical and horizontal directions may be whatever magnitude they must be to insure the net force on the board is zero, 2

3 R (reaction force) D x Axis mg as long as the structural integrity of the hinge isn t reached. However, the net torque on the board (about any and every point) must also be zero. Let us consider the axis about which the board is hinged; note, this may not be right at the end of the board. Since the reaction forces on the board due to the hinge act on a line passing directly through this axis, these forces can exert no torque on the board about the hinge point. This leaves two forces which may exert torques on the board, the weight force on the board due to the earth, and the vertical force up on the board from underneath. In accordance with equation, the torques due to these forces must be of equal magnitudes and in opposite directions so that when adding these torques together we will have a net torque of zero and the board will be in equilibrium. We can find the torque on an object about some point by an external force acting on the object a distance r away from that point using the following relationship, τ = r τ = τ = r sin(ϕ) (8) where r is the vector extending from the chosen point to where the force is exerted on the object, is the force under consideration, and ϕ is the angle between these two vectors. Notice if ϕ is 0 or 80 degrees then the torque due to the force is zero. The right hand rule for the vector cross product of r into will give the direction of the torque due to the force. You may do this by pointing the fingers of your right hand in the direction of r, curling them towards the direction of, and then your thumb will point in the direction of the torque due to the force. Inspection of the picture, or use of the right hand rule, should convince you that the weight force on the board and the vertical force from underneath ( ) are indeed trying to rotate the board about the hinge point in opposite directions. Setting the magnitudes of these two forces equal and using equation 8 yields x sin(90 ) = mgd sin(90 ) or x = mgd (9) where g is the magnitude of the gravitational field and D is the distance from the hinge axis tot he center of mass of the board. D would remain constant unless the hinge is changed. 4 Procedure 4. Part. orce Table. Set up the force table as described below. or part one you will need three pulleys and a string connected to the center metal loop and passing over each pulley. Connect mass hangers to two of the strings and your unknown mass to the third string. The unknown mass should roughly be larger than 00g but less than 00g. Place the string connected to the unknown mass at the angle specified for trial. Add mass to one hanger until the total mass, including the 50 hanger, is the specified amount and place that pulley at the listed angle. Now, vary the attached mass and angular location of the second string until the metal loop is in a state of equilibrium. Equilibrium can be recognized when the metal loop is placed centered around the center rod in the table and it remains there even when you lightly tap on the force table

6 7 Equilibrium Data Sheet Part - Equilibrium with three pulleys Trial Mass Theta Unknown θ Mass 2 Theta 2 δm2 (g) δθ2 (deg) Units Unknown mass measured using the triple beam balance Measured Unknown Mass (g) 6

7 Equilibrium Data Sheet cont d Part 2 - orce Probe Trial x Units Measured Board Length Measured Board Mass 2 7

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