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Crate on a ramp: normal force, friction, acceleration, net force

SPH3U Grade 11 Physics · Forces

At a tile depot, a flat-deck truck is unloaded down its steel loading ramp, which meets the driveway at 25.0° — an angle the crew checks each morning with a digital level. A crate of quarry tile, its shipping manifest listing the mass as 1.20 t, is eased down the ramp on hardwood skids, held from above by a restraining strap paying out from a hand winch. Partway down, the strap's hook lets go, and the crate begins to slide on its own, skids hissing over the steel deck. Earlier testing of the same skids on the same ramp put the coefficient of kinetic friction between the crate's skids and the ramp at 0.30. Find the crate's weight, the normal force the ramp exerts, the friction force resisting the slide, the component of the weight pulling it down the slope, the acceleration it picks up, and the net force acting on it.

1.20 t of quarry tilea = 1.48 m/s²25.0°friction 3.20 kNμₖ = 0.30

Every number in this problem is editable — change any value below and the whole chain recalculates.

Given
  • m = 1.2 t — Crate of quarry tile
  • θ = 25 ° — Ramp angle above the driveway
  • μₖ = 0.3 — — Kinetic friction, skids on steel
Determine
  1. (a)the crate's weight
  2. (b)the normal force the ramp exerts
  3. (c)the friction force resisting the slide
  4. (d)the component of the weight pulling down the slope
  5. (e)the acceleration the crate picks up
  6. (f)the net force acting on it
Step 1 of 6(a) · solve for Weight

Start with the one force that is not a component of anything: gravity pulls straight down with mg, whatever the ramp is doing. Freight is billed by the tonne and mg wants kilograms, so 1.20 t goes in as 1,200 kg — and everything on the slope is carved out of the 11.77 kN that comes back.

mW
Rearranged for W
W=mgW = m g
Your values, in your units
W=(1.2 t) (9.80665 m/s2)W = \left( 1.2\ \text{t} \right) \, \left( 9.80665\ \text{m/s}^{2} \right)
Converted to base units
W=(1,200 kg) (9.80665 m/s2)W = \left( 1{,}200\ \text{kg} \right) \, \left( 9.80665\ \text{m/s}^{2} \right)
Answer
W=11.768 kNW = 11.768\ \text{kN}

Carried onward at full precision, not this rounded figure.

Open the Weight (W = mg) solver →

Step 2 of 6(b) · solve for Normal force

The ramp can only push perpendicular to itself, so it carries the cos θ share of the weight — 10.67 kN, not the full 11.77 kN. Using the whole weight here is the standard wreck.

mmgNθ
Rearranged for N
N=mgcos⁡θN = m g \cos\theta
Your values, in your units
N=(1.2 t) (9.80665 m/s2) cos⁡(25 ∘)N = \left( 1.2\ \text{t} \right) \, \left( 9.80665\ \text{m/s}^{2} \right) \, \cos \left( 25\ ^{\circ} \right)
Converted to base units
N=(1,200 kg) (9.80665 m/s2) cos⁡(25 ∘)N = \left( 1{,}200\ \text{kg} \right) \, \left( 9.80665\ \text{m/s}^{2} \right) \, \cos \left( 25\ ^{\circ} \right)
Answer
N=10.665 kNN = 10.665\ \text{kN}

Carried onward at full precision, not this rounded figure.

Open the Normal Force on an Incline (N = mg cos θ) solver →

Step 3 of 6(c) · solve for Kinetic friction force

Friction is proportional to how hard the surfaces are squeezed together, so it is built on step 2's normal force and never on the weight.

Nfkμk
Rearranged for fk
fk=μkNf_k = \mu_k N
10.665 kNcarried from step 2
Your values, in your units
fk=(0.3) (10,665.4 N)f_k = \left( 0.3 \right) \, \left( 10{,}665.4\ \text{N} \right)
Answer
fk=3.1996 kNf_k = 3.1996\ \text{kN}

Carried onward at full precision, not this rounded figure.

Open the Kinetic Friction Force (f = μₖN) solver →

Step 4 of 6(d) · solve for Down-slope force

The other slice of the weight, the sin θ one, is what actually drives the crate down. Compare it with step 3: 4.97 kN of pull against 3.20 kN of friction, so the crate does slide.

mmgF∥θ
Rearranged for F∥
F∥=mgsin⁡θF_{\parallel} = m g \sin\theta
Your values, in your units
F∥=(1.2 t) (9.80665 m/s2) sin⁡(25 ∘)F_{\parallel} = \left( 1.2\ \text{t} \right) \, \left( 9.80665\ \text{m/s}^{2} \right) \, \sin \left( 25\ ^{\circ} \right)
Converted to base units
F∥=(1,200 kg) (9.80665 m/s2) sin⁡(25 ∘)F_{\parallel} = \left( 1{,}200\ \text{kg} \right) \, \left( 9.80665\ \text{m/s}^{2} \right) \, \sin \left( 25\ ^{\circ} \right)
Answer
F∥=4.9734 kNF_{\parallel} = 4.9734\ \text{kN}

Carried onward at full precision, not this rounded figure.

Open the Weight Component Along an Incline (mg sin θ) solver →

Step 5 of 6(e) · solve for Acceleration

Divide that surplus by the mass and the mass cancels: a = g(sin θ − μₖ cos θ). A 12 t crate on the same ramp would accelerate identically, and so would a 12 kg one.

aμkθ
Rearranged for a
a=g(sin⁡θ−μkcos⁡θ)a = g\left(\sin\theta - \mu_k \cos\theta\right)
Your values, in your units
a=(9.80665 m/s2)(sin⁡(25 ∘)−(0.3) cos⁡(25 ∘))a = \left( 9.80665\ \text{m/s}^{2} \right)\left(\sin \left( 25\ ^{\circ} \right) - \left( 0.3 \right) \, \cos \left( 25\ ^{\circ} \right)\right)
Answer
a=1.4781 m/s2a = 1.4781\ \text{m/s}^{2}

Carried onward at full precision, not this rounded figure.

Open the Acceleration Down an Incline with Friction solver →

Step 6 of 6(f) · solve for Force

Newton's second law turns the acceleration back into the net force. It must equal step 4 minus step 3 — 4,973.4 − 3,199.6 = 1,773.7 N — and that agreement is the check the whole free-body diagram was for.

mFa
Rearranged for F
F=maF = m a
1.4781 m/s²carried from step 5
Your values, in your units
F=(1.2 t) (1.47812 m/s2)F = \left( 1.2\ \text{t} \right) \, \left( 1.47812\ \text{m/s}^{2} \right)
Converted to base units
F=(1,200 kg) (1.47812 m/s2)F = \left( 1{,}200\ \text{kg} \right) \, \left( 1.47812\ \text{m/s}^{2} \right)
Answer
F=1.7737 kNF = 1.7737\ \text{kN}

Carried onward at full precision, not this rounded figure.

Open the Newton's Second Law solver →

Answer

Therefore the crate weighs 11.77 kN, the ramp pushes back with 10.67 kN of normal force, friction drags at 3.20 kN against a 4.97 kN down-slope pull, and the freed crate accelerates at 1.48 m/s² under a net force of 1,774 N.

Why this order

Every incline problem is the same decomposition: gravity points straight down, the surface can only push at right angles to itself, so the weight has to be split into a piece the ramp cancels (mg cos θ) and a piece nothing cancels (mg sin θ). Getting sine and cosine the right way round is the one thing worth memorising, and the way to check is to flatten the ramp in your head: at θ = 0° the down-slope pull must vanish and the normal force must equal the full weight, which sin 0° = 0 and cos 0° = 1 deliver. Steps 2 and 3 have to run in that order because friction is defined off the normal force — μₖ multiplied by the weight would overstate it here by about 10%, enough to predict a crate that stays put when it will in fact slide.

Step 6 is the payoff and the cross-check. The packaged formula in step 5 hides the reasoning inside a bracket, so recomputing F = ma and matching it against the down-slope force minus the friction force reconstructs the argument the bracket came from. The mass cancelling in step 5 surprises most students, and it is worth sitting with: on a given slope with a given surface pairing, every crate accelerates alike, because doubling the mass doubles the pull and doubles the friction at once. Tilt the ramp until tan θ = μₖ — about 16.7° here — and the two exactly balance; that angle is the angle of repose, and it is why a gravel pile always settles into the same cone no matter how much gravel you dump on it.

Carried values move at full precision, not the rounded figure shown — chaining rounded numbers compounds error.