Final answer:
To find the maximum value of force Q keeping the system at rest, the static friction must be calculated; however, with a force P already at 60 N, it seems there might be an inconsistency as this exceeds the calculated maximum static friction force of 58.8 N. If P was within limits, Q would need to be less than or equal to 58.8 N.
Step-by-step explanation:
Maximum Value of Force Q for Static Equilibrium
To determine the approximate maximum value of force Q at which the system remains at rest, we need to analyze the forces acting on block B and apply the concept of static friction. The static friction force will be the force that keeps block B from moving until it reaches a certain threshold. This threshold is determined by the static friction coefficient (μ_s = 0.40) and the weight of block B (15 kg × 9.8 m/s²).
First, we calculate the maximum static friction force (f_s_max) for block B:
- f_s_max = μ_s × Normal force on B
- f_s_max = 0.40 × (15 kg × 9.8 m/s²)
- f_s_max = 0.40 × 147 N = 58.8 N
Therefore, the maximum force Q before block B moves is the force at which the static friction is at its maximum, which is 58.8 N. Since we already have a force P of 60 N acting on block B horizontally, it exceeds the static friction limit, indicating that the details provided in the question might contain some inconsistencies.
However, if the details were consistent and P did not exceed f_s_max, the force Q would have to be less than or equal to the maximum static friction force so that P + Q ≤ f_s_max for the system to remain at rest.