Example 1
F = 20 N, r = 0.3 m, α = 90° → M = 6.
Type F [N], r [m] and angle α. The calculator computes M = F r sinα: 20 N, 0.3 m and 90° is 6 N·m, and at 30° you get 3 N·m. N·m of torque is not a joule of work.
Lever: F₁r₁ = F₂r₂. Point I: I = m·r². Energy: Eₖ = ½Iω².
Force F, Arm r and Angle α (degrees). The result shows up here.
Torque is M = F r sinα. At 20 N, 0.3 m and 90°: sin 90° = 1, so M = 6 N·m. The same F and r at 30° give 3 N·m, because sin 30° = 1/2. At 0° or 180° torque is zero: the force runs along the arm and does not twist. 10 N on 0.6 m at 90° is 6 N·m again.
The form has three fields: F in newtons, r in meters, α in degrees. The result is in N·m. That is not a joule of work, even though the dimensions match. Torque is twist. r must be positive. A comma and a period are the same r: 0,3 and 0.3.
Typed 0° gives M = 0. Typed 0 in F also gives M = 0. All three fields need a number before 6 N·m can appear. Measure α from the arm to the force, not from the floor, unless the problem says otherwise.
Two arms live on lever equilibrium: F₁ r₁ = F₂ r₂. Here one torque. Point I and rotational Eₖ take other inputs.
F⊥ = F sinα is the perpendicular component. The calculator multiplies it by r. A 0.3 m wrench at 20 N and 90° is the 6 N·m example.
Type 20, 0.3 and 90, click Calculate, and match 6 N·m. Then change the angle to 30 and see 3 N·m. The header symbol does not tighten a bolt.
M = F·r·sinα
F⊥ = F sinα
Unit: N·m.
One torque: M = F·r·sinα. 20 N, 0.3 m and 90° is 6 N·m. At 30° you get 3 N·m. N·m here is not a joule.
F = 20 N, r = 0.3 m, α = 90° → M = 6.
Same F and r, α = 30° → half M.
F = 10 N, r = 0.6 m, α = 90°.
F = 5 N at handle r = 0.8 m, α = 90°.
F = 150 N, r = 0.15 m, α = 90°.
F = 40 N, r = 0.4 m, α = 10°.
Force along the arm: M = 0.
F = 50 N, r = 0.25 m, α = 90°.
F = 30 N, r = 0.2 m, α = 60°.
Torque is 6 N·m. Sin 90° is one, so you compute 20 × 0.3. At 30° you get half, 3 N·m.
Force F [N], radius r [m], angle α in degrees. Result M [N·m]. That is not a joule, even though 1 N·m = 1 J in work.
Torque is 0 N·m. The force runs along the arm and does not twist. Sin 0° and sin 180° are zero.
On the lever-balance page. There two torques cancel. Here you compute one M = F r sinα.
In work, yes, 1 N·m = 1 J. For torque you keep N·m, so twisting is not confused with energy.
Yes. 0.3 and 0,3 are the same r [m]. A comma and a period mean the same value.
It is the distance from the axis to the point where F is applied. On a wrench that is often the handle length, but only if you measure to the nut axis.
Torque is 6 N·m. Same product as 20 N on 0.3 m: half the force, twice the arm.
The force component perpendicular to the arm, F sinα. At 90° F⊥ = F. At 30° F⊥ is half of F.
No. α is the angle between vector F and the arm, not the angle to the floor. 90° means the force is perpendicular to r.
The formula is the school one. Units follow SI; NIST SP 330 and BIPM define the measures, not your result.
Page updated in 2026.