Equilibrium of a rigid body lab report answers

There are three documents. One named PHYS 2425 Exp 9 Manual - Virtual (2)(1) (1).pdf which shows the steps and includes the questions. Other named Virtual Equilibrium - Set 1 (Locked) (1)(1) 2.pdf. It has the numbers and last one Standard Sheet 2.pdf you need to completeEXPERIMENT 9 Torque, Equilibrium & Center of Gravity ! Produced by the Physics Staff at Collin College Copyright © Collin College Physics Department. All Rights Reserved. University Physics, Exp 9: Torque, Equilibrium & Center of Gravity Page ! 1 Purpose In this experiment, you will investigate torques on rigid bodies and static equilibrium. Equipment • • • • • • 1 Lab Balance 1 Meter stick 1 Balance Stand for meter stick 1 Set of Mass Hangers for meter stick 1 Hooked Mass Set 1 Small, Unknown Metal Mass Introduction Consider, for example, an ordinary chair. It appears to be a rigid body. Is it in equilibrium as it sits on the floor? If, as often happens, one of its legs is a bit short, it will not be in equilibrium; instead, it will wobble about the two diagonally-opposite longer legs. But if an elephant sits on the chair, it will most definitely be in equilibrium with all four legs resting solidly on the floor. What has changed? Only the chair’s shape. The elephant has distorted it. It is not a rigid body under these conditions. In this experiment, you will investigate truly rigid bodies and static equilibrium. The static equilibrium condition is very important in civil engineering, applying to bridges, dams, buildings, statues, and balconies, and in our daily lives, applying to our ability to stand up, to drive around corners without overturning, and to slide a stein of beer the length of the bar without spilling it. Analyzing static equilibrium conditions is an essential part of architectural engineering. The designer needs to identify all the forces and torques that act on a structural element, and to ensure through design and materials selection that the element will safely tolerate the loads to be exerted on it. Two conditions must be met for a rigid object subject to a combination of external forces to be in mechanical equilibrium: 1. The vector sum of all the external forces acting on it must be zero. Translational → equilibrium: ∑ ! F = 0 2. The vector sum of all the torques about an arbitrary axis must be zero. Rotational τ = 0. To be in static equilibrium, a rigid object must also be in equilibrium: ∑ !→ rotational equilibrium. University Physics, Exp 9: Torque, Equilibrium & Center of Gravity Page ! 2 The concept of center of mass is what allows us to study the motion and equilibrium of extended (real world) objects as if they were point objects. By considering the translational motion of an object’s center of mass (the motion of a point mass), and the rotational motion of the object about its center of mass, we can determine the complex motion of any extended object. In this experiment, you will examine torques, rotational equilibrium, and center of mass as they apply to a rigid object. The rigid object will be an ordinary wooden meter stick. By measuring the forces and calculating the torques acting on this meter stick in different situations, you will experimentally verify the two equilibrium equations. In doing this, you will learn to 1. Describe mechanical equilibrium of a rigid object 2. Explain the center of mass concept 3. Explain how a laboratory balance measures mass Theory Equilibrium A rigid body in static equilibrium must necessarily be in rotational equilibrium. Torque about some axis of rotation (also called moment of force) results from a force being exerted at a point not on the axis. Torque is defined as the vector product of the force and the displacement to the axis: → → r ×F !τ =→ Therefore, the magnitude of the torque is τ! = rFsinθ The measurement unit for torque is the newton-meter (Nm), which you should not confuse with the unit of work (1 newton-meter = 1 joule). Torque is a vector quantity. Its direction is normal to the plane containing r and F. When you cross r into F using the right-hand rule, your thumb points in the direction of the torque. For convenience, torques are designated by the circular directions of motion that they tend to cause (clockwise and counter-clockwise) A rigid body can rotate about a specific axis in only two directions, clockwise or counterclockwise. Clockwise torques produce clockwise rotational motion and counterclockwise torques cause counterclockwise rotational motion. Rotation will not begin or change if the applied torques are balanced (if the system is in rotational equilibrium). The condition for rotational equilibrium is University Physics, Exp 9: Torque, Equilibrium & Center of Gravity Page ! 3 ∑ !→ ! τ ccw + ∑ → ! τ cw = 0 τ =∑ → τ ccw and !→ τ cw are the counterclockwise and clockwise torques. Conventionally, the where !→ counterclockwise direction is positive, and the clockwise direction is negative. ! For the system in Figure 10.1, the condition for rotational equilibrium becomes: ! F1 r1 = F2 r2 + F3 r3 In this case the counterclockwise torque is F1r1 and the clockwise torques are F2r2 and F3r3.