torque equilibrium and center of gravity lab report

Next we measured the weight of the suspension clip and The data was then recorded on Table 19.įor the mass and Table 19 for the unknown mass. Weighing the meterstick and the mass on the triple beam balance. The experiment was to determine the unknown mass and calculate the torque of a force. Weight and determine the center of gravity of the uniform and non-uniform meterstick. Uniform meterstick, the equilibrium in the horizontal position. We use the Second Condition to observe the vertical forces on the

However for a non-uniform object we can find the center of Meterstick is located at its geometrical center. In other words, the center of gravity of the uniform Of gravity is the geometrical center of the body. We can say that for an object of a uniform mass distribution and a geometrical shape, the center Object is the point through which the line of action of the single force, representing the weight of the object We can say that the center of gravity of the The weight force of the objectĪcts at a single point known as the body’s center of gravity. Torque due to the weight of the object in the rotational equilibrium condition. Now we cannot ignore the gravitational force- total weight, an object undergoes. This condition is referred as the Second Condition for Equilibrium. This equation states that the sum of all torques acting on the object Rotational equilibrium the sum of all torques is zero. From this we derive that net torque isĮqual to the difference between the counterclockwise and clockwise torques so when the object is in Torques must be equal to the sum of all counterclockwise torques. Similar to the forces acting on each other, the sum of all clockwise If the force produces a clockwise rotation. That because torque is positive if the force produces a counterclockwise rotation about the axis and negative Object, we know that an object can only rotate in two directions, clockwise and counterclockwise. Now to further get into the rotation of an Therefore we can say that the lever arm is the perpendicularĭistance from the axis of rotation to the line of action of the force. Line is called the line of action of the force. While a line along which the force acts, is in the direction of the force, this

Of rotation to the force, r, times the magnitude of the force, F This equation states that torque is equal to the distance from the axis Measure of the ability of a force to cause rotation. Now we know that an object will notĪccelerate but we cannot say it will not rotate. We refer to this as the First Condition for Equilibrium. This equation states that the sum of all forces, F, is equal to zero. Therefore the sums of all forces acting on the object must be zero. We start with establishing that when an object is in equilibrium, it must not be accelerating, We willĬonclude with the calculation of the torque of a force. Furthermore, with the unknown mass we are to locate the center of gravity. This rotational equilibrium we are being asked to use the Second Condition for Equilibrium to find an The objective of this experiment is to study objects that are in rotational equilibrium. New York City College of Technology The City University of New York General Physics I 1433 D Experiment 9: Rotational Equilibrium and Center of Gravity Submitted by Date of Experiment: Authors: Cristhian Urgiles NovemDavit Khomasuridze Melany Novas Karla Peña Objective

Torque equilibrium and center of gravity lab report