Question 4.SP.13: The movable bracket shown may be placed at any height on the...
The movable bracket shown may be placed at any height on the 3-in.-diameter pipe. If the coefficient of static friction between the pipe and bracket is 0.25, determine the minimum distance x at which the load W can be supported. Neglect the weight of the bracket.
STRATEGY: In this variation of the third type of friction problem, you know the coefficient of static friction and that motion is impending. Because the problem involves consideration of resistance to rotation, you should apply both moment equilibrium and force equilibrium.
MODELING:
Free-Body Diagram. Draw the free-body diagram of the bracket (Fig. 1). When W is placed at the minimum distance x from the axis of the pipe, the bracket is just about to slip, and the forces of friction at A and B have reached their maximum values:
\begin{aligned}&F_A=\mu_s N_A=0.25 N_A \\&F_B=\mu_s N_B=0.25 N_B\end{aligned}ANALYSIS:
Equilibrium Equations.
\begin{aligned}\underrightarrow{+}\Sigma F_x=0: \quad N_B-N_A &=0 \\N_B &=N_A \\+\uparrow \Sigma F_y=0: \quad F_A+F_B-W &=0 \\0.25 N_A+0.25 N_B &=W\end{aligned}Because N_B is equal to N_A,
\begin{gathered}0.50 N_A=W \\N_A=2 W \\+\circlearrowleft \Sigma M_B=0: \quad N_A(6 \text { in.})-F_A(3 \text { in.})-W(x-1.5 \text { in.})=0 \\6 N_A-3\left(0.25 N_A\right)-W_x+1.5 W=0 \\6(2 W)-0.75(2 W)-W_x+1.5 W=0\end{gathered}Dividing through by W and solving for x, you have
x = 12 in.
REFLECT and THINK: In a problem like this, you may not figure out how to approach the solution until you draw the free-body diagram and examine what information you are given and what you need to find. In this case, because you are asked to find a distance, the need to evaluate moment equilibrium should be clear.
