Find normal force, available friction, and the force needed to move a block on an incline. Compare static and kinetic friction with worked steps.
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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)
0
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)
1
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)
0
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.