A 30° Ramp in a Packing Godown: Will the Crate Move Before the Rope Arrives?
A 20 kg wooden packing crate is placed on a wooden ramp inside a godown. The ramp is set at 30° to the horizontal. The coefficient of static friction between crate and ramp is 0.45 and the coefficient of kinetic friction is 0.35. Take g = 9.8 m/s². A loader releases the crate from rest, expecting it to stay put while he walks to the far end to fetch a rope. Before he returns, the crate has begun to move and has already travelled some distance down the ramp. Later, the supervisor wants the crate held in place by a rope tied parallel to the ramp surface, and he also wants the ramp raised to 45° so that crates slide down faster during unloading. No one in the team has worked out whether the crate was ever in equilibrium at 30°, nor how much force the rope must actually take.
1. Show by calculation whether the crate remains at rest on the 30° ramp. If it moves, find its acceleration.
Read the worked answer
Weight component down the incline = mg sin θ = 20 × 9.8 × 0.5 = 98 N. Normal reaction = mg cos θ = 20 × 9.8 × 0.866 = 169.7 N. Maximum static friction available = μsN = 0.45 × 169.7 = 76.4 N.
Compare the driving force with the maximum resisting force: 98 N > 76.4 N, so static friction cannot hold the crate and it slides. This is why the loader's assumption fails - the crate was never in equilibrium.
Once sliding starts, kinetic friction applies: fk = 0.35 × 169.7 = 59.4 N. Net force = 98 − 59.4 = 38.6 N down the incline. Acceleration a = 38.6 / 20 = 1.93 m/s².
The judgement point is that the answer depends on comparing forces, not on quoting a formula for acceleration. Also note the crate does not accelerate at g sin θ, because friction removes a substantial share of the driving force.
2. What tension in a rope tied parallel to the ramp is needed to just hold the crate at 30°? Explain why the answer is smaller than the weight component of 98 N.
Read the worked answer
For equilibrium at the point of slipping, friction acts up the incline at its maximum value and the rope shares the remaining load: T + μsN = mg sin θ T = 98 − 76.4 = 21.6 N.
The tension is far below 98 N because static friction is already resisting most of the downward pull. A common error is to set T = mg sin θ = 98 N; that would be correct only on a frictionless ramp. Equally, the rope does not need to overcome the full 98 N because friction does useful work here.
A further judgement: 21.6 N is the threshold value. Any tension above it keeps the crate safe, but the margin is thin - a small nudge, vibration from a truck, or damp wood lowering μs to about 0.3 would raise the required tension to 98 − 50.9 = 47.1 N. In practice, ropes are chosen with a safety factor, not sized at the bare threshold.
3. The ramp is raised to 45°. Without recalculating everything from scratch, predict and then verify how the motion changes. Why does the situation worsen in two ways at once?
Read the worked answer
Prediction: raising the angle increases the driving component while simultaneously reducing the normal reaction, so friction falls. Both effects push the crate faster.
Verification at 45°: Down-slope component = 196 × sin 45° = 196 × 0.707 = 138.6 N. Normal reaction = 196 × 0.707 = 138.6 N (equal at 45°). Maximum static friction = 0.45 × 138.6 = 62.4 N, which is far below 138.6 N. Kinetic friction = 0.35 × 138.6 = 48.5 N. Net force = 138.6 − 48.5 = 90.1 N; a = 90.1 / 20 = 4.5 m/s².
So acceleration more than doubles, from 1.93 to 4.5 m/s². The general lesson: on an incline, sin θ rises while cos θ falls, so the safety margin against sliding shrinks faster than most people expect. Only at θ = 0 does friction get its full mg value.