Holding a Cart on a Ramp: Strap Angle and Rating
A 900 kg equipment cart must be parked on an inclined cargo ramp while an aircraft is loaded. Rolling friction is negligible. The cart's parking brake is unserviceable, so a single tie-down strap running from the cart up to a deck fitting will hold it. The strap is rated to 1,200 N and, as currently rigged, makes an angle of 15° with the ramp surface. The ramp is set to 6°. The loadmaster wants to raise the ramp to 10° for loading and proposes to 'angle the strap flatter' if the rating becomes a problem. Use g = 9.81 m/s² and treat the strap as a two-force member.
1. Show whether the strap holds the cart on the 6° ramp.
Read the worked answer
Weight: 900 x 9.81 = 8,829 N
Down-slope component at 6°: 8,829 x sin 6° = 8,829 x 0.10453 = 923 N
Strap's up-slope component: 1,200 x cos 15° = 1,200 x 0.96593 = 1,159 N
Since 1,159 N > 923 N, the strap holds at 6°, with a margin of about 236 N (roughly 26% above the demand).
Two cautions belong in the answer. First, the strap's useful contribution is its component along the ramp; multiplying 1,200 N by cos 15° rather than by 1 is the whole point of the angle. Second, the strap also pulls the cart slightly off the ramp (1,200 x sin 15° = 311 N), which would reduce normal force and hence any friction if friction were present - it is stated negligible here, so the result stands.
2. Does the strap hold at 10°, and can a flatter strap angle rescue the situation?
Read the worked answer
Down-slope component at 10°: 8,829 x sin 10° = 8,829 x 0.17365 = 1,533 N
Available along-ramp pull is still 1,159 N, so the strap is short by 374 N - it fails, and not marginally.
The flatter-strap idea does not work. The best case for any strap angle is a strap lying parallel to the ramp, where the full 1,200 N acts up-slope. That is still less than 1,533 N. Flattening the strap from 15° to 0° recovers only 1,200 - 1,159 = 41 N, against a 374 N deficit.
The loadmaster has confused 'get more out of the strap' with 'get more strap'. Above roughly cos⁻¹(1,533/1,200) - which has no solution, since the ratio exceeds 1 - no geometry helps. The fix must be a second strap, a higher-rated strap, or a lower ramp angle.
3. What is the steepest ramp angle this single strap can hold, and what does that imply for operations?
Read the worked answer
Set the along-ramp pull equal to the down-slope component:
1,200 x cos 15° = 8,829 x sin θ 1,159 = 8,829 sin θ sin θ = 0.1313 θ = 7.5°
So the rig works at 6° and fails above about 7.5°; a 10° ramp is out of limits by a wide margin.
Operational implications
- The existing margin at 6° is real but small in angle terms - only about 1.5° of ramp travel separates 'held' from 'released'.
- The most dangerous moment is not the static hold: any bouncing, wind load, or a loader leaning on the cart adds dynamic load the static calculation ignores.
- Defensible action is to fit a second strap or a higher-rated one before raising the ramp at all, rather than setting a working limit at the computed failure angle. A computed maximum is a line not to be approached, and 6° with an unserviceable brake is already close to it.