components of the forces ? Because this is a statics problem, these forces will sum to zero.Use the sign convention indicated in the figure, and express your answer in terms of , , , , , , and/or . Note that not all of these quantities will appear in your answer.ANSWER:Hint 4. Forces: ycomponentsAssuming that the tensions in the left and right wires are and , respectively, what is the sum of the ycomponents of the forces ? Because this is a statics problem, these forces will sum to zero.Use the sign convention indicated in the figure, and express your answer in terms of , , , , , , and/or . Note that not all of these quantities will appear in your answer.ANSWER:Hint 5. Eliminate weight from your equationsYou should have found three equations by now. It is possible to eliminate two variables and solve for in termsof the others. As an intermediate step, solve your torque equation for in terms of , , , etc. and thensolve your y-component force equation for and substitute back into your expression for . In other words, findan expression for .Answer in terms of , , , , and .ANSWER:Hint 6. A useful trig identityThe dimensions for the expression you just found for are correct, since the units of the tensions cancel out,leaving the units of length in the numerator. If you now solve the x-component force equation for in terms of , and substitute into your equation for , you should find the following trig identity useful:.
4/24/2014Chapter 11 and 13 Homeworkhttp://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=28541271/31
Chapter 11 and 13 Homework
Due: 10:00pm on Wednesday, April 23, 2014
You will receive no credit for items you complete after the assignment is due. Grading Policy
A 0.100-, 40.2--long uniform bar has a small 0.080- mass glued to its left end and a small 0.150- mass gluedto the other end. You want to balance this system horizontally on a fulcrum placed just under its center of gravity.
How far from the left end should the fulcrum be placed?ANSWER:
Introduction to Static Equilibrium
To understand the conditions necessary for static equilibrium.Look around you, and you see a world at rest. The monitor, desk, and chair—and the building that contains them—arein a state described as
. Indeed, it is the fundamental objective of many branches of engineering tomaintain this state in spite of the presence of obvious forces imposed by gravity and static loads or the moreunpredictable forces from wind and earthquakes.The condition of static equilibrium is equivalent to the statement that the bodies involved have neither linear nor angularacceleration. Hence static mechanical equilibrium (as opposed to thermal or electrical equilibrium) requires that theforces acting on a body simultaneously satisfy two conditions:and ;that is, both external forces and torques sum to zero. You have the freedom to choose any point as the origin about which to take torques.Each of these equations is a vector equation, so each represents three independent equations for a total of six. Thus tokeep a table static requires not only that it neither slides across the floor nor lifts off from it, but also that it doesn't tiltabout either the
axis, nor can it rotate about its vertical axis.
Frequently, attention in an equilibrium situation is confined to a plane. An example would be a ladder leaningagainst a wall, which is in danger of slipping only in the plane perpendicular to the ground and wall. By orienting aCartesian coordinate system so that the
axes are in this plane, choose which of the following sets ofquantities must be zero to maintain static equilibrium in this plane.
Simplifying the equations