Question 9.1.4: End C of the frame in Figure 1 rests against a rough surface...

Goal Find the loads at the supports A and C (a) and determine whether or not the frame will slide (b).
Given Information about the geometry of the structure, the loading placed on it, and the value of the coefficient of friction.
Assume The system is planar because the members lie in a single plane and all the loads act in that same plane. The frame is stationary where it rests on the ground at C (therefore, it is not sliding), which assumes that the component of force acting on the frame parallel to the surface at C is less than or equal to the product of the coefficient of friction and the normal force at C. We will check this assumption in (b).
Draw We draw the free-body diagram for the entire frame (Figure 2) and for member AB (Figure 3).
Formulate Equations and Solve (a) For member AB (Figure 3):

\sum{M_{z @ B} }\left(\curvearrowleft + \right) =- (6  ft)F_{Ay}(150  lb)(3  ft)= 0
F_{Ay}=150   lb\left\lgroup\frac{3  ft}{6  ft} \right\rgroup \Rightarrow F_{Ay}=75.0  lb        (1)

For the entire frame (Figure 2): Refer to Figure 4 in Example 9.1.3 as an aid in calculating the moment arms:

\sum{M_{z @ C} }\left(\curvearrowleft + \right) =- F_{Ay}(6  ft + 2  ft)- F_{Ax}(3.46  ft)
+ 150 lb(3 ft + 2 ft) + 10 lb (2 ft) sin 60° = 0

We substitute for F_{Ay}  from (1) to get

−75.0  lb (8  ft) − F_{Ax} (3.46  ft) + 150  lb (5  ft) + 10 lb (1.73  ft) = 0
F_{Ax}=\frac{167.3  lb.ft}{3.46  ft} =48.4  lb          (2)
\sum{F_{x} \left(\rightarrow + \right) } =F_{Ax}- F_{Cx} – 10  lb=0

We substitute for F_{Ax}from (2) to get

F_{Cx}=38.4  lb
\sum{F_{y}\left(\uparrow + \right) } =F_{Ay}+ F_{Cy}-150  lb = 0

We substitute for F_{Ay} from (1) to get

F_{Cy}= 75.0 lb

(b) We confirm that sliding is not predicted at C. According to the Coulomb friction law as described in detail in Chapter 7, we need to confirm: \left|F_{Cx}\right| (friction force) \leq \mu \left|F_{Cy}\right| (max allowable friction force without sliding). In other words, if \left|F_{Cx}\right| \leq \mu \left|F_{Cy}\right| , then the support will not slide at C. Otherwise, the support will slide.

F_{Cx}=(38.4  lb) \lt \mu F_{Cy} = (0.6)(75.0  lb) = 45.0  lb

We conclude that the frame is stationary at C and we have properly modeled the presence of friction. If we had found that \left|F_{Cx}\right| \gt \mu \left|F_{Cy}\right| we would have concluded that the frame slides along the ground and the frame is therefore not in equilibrium (and the analysis we carried out in (a) is not valid).
Check We could check the moment equilibrium with the results above using an alternate moment center. For example, if we choose the moment center to be point D (Figure 4) we find

\sum{M_{z @ D} }\left(\curvearrowleft + \right) =- 75.0  lb(3 ft) – 10  lb\left\lgroup\frac{3.46  ft}{2} \right\rgroup
+75.0 lb (3  ft + 2  ft) − 38.4 lb (3.46 ft) = 0

The left-hand side of the equation sums to −0.16 lb.ft. This is very close to zero, and we can say our answer is correct. (If we were to carry one more significant digit, using F_{Ay}= 38.35 lb, then the left hand side would decrease to 0.009.)