Civil Engineering:
Analysis of Static Friction; Geometrical Properties of Areas
The bar in Fig.2a weighs 100 lb and is acted on by a force P that makes an angle
of 55 degrees with the horizontal. The coefficient of friction between the bar
and the inclined plane is 0.20. Compute the minimum value of P required (a) to
prevent the bar from sliding down the plane; (b) to cause the bar to move upward
along the plane.
Fig.2 Body on inclined plane:
Calculation Procedure:
1. Select coordinate axes
Establish coordinate axes x and y throught the center of the bar, parallel and
perpendicutlar to the plane, respectively.
Fig. 2
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2. Draw a free body diagram of the system
In Fig. 2b, draw a free-body diagram of the bar. The bar is acted on by its
weight W, the forece P, and the reaction R of the plane on the bar. Show R
resolved into its x and y components, the former being directed upward.
3. Resolve the foreces into their components
The foreces W and P are the important ones in this step, and they must be
resolved into their x and y components. Thus;
Wsubx = -100 sin 40 degrees = -64.3 lb Wsuby = -100 cos 40 degrees = -76.6 lb
Px = P cos 15 degrees = 0.966P Py = P sin 15 degrees = 0.259P
4. Apply the equations of equilibrium
CConsider that the bar remains at rest and apply the equations of equilibrium.
Thus:
Sum Fsubx = Rsubx + 0.966P -64.3 = 0 Rsubx = 64.3 - 0.966P
Sum Fsuby = Rsuby + 0.259P -76.6 = 0 Rsuby = 76.6 - 0.259P
5. Assume maximum friction exists and solve for the applied force
Assume that Rsubx, which represents the frictional resistance ot motion, has
its maximum potential value. Apply Rsubx = uRsuby, where u = coefficient of
friction. Then, Rsubx = 0.20Rsuby = 0.20(76.6-0.259P) = 15.32 - 0.052P.
Substituting for Rsubx from Step 4, 64.3 - 0.966P = 15.32 - 0.052P; P = 53.6 lb.
6. Draw a second free-body diagram
In Fig.2c, draw a free-body diagram of the bar, Rsubx being directed downward.
7. solve as in steps 1 through 5
As before, Rsuby = 76.6 - .259P. Also, the absolute value of Rsubx = 0.966P- 64.3.
But Rsubx = 0.20Rsuby = 15.32 * 0.052P. Then, 0.966P - 64.3 = 15.32 - 0.052P;
P = 78.2 lb.
