How do I know which way friction acts on the incline?

Assume the block stays put and add up every force along the slope except friction: the weight component mg sinθ, the pull component P cosα and any rope tension. Static friction is equal and opposite to that sum, as long as it does not exceed μsN. If it would have to, the block slides in the direction of the net force and friction becomes kinetic, μkN, opposing the motion. The tool follows exactly this order, so friction can point up or down the slope.

Why is the normal force not always mg cosθ?

The normal force comes from balance perpendicular to the slope: N + P sinα − mg cosθ = 0. It equals mg cosθ only when there is no applied force or the force is parallel to the slope. Pulling upward and outward reduces N (and the friction), pushing into the slope increases it. If the outward part of the pull exceeds mg cosθ the block leaves the surface, and the tool warns you.

Why are the tensions undetermined with three ropes?

A particle in a plane gives only two equations, ΣFx = 0 and ΣFy = 0, but three ropes mean three unknown tensions: the problem is statically indeterminate. The real split depends on how stretchy the ropes are and how tightly they were installed. The tool assumes three identical elastic ropes (stiffness inversely proportional to length) with small deflections and solves by the displacement method; a rope that would need to push is treated as slack and the system is solved again.

Where does the minimum safe ladder angle come from?

The ladder carries its own weight, the person, the floor normal force and friction, and the wall normal force and friction. Writing ΣFx = 0, ΣFy = 0 and torque balance about the foot, then setting both frictions to their limits f₁ = μ₁N₁ and f₂ = μ₂N₂, gives tanθmin = [(m/2 + Md/L)(1 + μ₁μ₂)/(m + M) − μ₁μ₂]/μ₁. For a smooth wall and nobody on it this reduces to the familiar tanθmin = 1/(2μ₁). The higher the person climbs, the steeper the ladder must be.

Parallelogram rule or tip-to-tail: what is the difference?

They give the same answer because both are vector addition. The parallelogram rule suits two forces: draw them as adjacent sides and the diagonal is the resultant. With three or more forces tip-to-tail (the polygon rule) is easier: chain the forces one after another and the arrow from the start to the end is the resultant. The equilibrant has the same size as the resultant but points the opposite way, so adding it makes the net force zero.

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🏋️Free-Body Diagram & Force Calculator - Incline, Tension, Ladder

This tool turns the classic force problems from high-school and first-year physics into diagrams you can drag: adding and resolving several forces, a block on an incline with friction and a hanging mass over a pulley, a weight hung from ropes, and a ladder leaning on a wall. Change any value and the free-body diagram, the verdict and the ΣFx, ΣFy and Στ equations all update together.

Free · No sign-up · Runs in your browserLast updated: 2026-10-29

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How to use Free-Body Diagrams & Force Vectors

1Pick a mode at the top: vectors, incline or static equilibrium
2Drag arrow tips or anchor points, or use the sliders for angles, masses and friction coefficients
3Read the free-body diagram and the verdict (at rest, sliding up, sliding down, stable or slips)
4Compare your own working with the equations and substituted numbers listed below

FAQ

How do I know which way friction acts on the incline?: Assume the block stays put and add up every force along the slope except friction: the weight component mg sinθ, the pull component P cosα and any rope tension. Static friction is equal and opposite to that sum, as long as it does not exceed μsN. If it would have to, the block slides in the direction of the net force and friction becomes kinetic, μkN, opposing the motion. The tool follows exactly this order, so friction can point up or down the slope.
Why is the normal force not always mg cosθ?: The normal force comes from balance perpendicular to the slope: N + P sinα − mg cosθ = 0. It equals mg cosθ only when there is no applied force or the force is parallel to the slope. Pulling upward and outward reduces N (and the friction), pushing into the slope increases it. If the outward part of the pull exceeds mg cosθ the block leaves the surface, and the tool warns you.
Why are the tensions undetermined with three ropes?: A particle in a plane gives only two equations, ΣFx = 0 and ΣFy = 0, but three ropes mean three unknown tensions: the problem is statically indeterminate. The real split depends on how stretchy the ropes are and how tightly they were installed. The tool assumes three identical elastic ropes (stiffness inversely proportional to length) with small deflections and solves by the displacement method; a rope that would need to push is treated as slack and the system is solved again.
Where does the minimum safe ladder angle come from?: The ladder carries its own weight, the person, the floor normal force and friction, and the wall normal force and friction. Writing ΣFx = 0, ΣFy = 0 and torque balance about the foot, then setting both frictions to their limits f₁ = μ₁N₁ and f₂ = μ₂N₂, gives tanθmin = [(m/2 + Md/L)(1 + μ₁μ₂)/(m + M) − μ₁μ₂]/μ₁. For a smooth wall and nobody on it this reduces to the familiar tanθmin = 1/(2μ₁). The higher the person climbs, the steeper the ladder must be.
Parallelogram rule or tip-to-tail: what is the difference?: They give the same answer because both are vector addition. The parallelogram rule suits two forces: draw them as adjacent sides and the diagonal is the resultant. With three or more forces tip-to-tail (the polygon rule) is easier: chain the forces one after another and the arrow from the start to the end is the resultant. The equilibrant has the same size as the resultant but points the opposite way, so adding it makes the net force zero.

Features

Drag arrow tips to set magnitude and direction; tip-to-tail or parallelogram construction
Resolve any force along rotated axes such as the slope direction, and show the equilibrant
Incline with static and kinetic friction, a pull at any angle and a pulley with a hanging mass: decides whether it slides, gives acceleration and tension
Weight on two or three ropes with draggable anchors; the indeterminate three-rope case is solved with elastic ropes
Ladder against a wall with floor and wall friction: minimum safe angle and how high a person can climb
ΣFx, ΣFy and Στ equations with the numbers substituted for every case

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