Static equilibrium is especially easy for students to they stick to the basics. I encourage my students to start all solutions of the problem in the same way. They told that after identification of all forces, and draw a free body diagram, they should start all solutions of problem with the three declarations:
....Sum (FX) = 0... SUM (Fy) = 0... SUM (T) = 0.
Once they buy in this message, they have no difficulty in static equilibrium problems. Let's look at some examples.
Note: due to the limitations of the text editor, I use a rather unusual rating. This notation is summarized in article Ezine "teaching dynamic rotation.
Problem. The upper limit of a uniform scale weight 300 N is placed against a wall. The other end of the scale is based on the floor. The angle between the scale and the floor is 45?. The floor exerts a force of friction at the bottom of the scale. However, the upper limit of the ladder rests against a slippery (no friction) vertical wall. to determine the forces of the wall and the floor on the scale. (b) what minimum coefficient of friction us is necessary to keep at the bottom of sliding scale?
Analysis. a scale is supported by word (normal force s and f friction force). Upper end of the scale is resting against the smooth wall (normal force p with no friction force). Scale weight is 300 N and acts in its geometric center. We will assume that the scale length is l. We calculate couples axis from the base of the ladder (where it touches the ground). Weapons of the moment of the force P, weight 300 - N force s and friction force f are Lsin45?, Lcos45?/2, 0 and 0, respectively. With the help of a free body diagram, we apply the static equilibrium conditions:
...................Static equilibrium conditions
.Sum (FX) = 0... SUM (Fy) = 0... SUM (T) = 0
... .f - P = 0............ S-300 N = 0.. P (Lsin45?)-(300_N) (Lcos45?/2) = 0
Length l is a common factor in the equation for pair and can therefore be set aside. This leaves an equation for one unknown force P. solving p, we find
.....................................P = 150 N.
Now both strength equations are easily solved for f and S:
.............f = P = 150 N........................S = 300N.
(b) the scale does not slip out as long as the maximum static friction force (UsS) is greater than the actual force of friction (f):
...............................UsS > f.
......................US (300 N) > 150 N,
so.............................We > 0.50.
Problem. The left end of a uniform horizontal beam of mass M = 200 kg is connected by a hinge of the wall and the right end of the beam is connected to a cable of light. The other end of the cable is attached to the wall so oriented to 30? horizontally. The length of the beam is L = 8.0 Mr. man mass m = 80 kg walks along the beam of the hinge outward. If cable breakdown voltage is 2500 N, l distance along the beam, can the man walk before cable snaps?
Analysis. We will investigate the beam. The beam is touch: (i) the hinge (forces Hx Hy), feet (ii) human (mG weight downward), (iii) cable (tension T). (iv) the weight of the beam is MG. With the help of a free body diagram, we apply static equilibrium conditions in the beam. With couples taken around the left from the hinge beam, we have
............................................Static equilibrium conditions
....................................Sum (FX) = 0... (Fy) SUM = 0
................................HX - Tcos30? = 0... Hy + Tsin30? - MG - mG = 0
..........................................................SUM (T) = 0
.............................................-mGl-MGL/2 + Tsin (30?) (L) = 0
Since we want to the point where the cable breaks, we set T = 2500 n. After this insertion with m = 80 kg, M = 200 kg, L = 8.0 m and G = 9,8 m/s? in three equations, find us easily that
..............................HX = 2165 n...Hy = 1494 n... l = 2.8 m.
When mankind reaches a point of 2.8 m of the hinge, the cable breaks.
Each problem of static equilibrium I covers is approached in the same way. I never solve a problem without starting with the three conditions for equilibrium. As each subject covers I do my best to highlight how easily Physics problems are resolved if they are just approached in terms of basic ideas.
Dr. William Moebs is physical retired professor who has taught at both universities: Indiana - Purdue Fort Wayne and Loyola Marymount University. You can see hundreds of examples illustrating how stressed the fundamental principles by visiting physical support.
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