Weight and Normal Force

Weight and normal force often appear together in introductory mechanics, but they arise from different interactions and obey different rules. Weight is the gravitational force exerted by Earth on a body. The normal force is a contact force exerted by a surface and directed perpendicular to that surface.

The most important distinction is

(1)

near Earth’s surface, while the magnitude of the normal force usually has to be determined from Newton’s second law:

(2)

There is no universal formula N = mg. That equality is only a result of particular force balances.

1 Weight near Earth’s surface

Near Earth’s surface, when The Gravitational Field can be treated as uniform, the weight of a body of mass m is

(3)

If the positive y direction is upward, then

(4)

so

(5)

The magnitude is

(6)

Mass and weight are not the same quantity. Mass is measured in kilograms, while weight is a force measured in newtons. The same body can have the same mass in two locations but different weight if the local value of g differs.

2 The normal force

When a body is in contact with a surface, the surface can exert a contact force. The component perpendicular to the surface is called the normal force.

For a frictionless contact, the contact force is purely normal:

(7)

where e_n is the unit vector pointing away from the surface and into the body.

The direction of N is therefore determined by geometry. Its magnitude is determined by the motion and by the other forces acting on the body.

A simple contact surface can push on a body but cannot pull it. For that reason,

(8)

If an equation of motion would require N < 0, the assumed contact cannot be maintained. The body loses contact and the correct model changes to N = 0 until contact is restored.

3 Why N = mg is not a general law

Consider a body resting on a horizontal floor with no other vertical forces. Taking upward as positive,

(9)

If

(10)

then

(11)

The equality follows from this particular acceleration and force set. Change the acceleration or add another vertical force and the result changes.

For example, if an external force F pulls upward at angle α above the horizontal while a_y = 0, then

(12)

so

(13)

An upward component reduces the normal force. A downward applied component increases it.

4 Weight and normal force are not a Newton third law pair

It is common to see N and mg drawn as equal and opposite on a stationary body and then incorrectly call them a Newton third law pair. They are not.

The force

(14)

is Earth’s gravitational force on the body. Its Newton third law partner is the body’s gravitational force on Earth.

The force

(15)

is the surface’s contact force on the body. Its Newton third law partner is the body’s contact force on the surface.

The normal force and weight can balance on one body, but a third law pair always acts on two different bodies.

5 Apparent weight and scale readings

A bathroom scale does not directly measure the gravitational force mg. It measures a contact force associated with the deformation of the scale. In the usual model this reading is the normal force magnitude N.

This measured support force is often called the apparent weight.

For a person of mass m in an elevator, choose upward as positive. Newton’s second law gives

(16)

Hence

(17)

This one signed equation covers all vertical elevator cases.

If the acceleration is upward,

(18)

If the acceleration is zero,

(19)

If the acceleration is downward but smaller in magnitude than g,

(20)

The direction of the velocity does not determine the scale reading. An elevator can be moving upward while slowing down; its acceleration is then downward and the scale reads less than mg.

6 Free fall and apparent weightlessness

If the elevator and person are both in ideal free fall, then

(21)

Therefore

(22)

which gives

(23)

The person is not without gravity. Gravity is precisely what produces the free fall. The apparent weight is zero because there is no supporting contact force.

This distinction is important in orbital motion as well: astronauts in orbit experience gravity but are in continuous free fall, so local support forces can be very small.

7 Normal force on an inclined plane

For a body on a plane inclined by angle 𝜃, choose one axis parallel to the plane and one axis normal to it. The normal component of weight is

(24)

while the component parallel to the plane is

(25)

If the body has no acceleration normal to the plane and no other force has a normal component, then

(26)

so

(27)

Again, this is not a universal normal force formula. It follows from the stated geometry, force set, and normal acceleration.

8 Normal force on a curved path

A contact surface can constrain a body to follow a curved path. Even if the speed is constant, the body can have normal acceleration

(28)

where R is the local radius of curvature.

The normal force must then participate in whatever inward net force is required.

At the crest of a convex hill, the inward direction is downward. For a vehicle whose only vertical forces are weight and the road’s normal force,

(29)

Therefore

(30)

At the bottom of a concave valley, inward is upward, so

(31)

and

(32)

Thus a rider generally feels lighter at the top of a hill and heavier at the bottom of a valley.

9 Loss of contact

At the crest of a hill,

(33)

As the speed increases, the required normal force decreases. The limiting condition for contact is

(34)

Therefore

(35)

and the threshold speed is

(36)

Above this value, the equation for maintained contact would demand a negative normal force. An ordinary road surface cannot pull the vehicle downward, so the vehicle leaves the surface instead.

10 A systematic procedure

Normal force problems become much easier if the geometry is handled before the algebra.

1. Choose the body or system.
2. Draw only the external forces acting on it.
3. Identify the contact surface and draw N perpendicular to it.
4. Choose an axis normal to the surface or inward toward the local center of curvature.
5. Determine the acceleration component along that axis.
6. Apply Newton’s second law along that direction.
7. Check that the final result satisfies N ≥ 0 when contact is assumed.

11 Worked example 1: scale reading in an accelerating elevator

A person of mass

(37)

stands on a scale in an elevator accelerating upward at

(38)

Taking upward as positive,

(39)

Therefore

(40)

Using g = 9.81 m∕s²,

(41)

Thus the scale reads

(42)

The person’s gravitational force is still

(43)

The larger scale reading is caused by the upward acceleration, not by a change in gravity.

12 Worked example 2: block on an incline

A 12.0 kg block rests on a frictionless plane inclined at

(44)

There is no acceleration normal to the plane. Therefore

(45)

Substituting,

(46)

so

(47)

Notice that the normal force is smaller than the full weight mg ≃ 117.7 N because only the component of weight perpendicular to the plane is balanced by N.

13 Worked example 3: vehicle at the crest of a hill

A 1200 kg vehicle crosses the crest of a hill whose local radius of curvature is

(48)

at speed

(49)

At the crest, inward is downward. Thus

(50)

The mass cancels if we solve symbolically first:

(51)

Therefore

(52)

which gives

(53)

The vehicle’s weight is about 1.18 × 10⁴ N, so the road supports it with substantially less force at the crest.

The contact threshold is

(54)

At that speed the required normal force would fall to zero.

14 Worked example 4: rider at the bottom of a valley

A 60.0 kg rider passes through the bottom of a circular valley of radius

(55)

at speed

(56)

At the bottom, inward is upward. Therefore

(57)

Hence

(58)

Substituting,

(59)

so

(60)

The rider’s ordinary weight is only

(61)

so the support force at the bottom is nearly twice as large.

15 Practice problems

1. A 5.0 kg book rests on a horizontal table with no other vertical forces. Find its weight magnitude and the normal force magnitude.
2. A 65 kg passenger stands on a scale in an elevator accelerating upward at 2.0 m/s². Find the scale reading.
3. The same passenger is moving upward but slowing at 2.0 m/s². Find the scale reading and explain why the direction of motion is not enough to determine the result.
4. A 10 kg block rests on a frictionless 30^∘ incline. Find the normal force.
5. A 20 kg crate is on a horizontal floor. A rope pulls with force 80 N at 35^∘ above horizontal, and there is no vertical acceleration. Find the normal force.
6. A 1000 kg CAR crosses the crest of a hill of radius 50 m at 18 m/s. Find the normal force from the road.
7. For the hill in Problem 6, find the speed at which the car just loses contact with the road.
8. A 75 kg rider passes through the bottom of a circular dip of radius 20 m at 12 m/s. Find the support force on the rider.
9. A person stands on a scale inside an elevator in ideal free fall. What does the scale read? Is the gravitational force zero?
10. A block remains in contact with a moving platform. Explain how a calculated value N < 0 should be interpreted physically.

16 Answer check

1. W = 49.1 N and N = 49.1 N.
2. N = 65(9.81 + 2.0) = 768 N.
3. The acceleration is downward, so N = 65(9.81 − 2.0) = 508 N. Velocity direction does not set the force balance.
4. N = mg cos 30^∘ = 85.0 N.
5. N = mg − F sin 35^∘ ≃ 150 N.
6. N = m(g − v²∕R) = 3.33 × 10³ N.
7. v = = 22.1 m/s.
8. N = m(g + v²∕R) = 1.28 × 10³ N.
9. N = 0. Gravity is not zero; the person and scale are both in free fall.
10. An ordinary contact surface cannot supply a negative normal force. Contact is lost and the correct model sets N = 0 until the bodies meet again.

17 Summary

Weight and normal force are fundamentally different interactions. Near Earth’s surface,

(62)

while the normal force is determined from the contact geometry and Newton’s second law.

For common situations,

(63)

for a vertical elevator when upward is positive,

(64)

for a simple incline with zero normal acceleration, and

(65)

at the crest or bottom of a vertical curved path when weight and the support force are the only forces in the normal direction.

The common theme is not memorizing separate formulas. It is choosing the correct normal direction and applying

(66)

References