Example 1
T = 39.24 N, N = 98.1 N → μ = 0.4.
Type friction force T [N] and normal force N [N]. The calculator computes μ = T/N: 39 N and 98 N give 0.4. That is a number with no unit, not a force in newtons. Zero normal force does not divide.
T from a known μ: friction force. Sliding: incline with friction.
Friction force T and Normal force N. The result shows up here.
The friction coefficient is μ = T/N. From a lab or a problem: T = 39.24 N and N = 98.1 N give μ = 0.4. 20 N on 400 N is μ = 0.05. 60 N on 200 N is 0.3. 15 N on 80 N is 0.1875. 8.49 N on 42.45 N is 0.2. 30 N on 250 N is 0.12. 40 N on 100 N is 0.4 again. 9000 N on 12,000 N is 0.75. The result has no unit. 0.4 is four tenths, not 0.4 N.
The form has two fields: T and N in newtons, or the labeled unit. The result is dimensionless. A comma and a period are the same T: 39,24 and 39.24. N must be greater than zero. Zero normal force does not divide. Typed 0 in T gives μ = 0: no friction at a positive N. Both fields need a number before 0.4 can appear.
The problem says whether that is μ_s or μ_k. The calculator computes one μ from T/N. It does not split static from kinetic. From a known μ and a mass you compute T next door on friction force: 0.4 and 10 kg on the flat give about 39 N again.
A slide with this μ lives on incline with friction. There a = g(sin α - μ cos α), and at 30° with 0.2 you get about 3.21 m/s². Here you go from T and N to μ, not the other way. The incline angle is not a field in this calculator.
μ is not a force and not an angle. The angle belongs on the incline pages. This calculator ends at the ratio. The unit switch does not give μ a unit. The header symbol does not hold the block.
Type 39.24 and 98.1, click Calculate, and match 0.4. Then 20 and 400 for 0.05. Treat 0.4 as a number from the problem, not as newtons on a scale.
μ = T / N
Dimensionless.
The friction coefficient here is μ = T/N. 39 N on 98 N is 0.4. The result has no unit. Zero N does not divide.
T = 39.24 N, N = 98.1 N → μ = 0.4.
T = 20 N, N = 400 N.
T = 9000 N, N = 12000 N.
T = 60 N, N = 200 N.
T = 15 N, N = 80 N.
T = 8.49 N, N = 42.45 N.
T = 30 N, N = 250 N.
T = 40 N, N = 100 N.
T = 50 N, N = 200 N.
The coefficient is 0.4. That is 39.24 / 98.1. 40 N on 100 N also gives 0.4, the same order.
No. μ has no unit. 0.4 is four tenths, not 0.4 N. T and N are in newtons; the quotient is not.
Friction T [N] and normal force N [N]. Result μ with no unit. Multiply kilograms by g first if the problem gives mass.
You get μ = 0.05. At 60 N on 200 N that is 0.3. At 15 N on 80 N that is 0.1875.
Here T and N give μ. There μ and m give T. 0.4 and 10 kg on the flat give about 39 N, the inverse unknown.
No. Zero normal force does not divide. The normal force must be positive. Both fields need a number.
The calculator computes one μ from T/N. The problem says whether it is static or kinetic. The fields do not split them.
Yes. 39.24 and 39,24 are the same T [N]. A comma and a period mean the same value.
On the incline-with-friction page. At 30° and μ = 0.2, a is about 3.21 m/s². Here μ is the result, there it is a given.
The coefficient is 0.75. At 8.49 N and 42.45 N you get 0.2. Same quotient, another scale.
The formula is the school one. Units follow SI; NIST SP 330 and BIPM define the measures, not your result.
Page updated in 2026.