Will the crate slide? Static friction on an incline with a horizontal push
An 18 kg crate rests on a rough plane inclined at 25° to the horizontal. The coefficients of static and kinetic friction are 0.55 and 0.40. A horizontal force P is applied to the crate, pushing it up the slope. A student sets P = 120 N and claims that the crate must slide because the component of P up the slope is greater than the component of weight down the slope. The student ignores the fact that P also presses the crate into the plane. In fact the crate remains at rest. The student then asks what minimum horizontal force would just start it moving, and what acceleration would result if the force were increased to 300 N. Assume g = 9.81 m/s² and that the crate does not tip.
1. Show whether P = 120 N causes motion. Determine the friction force required and compare it with limiting friction.
Read the worked answer
Resolve perpendicular to the plane. The normal reaction is increased by the inward component of P:
N = mg cos25° + P sin25° = 18(9.81)(0.9063) + 120(0.4226) = 160.0 + 50.7 = 210.7 N.
Limiting static friction is
f_max = μ_s N = 0.55(210.7) = 115.9 N.
The force up the slope from P is P cos25° = 108.8 N. The weight component down the slope is mg sin25° = 74.6 N. Without friction the net force would be 34.2 N up the slope, so friction must act down the slope with magnitude 34.2 N.
Since 34.2 N < 115.9 N, the crate is in static equilibrium and does not move. The student’s error is to compare only the along-slope components; the horizontal push also increases N and therefore increases the maximum available friction.
2. Calculate the minimum horizontal force needed to just start the crate moving up the slope.
Read the worked answer
At limiting equilibrium, friction acts down the slope at its maximum value:
P cos25° = mg sin25° + μ_s(mg cos25° + P sin25°).
Collect P terms:
P(cos25° − μ_s sin25°) = mg(sin25° + μ_s cos25°).
Numerically,
cos25° − 0.55 sin25° = 0.9063 − 0.2324 = 0.6739,
and
mg(sin25° + 0.55 cos25°) = 176.6(0.4226 + 0.4985) = 176.6(0.9211) = 162.7 N.
Therefore
P = 162.7 / 0.6739 ≈ 241 N.
A slightly larger force starts motion. The result is not simply the weight component divided by cos25° because increasing P also increases the normal reaction and hence the limiting friction.
3. If P is increased to 300 N, calculate the acceleration of the crate, stating any assumption.
Read the worked answer
Once moving, use kinetic friction. The normal reaction is
N = 160.0 + 300(0.4226) = 160.0 + 126.8 = 286.8 N.
Kinetic friction is
f_k = 0.40(286.8) = 114.7 N.
The up-slope driving component is
300 cos25° = 271.9 N.
The down-slope weight component is 74.6 N, so the net up-slope force is
F_net = 271.9 − 74.6 − 114.7 = 82.6 N.
The acceleration is
a = F_net/m = 82.6/18 ≈ 4.6 m/s² up the slope.
This assumes the crate is already moving up the slope and that μ_k remains constant. If the crate were initially at rest, the force would certainly exceed the starting threshold, but the initial acceleration would still be described by the same kinetic-friction model once slipping begins.