Statics & Mechanics of Materials

Friction on an Incline

The classic friction problem: a block of given weight (or mass) on a flat or inclined surface with coefficient of friction. Finds the normal force, the friction force available, whether the block slides on its own, the force parallel to the surface needed to START it moving up-slope (and to KEEP it moving with μk), and — when an applied force is given — whether that force suffices. Use for: 'will the block slide', 'force required to push the crate up the ramp', minimum μ questions. Not for: accelerating blocks (kinematics/work-energy).

Direct solving is free and needs no account. Have a word problem instead? Submit it as text.

Inputs

Units

Load a sample problem:

weight (optional)Block weight W (give weight OR mass) [dimension: force]mass (optional)Block mass — alternative to weight (W = m·g) [dimension: mass]incline angle deg (optional)Surface angle above horizontal in degrees (0 = flat ground)static frictionCoefficient of static friction μs between block and surfacekinetic friction (optional)Coefficient of kinetic friction μk (for the keep-it-moving force)applied force (optional)Externally applied force, if the question asks whether THIS force moves the block [dimension: force]applied angle deg (optional)Angle of the applied force above the surface direction, degrees (0 = parallel to surface, pushing up-slope)

Worked examples

These are the solver's own reference problems — the answers come from published sources or independent hand computation, never from the solver itself. The solution below is the live solver output for each.

Example 1

flat ground: W = 100 N, μs = 0.3

Given

weight: 100 N
incline angle deg: 0
static friction: 0.3

Assumptions

Rigid block on a plane surface; friction follows the Coulomb model f ≤ μs·N; the applied force (if any) acts in the vertical plane of the slope.

Solution steps

Resolve the weight on the incline
The weight splits into a component pressing into the surface and one pulling down the slope.
N0 = W·cosθ; W_par = W·sinθ
N0 = 100 N; W_par = 0 N
Does it slide by itself?
The block stays put while the down-slope pull is within the friction available: tanθ ≤ μs.
slides when tanθ > μs
tanθ = 0 vs μs = 0.3 — the block holds (no sliding)
Force to start moving up the slope
Pushing up-slope must beat both the weight component and full static friction.
P_start = W·sinθ + μs·W·cosθ
P_start = 30 N
= 30 N

Results

Normal force N (no applied force)

0.1kN

Weight component along slope

0kN

Slides without help? (1 = yes, 0 = no)

Force (parallel) to start up-slope

0.03kN

Where this answer was checked

source: f_max = μs·N on level ground — force to start sliding
verified by: hand-recomputed
derivation: N = 100 N; P_start = 0 + 0.3·100 = 30 N.

Example 2

steep slope slides on its own: θ = 30°, μs = 0.5

Given

weight: 200 N
incline angle deg: 30
static friction: 0.5

Assumptions

Rigid block on a plane surface; friction follows the Coulomb model f ≤ μs·N; the applied force (if any) acts in the vertical plane of the slope.

Solution steps

Resolve the weight on the incline
The weight splits into a component pressing into the surface and one pulling down the slope.
N0 = W·cosθ; W_par = W·sinθ
N0 = 173.2 N; W_par = 100 N
Does it slide by itself?
The block stays put while the down-slope pull is within the friction available: tanθ ≤ μs.
slides when tanθ > μs
tanθ = 0.5774 vs μs = 0.5 — the block SLIDES on its own
Force to start moving up the slope
Pushing up-slope must beat both the weight component and full static friction.
P_start = W·sinθ + μs·W·cosθ
P_start = 186.6 N
= 186.6 N

Results

Normal force N (no applied force)

0.1732kN

Weight component along slope

0.1kN

Slides without help? (1 = yes, 0 = no)

Force (parallel) to start up-slope

0.1866kN

Where this answer was checked

source: Slides when tanθ > μs — self-sliding check
verified by: hand-recomputed
derivation: tan30° = 0.5774 > 0.5 → slides. N = 200·cos30 = 173.2 N; W_par = 100 N.

Example 3

mass input with both coefficients: 50 kg at 20°, μs = 0.4, μk = 0.3

Given

mass: 50 kg
incline angle deg: 20
static friction: 0.4
kinetic friction: 0.3

Assumptions

Rigid block on a plane surface; friction follows the Coulomb model f ≤ μs·N; the applied force (if any) acts in the vertical plane of the slope.

Solution steps

Resolve the weight on the incline
The weight splits into a component pressing into the surface and one pulling down the slope.
N0 = W·cosθ; W_par = W·sinθ
N0 = 460.8 N; W_par = 167.7 N
Does it slide by itself?
The block stays put while the down-slope pull is within the friction available: tanθ ≤ μs.
slides when tanθ > μs
tanθ = 0.364 vs μs = 0.4 — the block holds (no sliding)
Force to start moving up the slope
Pushing up-slope must beat both the weight component and full static friction.
P_start = W·sinθ + μs·W·cosθ
P_start = 352 N
= 352 N
Force to keep it moving
Once moving, kinetic friction (μk) applies.
P_keep = 305.9 N

Results

Normal force N (no applied force)

0.4608kN

Weight component along slope

0.1677kN

Slides without help? (1 = yes, 0 = no)

Force (parallel) to start up-slope

0.352kN

Force (parallel) to keep moving up-slope

0.3059kN

Where this answer was checked

source: P_start = W·sinθ + μs·W·cosθ; P_keep with μk — start vs keep-moving forces
verified by: hand-recomputed
derivation: W = 50·9.80665 = 490.33 N. cos20 = 0.93969, sin20 = 0.34202. N = 460.76; W_par = 167.70. P_start = 167.70 + 0.4·460.76 = 352.00 N. P_keep = 167.70 + 0.3·460.76 = 305.93 N. tan20 = 0.364 < 0.4 → holds.