For a block on an incline to remain at rest considering static friction μ_s, which inequality must hold?

For a block on an incline to remain at rest considering static friction μ_s, which inequality must hold?

• A. tan θ ≤ μ_s
• B. tan θ > μ_s
• C. sin θ ≤ μ_s
• D. θ ≤ 90 degrees

Answer: (A)

Static friction can adjust up to a maximum value to prevent motion. On an incline, gravity pulls the block down the slope with a component m g sin θ, while the normal force is N = m g cos θ. The friction force that prevents slipping acts up the slope and can supply any value up to μ_s N. For the block to stay at rest, the friction needed to balance the downslope pull must be less than or equal to the maximum friction: m g sin θ ≤ μ_s m g cos θ. Assuming the angle is below 90 degrees so cos θ is positive, you can divide by m g cos θ to get tan θ ≤ μ_s. If the angle is large enough that tan θ exceeds μ_s, the required friction would exceed what static friction can provide, and the block will start to slide. The angle where sliding starts is when tan θ = μ_s, the angle of repose.

Static friction can adjust up to a maximum value to prevent motion. On an incline, gravity pulls the block down the slope with a component m g sin θ, while the normal force is N = m g cos θ. The friction force that prevents slipping acts up the slope and can supply any value up to μ_s N.

For the block to stay at rest, the friction needed to balance the downslope pull must be less than or equal to the maximum friction: m g sin θ ≤ μ_s m g cos θ. Assuming the angle is below 90 degrees so cos θ is positive, you can divide by m g cos θ to get tan θ ≤ μ_s.

If the angle is large enough that tan θ exceeds μ_s, the required friction would exceed what static friction can provide, and the block will start to slide. The angle where sliding starts is when tan θ = μ_s, the angle of repose.