Connected masses — effect of friction on string tension: A 20 kg block on a rough horizontal plane is connected by a light string over a smooth pulley to a 5 kg hanging mass (free to move vertically). As the coefficient of friction between the block and plane increases, how does the string tension change?

Introduction / Context:
This classic two-body system illustrates how friction alters internal force (tension) and acceleration. Understanding the direction and magnitude trends is crucial for conveyor, hoist, and brake problems in mechanical design.

Given Data / Assumptions:

• Block on horizontal plane: m1 = 20 kg; hanging mass: m2 = 5 kg.
• Smooth pulley, light inextensible string.
• Kinetic/limiting friction modeled as F = μ * N = μ * m1 * g.
• Hanging mass tends to move downward; block tends to move toward the pulley.

Concept / Approach:
Apply Newton’s second law to each mass and eliminate acceleration to express tension T as a function of μ. Then inspect how T varies with μ. The algebra reveals monotonic behavior here.

Step-by-Step Solution:
For m2 (down positive): m2g − T = m2a.For m1 (toward pulley positive): T − μm1g = m1a.Eliminate a: add equations ⇒ m2g − μm1g = (m1 + m2)a ⇒ a = (m2g − μm1g)/(m1 + m2).Back-substitute into T = m1a + μm1g ⇒ T = m1m2g(1 + μ)/(m1 + m2).Hence, dT/dμ > 0 ⇒ tension increases with μ.

Verification / Alternative check:
At μ = 0 (frictionless plane), T = m1m2g/(m1 + m2). As μ grows, friction resists motion of m1 more, so the string must carry more load, raising T.

Why Other Options Are Wrong:

• decrease: Opposite of derived trend.
• not be affected: Contradicted by the explicit dependence T ∝ (1 + μ).
• indeterminate: The model yields a clear analytic dependence on μ.

Common Pitfalls:

• Assuming friction always reduces internal forces; here it reduces acceleration but increases tension.
• Sign mistakes when writing the equations of motion for each mass.

Final Answer:
increase