Normal Force Calculator

The normal force depends on the situation. On a flat surface with no other vertical force, N = m g. With a force F at angle theta, N = m g - F sin(theta) when pulling up and N = m g + F sin(theta) when pushing down. On a slope of angle alpha it is N = m g cos(alpha). In an elevator accelerating upward at a it is N = m (g + a).

Multiply the weight by the cosine of the slope angle: N = m g cos(alpha). A 10 kg box on a 30 degree ramp on Earth has N = 10 x 9.80665 x cos 30 = 84.93 N, while its weight is 98.07 N. The rest of the weight, m g sin(alpha) = 49.03 N, pulls the box down the slope.

No. It equals the weight only on a level surface with no vertical push or pull and no vertical acceleration. It is smaller than the weight on a slope or when you pull upward at an angle, and larger when you push down or when the surface accelerates upward. It is a reaction to what presses on the surface, not a copy of gravity.

The maximum friction force is the friction coefficient times the normal force, f = mu N. A larger normal force presses the surfaces together harder and allows more friction. Pulling a box upward at an angle reduces N and therefore the friction it has to overcome, which is one reason angled pulls work well.

N = m g - F sin(theta). A 10 kg box pulled with 50 N at 30 degrees above horizontal on Earth has N = 98.07 - 25.00 = 73.07 N. The vertical part of the pull holds up some of the weight so the surface has less to carry.

The formula would give a negative number, which a surface cannot provide because it can only push, never pull. The normal force becomes 0, the object leaves the surface, and friction vanishes. The calculator shows N = 0 and a note that the object leaves the surface when F sin(theta) is larger than m g.

It slides when tan(alpha) is larger than the static friction coefficient mu. This is the same as m g sin(alpha) exceeding mu m g cos(alpha), so the mass cancels out. With mu = 0.3 the critical angle is arctan(0.3) = 16.70 degrees, and any steeper slope makes the object slide.

For a sliding object, a = g (sin(alpha) - mu cos(alpha)) with the kinetic friction coefficient. A box on a 30 degree slope with mu = 0.3 accelerates at 9.80665 x (0.5 - 0.3 x 0.866) = 2.36 m/s2. If tan(alpha) does not exceed mu, the object stays at rest and the acceleration is 0.

What you feel is the normal force from the floor. To accelerate upward the floor must push with m (g + a), which is more than your weight m g. A 70 kg person accelerating up at 2 m/s2 feels a force of 826.47 N, about 1.204 g, and a scale would read 84.28 kg.

Zero. If the elevator accelerates downward at a = -g, then N = m (g + a) = 0. The floor no longer pushes on you, so you feel weightless even though gravity still acts. The same is true for astronauts in orbit, who are in continuous free fall.

Yes, through g. The weight m g and the normal force on a flat surface scale with g, so a 10 kg box has N = 98.07 N on Earth, 16.20 N on the Moon (g = 1.62 m/s2) and 37.21 N on Mars (g = 3.721 m/s2). The slope condition tan(alpha) > mu does not depend on g.

On a vertical wall alpha is 90 degrees and cos(90) is 0, so the weight itself produces no normal force. A normal force exists only if something else presses the object against the wall, and then friction can hold it up.