Final Answer:
(D) F = Pr/(2R), Q = Pr/(2R).
Explanation: The equilibrium condition for the disk on a rough surface indicates that the maximum horizontal forces, F and Q, are both equal to Pr/(2R) to balance the frictional force and maintain stability.Thus the correct option is:(D) F = Pr/(2R), Q = Pr/(2R)
Step-by-step explanation:
The correct answer is (D) F = Pr/(2R), Q = Pr/(2R). When a disk is in equilibrium on a rough horizontal surface, the maximum horizontal force before sliding occurs is given by the frictional force. The frictional force (F_friction) is equal to the coefficient of friction (μ) multiplied by the normal force (N), where N is the weight of the disk. Since the disk is in balance, the weight is distributed between points A and C, and the vertical distance between them is 'r'. Therefore, the normal force at each point is P/2.
The formula for frictional force is F_friction = μ * N. In this case, it's (0.5) * (P/2) = Pr/(4). To maintain equilibrium, the horizontal forces F and Q must balance the frictional force. So, F + Q = Pr/(4). Since the forces are applied at different distances from the center (A and C), their torque contributions need to be considered. The maximum torque occurs when the forces are applied at the edges of the disk, so the torque is F * R and Q * R, respectively. The condition for equilibrium is F * R + Q * R = 0.5 * Pr.
Solving these equations simultaneously, we find that F = Pr/(2R) and Q = Pr/(2R). Therefore, the correct answer is (D) F = Pr/(2R), Q = Pr/(2R).Thus the correct option is:(D) F = Pr/(2R), Q = Pr/(2R)