Introduction To Static Equilibrium?

[Briana]

Learning Goal: 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—are in a state described as static equilibrium. Indeed, it is the fundamental objective of many branches of engineering to maintain this state in spite of the presence of obvious forces imposed by gravity and static loads or the more unpredictable forces from wind and earthquakes.

The condition of static equilibrium is equivalent to the statement that the bodies involved have neither linear nor angular acceleration. Hence static mechanical equilibrium (as opposed to thermal or electrical equilibrium) requires that the forces acting on a body simultaneously satisfy two conditions:

\sum \vec{F} = 0 and \sum \vec{\tau}=0;
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 to keep a table static requires not only that it neither slides across the floor nor lifts off from it, but also that it doesn't tilt about either the x or y axis, nor can it rotate about its vertical axis.

http://session.masteringphysics.com/problemAsset/1007460/27/55544.jpg

Part A

Frequently, attention in an equilibrium situation is confined to a plane. An example would be a ladder leaning against a wall, which is in danger of slipping only in the plane perpendicular to the ground and wall. By orienting a Cartesian coordinate system so that the x and y axes are in this plane, choose which of the following sets of quantities must be zero to maintain static equilibrium in this plane.

/sigmaFx and /sigmaT_z and /sigmaF_y

Part B

As an example, consider the case of a board of length L and negligible mass. Take the x axis to be the horizontal axis along the board and the y axis to be the vertical axis perpendicular to the board. A mass of weight W is strapped to the board a distance x from the left-hand end. This is a static equilibrium problem, and a good first step is to write down the equation for the sum of all the forces in the y direction since the only nonzero forces of \sum \vec{F}=0 that exist are in the y direction.

What is \sum {F}_y? Your equation for the net force in the y direction on the board should contain all the forces acting vertically on the board.
Express your answer in terms of the weight W and the tensions in the two vertical ropes at the left and right ends T_L and T_R. Recall that positive forces point upward.

/sumF_y=0=T_L+T_R-W


The only relevant component of the torques is the z component; however, you must choose your pivot point before writing the equations. This point could be anywhere; in fact, the pivot point does not even have to be at a point on the body. You should choose this point to your advantage. Generally, the best place to locate the pivot point is where some unknown force acts; this will eliminate that force from the resulting torque equation

Part C



What is the equation that results from choosing the pivot point to be the point from which the mass hangs (where W acts)?
Express your answer in terms of the unknown quantities T_L and T_R and the known lengths x and L. Recall that counterclockwise torque is positive.

\sumTw_1z=0=-TL_x+(T_r)(L-x)

Part D



What is the equation that results from choosing the pivot point to be the left end of the plank (where T_L acts)?
Express your answer in terms of T_L, T_R, W, and the dimensions L and x. Not all of these variables may show up in the solution.

/sumTL_1z=0=-W(x)+T_rL

Part E

What is the equation that results from choosing the pivot point to be the right end of the plank (where T_R acts)?
Express your answer in terms of T_L, T_R, W, and the dimensions L and x. Not all of these variables may show up in the solution.

/sumR_1z=0=-T_LL+W(L-x)

_______________________________________________

Now these are the parts I don't understand...

Part F

Solve for T_R, the tension in the right rope.
Express your answer in terms of W and the dimensions L and x. Not all of these variables may show up in the solution.

T_R=????

Part G

Solve for T_L, the tension in the left rope.
Express your answer in terms of W and the dimensions L and x. Not all of these variables may show up in the solution.

T_L=?

Thanks!!!!!!!!!!!!

[debydete]

F) To find T_R just use the net torque eq. you wrote in part D;

T_RL - Wx = 0

T_R = W(x/L)

G) To find T_L use the net torque eq. you wrote in part E;

-T_LL + W(L-x) = 0

T_L = W[(L-x)/L]

You can check these by substituting in the Sum of Force = 0 eq you wrote in part B;

T_L + T_R - W = W[(L-x)/L] + W(x/L) - W
= W{1 - x/L + x/L -1}
= 0
It is indeed zero so you can have confidence in the solutions for T_R & T_L.