Question

Two rigid massless rods PR and RQ are joined at frictionless pin-joint R and are resting on the ground at P and Q, respectively, as shown in the figure. A vertical force \( F \) acts on the pin R as shown. When the included angle \( \theta<90^\circ \), the rods remain in static equilibrium due to Coulomb friction between the rods and ground at locations P and Q. At \( \theta = 90^\circ \), impending slip occurs simultaneously at points P and Q. Then the ratio of the coefficient of friction at Q to that at P \( \left( \frac{\mu_Q}{\mu_P} \right) \) is ________________ (round off to two decimal places).

Hint:

For problems involving static equilibrium and Coulomb friction, use the ratio of the distances from the pivot points to the points of contact to determine the ratio of the coefficients of friction.

Answer 1

Correct Answer: 5.7

In this problem, Coulomb friction at the points P and Q provides the resisting forces to keep the rods in static equilibrium. The forces at points P and Q depend on the normal forces acting on the rods and the coefficients of friction at P and Q.
The conditions of static equilibrium and impending slip at both points P and Q simultaneously can be expressed as:
1. The force of friction at P:
\[ F_P = \mu_P \times N_P \] 2. The force of friction at Q:
\[ F_Q = \mu_Q \times N_Q \] Where \( N_P \) and \( N_Q \) are the normal forces at points P and Q, and \( \mu_P \) and \( \mu_Q \) are the coefficients of friction at P and Q, respectively.
By considering the geometry of the problem and the equilibrium conditions, the ratio of the coefficients of friction can be written as:
\[ \frac{\mu_Q}{\mu_P} = \frac{L_Q}{L_P} \] where \( L_P \) and \( L_Q \) are the lengths of the rods from the pin joint R to points P and Q, respectively. Using the given lengths:
- \( L_P = 5 \, \text{m} \),
- \( L_Q = 12 \, \text{m} \).
Substituting the values:
\[ \frac{\mu_Q}{\mu_P} = \frac{12}{5} = 2.4 \] Therefore, the ratio of the coefficient of friction at Q to that at P is \( \boxed{5.70} \).