RBD O11

Solution Q11

Solution: Question 11

Diagram: Stick-Slip Dynamics

Bar Center C P (closer) v Q (farther) v N1 (Large) N2 (Small)
Step 1: Normal Force Distribution

The normal force exerted by a support is inversely proportional to its distance from the center of mass (C) to satisfy torque equilibrium (\( N_1 x_1 = N_2 x_2 \)).
The wedge closer to the center bears more weight (\( N \) is higher).

Step 2: Friction Threshold

The maximum static friction is \( f_{max} = \mu_s N \). Since \( N \) is higher for the closer wedge, the bar can sustain a larger horizontal force from the closer wedge before slipping.

Step 3: The “Stick-Slip” Mechanism

The bar will “stick” to the wedge with the stronger grip (the closer one) and “slip” on the farther one. As the closer wedge moves inwards, it carries the bar with it. Eventually, the center C passes the midpoint, making the other wedge closer. The sticking roles reverse.

Result (d): At any time, the wedge farther from C slides (slips) relative to the bar.

Result (a): This self-correcting process ensures the wedges converge exactly beneath the center C.

Conclusion: Correct Options are (a) and (d).