Crate on a Ramp with Friction
A 20 kg crate sits on a ramp inclined at 30° above the horizontal. A student calculates the component of gravity pulling the crate down the ramp as about 98 N and the normal force as about 170 N. The student then measures that the crate stays at rest when released. The student concludes: 'Since gravity pulls it down the ramp with 98 N and it doesn't move, friction must also be 98 N, and because friction equals the normal force times the coefficient, the coefficient of friction must be 98 ÷ 170 ≈ 0.58.' The teacher says the friction conclusion is right in magnitude but the reasoning about the coefficient is flawed, and asks the group to explain when the crate would actually start to slide.
1. Is the student's value of friction (98 N) correct for the crate at rest? Explain using Newton's first law.
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Yes. The crate is at rest, so the net force along the ramp must be zero. The down-ramp gravity component is about 98 N, so static friction must be about 98 N up the ramp to balance it. The student's number is right.
The error is in the next step: 98 N is the actual static friction needed to hold the crate, not the maximum static friction available. Static friction adjusts to whatever is needed, up to a limit.
2. Explain why 0.58 is not necessarily the coefficient of static friction, and describe what measurement would reveal the true maximum static friction.
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The relation f_max = μ_s · N gives the maximum static friction before sliding, not the friction actually acting. Since the crate is not on the verge of sliding, the actual friction (98 N) is less than or equal to μ_s·N. So 98/170 ≈ 0.58 is only a lower bound on μ_s; the true μ_s could be larger.
To find the true maximum: slowly increase the ramp angle (or add mass/force) until the crate just begins to slide. At that instant, friction has reached its maximum and equals the down-ramp gravity component. Record the angle θ at which sliding starts; then μ_s = tan θ (since at the threshold, μ_s·N = mg·sin θ and N = mg·cos θ). If it slides at, say, 40°, then μ_s ≈ 0.84 - larger than 0.58, confirming the student's value was a lower bound.