FIGURE 2. Equilibrant of force system.ANALYSIS OF STATIC FRICTION
The bar in Fig. 2a weighs 100 Ib (444.8 N) and is acted on by a force P that makes an an-
gle of 55° 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 slid-
ing down the plane; (b) to cause the bar to move upward along the plane.
Calculation Procedure:
- Select coordinate axes
Establish coordinate axes x and y through the center of the bar, parallel and perpendicular
to the plane, respectively. - 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
force P, and the reaction R of the plane on the bar. Show R resolved into its jc andy com-
ponents, the former being directed upward. - Resolve the forces into their components
The forces W and P are the important ones in this step, and they must he resolved into
their x and y components. Thus
Wx = -100 sin 40° = -64.3 Ib (-286.0 N)
Wy = -100 cos 40° = -76.6 Ib (-340.7 N)
Px = P cos 15°- 0.966P
P^ = Psml5° = 0.259P- Apply the equations of equilibrium
Consider that the bar remains at rest and apply the equations of equilibrium. Thus
ZFx = RX + 0.966P - 64.3 = 0 Rx = 64.3 - 0.966P
^Fy = Ry + 0.259P - 76.6 = 0 Ry = 76.6 - 0.259P- Assume maximum friction exists and solve for the applied force
Assume that Rx, which represents the frictional resistance to motion, has its maximum po-
tential value. Apply Rx = ^Ry, where JJL = coefficient of friction. Then Rx = Q.2QRy =
Bar