Normal Force

FIELDS OF STUDY: Classical Mechanics

ABSTRACT: This article defines normal force and describes how it relates to the weight of an object, frictional forces, and an inclined plane. Normal force is a contact force related to Newton’s third law of motion. The normal force measurement is used to determine the net force acting on an object.

Principal Terms

• acceleration: the rate of change of an object’s velocity over time.
• frictional resistance: the force created by the resistance to relative motion between solid surfaces; it is normally proportional to the roughness of the surfaces as well as the force squeezing the surfaces together.
• gravity: the attractive force of one body on another.
• net force: the vector sum of every force acting on an object.
• perpendicular: set at 90 degrees to a line or surface, forming a right angle.
• vector: a quantity that has both a magnitude and a direction.
• velocity: the rate of change of an object’s displacement over time.
• weight: the force due to gravity acting upon an object.

The Normal Force due to Gravity

Normal force is related to Newton’s third law of motion, which states that for every action there is an equal and opposite reaction. An object at rest on a table has an acceleration of zero, so its net force is also zero. The force of the table pushing up on the object and opposing the downward force of gravity is referred to as the "normal force." The normal force is always perpendicular to the surface on which an object is resting. The weight (W) of an object is equal to the object’s mass (m) in kilograms (kg) multiplied by gravity(g) in units of meters per second squared (m/s²):

W = mg

When an object is placed on a horizontal surface with no other forces involved, the normal force (N or Fn) is directly equal to the object’s weight:

N = mg

The weight and therefore the normal force are vector quantities measured in newtons (N). The greater the weight of an object placed on a table, the greater the normal force the table must exert to oppose being compressed by the object.

The normal force is not always directly equal to an object’s weight; it can be greater or less. For example, if someone is pulling a crate by a handle at angle above the horizontal, the force exerted to pull the crate has an upward component that counteracts part of the crate’s weight. In this case, the normal force would be less than the weight of the crate.

Normal Force on an Incline

For an object on an incline, the normal force is still perpendicular to the surface but is no longer vertical. When drawing a free-body diagram (a vector diagram that depicts all forces for a given situation) for object on an inclined surface, the x- and y-axes are typically oriented so that the x-axis is parallel and the y -axis is perpendicular to the surface. The normal force points in the positive y direction, perpendicular to the surface. The weight points downward at the same angle with respect to the negative y-axis in which the surface is inclined. The x and y components of weight are now:

W_x = Wsinθ = mgsinθ

W_y = −Wcosθ = −mgcosθ

Frictional Forces

Frictional resistance refers to forces that are parallel to a surface and oppose the direction of the object’s motion. Kinetic friction occurs when surfaces slide against each other at a certain velocity. The force of kinetic friction (f_k) is equal to the normal force (N) multiplied by the coefficient of kinetic friction (μ_k):

f_k = μ_kN

Static friction prevents two surfaces from moving against each other. The force due to static friction (fs) can have values between 0 and f_s,max, where f_s,max is a maximum limit to the force that can be delivered by static friction. If an applied force exceeds f_s,max, an object may start to slide and then kinetic friction takes over. Like kinetic friction, static friction is proportional to the normal force. The force of static friction (f_s) is equal to the normal force (N) multiplied by the coefficient of static friction (μ_s):

f_s,max = μ_sN

The coefficient for kinetic and static friction is a positive number typically between 0 and 1 and varies depending on the material.

Sample Problem

A 50 kg box is placed on a plank of wood. When the wood is tilted to an angle of 25 degrees, the box begins to slide. Find the coefficient of static friction between the wood and the box and the magnitude of the force of static friction on the box.

Answer:

First, draw a free-body diagram where with the coordinate system aligned with the incline of the wood. Choose the positive x direction to point down the incline and the positive y direction to be perpendicular to the wood.

There are three forces that act on the box, the normal force (N), the force from static friction (f_s), and the box’s weight (W). These forces are resolved into the following x and y components:

Since there is only a normal force in the y direction, the normal force in the x direction is set to zero.

N_x = 0

N_y = N

When an object is on the verge of slipping, the static friction is at its maximum value (f_s = f_s,max = μ_sN). Since there is only a force due to static friction acting in the x direction, the force due to static friction in the y direction is set to zero.

f_s,x = −f_s,max = −μ_sN

f_s,y = 0

The weight of the box has components in both the x and y directions.

W_x = mgsinθ

W_y = −mgcosθ

Since the box is at rest, the acceleration of the box is zero in both the x and y directions, and therefore the net force in both the x and y directions equals zero.

F_y = ma_y = 0

The net force in the y direction is a sum of all the forces acting on the box in the y direction and is equal to the y components of the normal force (N_y), static friction (f_s,y), and the weight (W_y):

F_y = N_y + f_s,y + W_y

Substitute in the values for static friction and weight and zero for F_y. Then solve for the normal force (N):

0 = N + 0 – mgcosθ

N = mgcosθ

The net force in the x direction also equals zero:

F_x = ma_x = 0

The net force in the x direction is a sum of all the forces acting on the box in the x direction and is equal to the x components of the normal force (N_x), static friction (f_s,x), and the weight (W_x):

F_x = N_x + f_s,x + W_x

Substitute in the values for static friction and weight and zero for F_xand N_x:

0 = 0 – μ_sN + mgsinθ

Now, mgcosθ can be substituted for N, giving

0 = 0 – μ_smgcosθ + mgsinθ

Then solve for the coefficient of static friction (μ_s):

Note that this result does not depend on the mass of the object. To find the force of static friction acting (f_s,max) on the box, plug in the coefficient (μ_s), the mass (m), the gravitational force (g), and the angle (cosθ). Then solve:

Normal Forces and Apparent Weight

Depending on the specific situation or other forces involved, the normal force can be less than or greater than the object’s weight. While riding on an elevator, one experiences apparent weight. If the elevator is moving with an upward acceleration, one feels heavier since the normal force is greater than one’s weight. If the elevator is moving with a downward acceleration, one feels lighter since the normal force is less than one’s weight. While training astronauts, the National Aeronautic and Space Administration uses this effect to simulate an experience of weightlessness.

Bibliography

Cutnell, John D., et al. Physics. 10th ed. Hoboken: Wiley, 2015. Print.

Halliday, David. Fundamentals of Physics. 10th ed. Hoboken: Wiley, 2014. Print.

Ohanian, Hans C., and John T. Markert. Physics for Engineers and Scientists. 3rd extended ed. New York: Norton, 2007. Print.

Tipler, Paul A., and Gene Mosca. Physics for Scientists and Engineers: Mechanics, Oscillations and Waves, Thermodynamics. 6th ed. New York: Freeman, 2008. Print.

"Types of Forces" Physics Classroom. Physics Classroom, n.d. Web. 20 June 2015.

Walker, James S. Physics. 5th ed. San Francisco: Pearson, 2010. Print.