SME
Statics
Friction
Use this for “A 400 N crate rests on a 25° slope (μₛ = 0.3). What push along the slope starts it up, and what push keeps it from sliding down?”
Diagram
Weight = 400 N
Down the slope: Weight sin θ = 400 × sin 25° = 169 N
Normal force = Weight cos θ = 400 × cos 25° = 363 N
f = μₛ × normal force = 0.3 × 363 = 109 N
Friction at its limit, 109 N, balances the push less the weight’s pull down the slope (278 − 169 = 109 N): the block is about to move up, with a net force of 0 N.
a = Net force × g/Weight = 0 × 9.81/400 = 0 m/s²
This is an example. Type your own values to start a new problem.
W
Weight
example
θ
Slope angle
example · °
μₛ
Static friction coefficient
example
N
Normal force
calculated
Wₓ
Weight down the slope
worked out (not typed)
Fₘₐₓ
Friction at its limit
worked out (not typed)
Pᵤₚ
Push to start it up
calculated
Phold
Push to hold it
calculated
Formulas
N
=
W
×
cos(
θ
°)
362.5231 = 400 × cos(25°)
Wₓ
=
W
×
sin(
θ
°)
169.0473 = 400 × sin(25°)
Fₘₐₓ
=
μₛ
×
N
108.7569 = 0.3 × 362.5231
Pᵤₚ = Wₓ + Fₘₐₓ
277.8042 = 169.0473 + 108.7569
Phold = Wₓ − Fₘₐₓ
60.2904 = 169.0473 − 108.7569
Assumptions
•
The push P acts along the slope; friction is at its limit, μₛN, as motion is about to start.
•
Across the slope nothing moves, so N = W cos θ; along it the forces balance.
•
Pushed up, friction acts down the slope; held, friction acts up the slope.
•
If W sin θ ≤ μₛN (tan θ ≤ μₛ), the crate holds by itself: Phold = 0.
Step-by-step
WE KNOW
W = 400 N, θ = 25°, μₛ = 0.3
FIND
normal force (N), weight down the slope (Wₓ), friction at its limit (Fₘₐₓ), push to start it up (Pᵤₚ), push to hold it (Phold)
Step 1 · Find normal force (N)
Use N = W × cos(θ°)
Nothing
moves
across
the
slope,
so
N
balances
the
part
of
the
weight
pressing
into
it.
N
=
400
×
cos(25°)
N
=
400
×
0.9063
N
=
362.5231
N
Step 2 · Find weight down the slope (Wₓ)
Use Wₓ = W × sin(θ°)
The part of the weight along the slope pulls the crate down it.
Wₓ = 400 × sin(25°)
Wₓ = 400 × 0.4226
Wₓ = 169.0473 N
Step 3 · Find friction at its limit (Fₘₐₓ)
Use Fₘₐₓ = μₛ × N
Static friction can grow up to μₛN; at the start of motion it reaches that limit.
Fₘₐₓ = 0.3 × 362.5231
Fₘₐₓ = 108.7569 N
Step 4 · Find push to start it up (Pᵤₚ)
Use Pᵤₚ = Wₓ + Fₘₐₓ
Moving up, the push must beat both the weight’s pull down the slope and friction, which acts down it.
Pᵤₚ = 169.0473 + 108.7569
Pᵤₚ = 277.8042 N
Step 5 · Find push to hold it (Phold)
Use Phold = Wₓ − Fₘₐₓ
Sliding down, friction acts up the slope and helps; the push makes up the rest.
169.0473 > 108.7569: it would slide down, so it needs a push to hold it
Phold = 169.0473 − 108.7569
Phold = 60.2904 N
Check
Put the numbers back in. Both sides should match.
362.5231 = 400 × cos(25°) ✓
169.0473 = 400 × sin(25°) ✓
108.7569 = 0.3 × 362.5231 ✓
277.8042 = 169.0473 + 108.7569 ✓
60.2904 = 169.0473 − 108.7569 ✓
Topic check
See what you know: 5 questions on this topic
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