How To Calculate Contact Force

Force, as a physics concept, is described by Newton's second law, which states that acceleration results when a force acts on a mass. Mathematically, this means:

\(F=ma\)

although it is important to note that acceleration and force are vector quantities (i.e., they have both a magnitude and a direction in three-dimensional space) whereas mass is a scalar quantity (i.e., it has a magnitude only). In standard units, force has units of Newtons (N), mass in measured in kilograms (kg), and acceleration is measured in meters per second squared (m/s²).

Some forces are non-contact forces, meaning that they act without the objects experiencing them being in direct contact with each other. These forces include gravity, the electromagnetic force, and internuclear forces. Contact forces, on the other hand, require objects to touch one another, be this for a mere instant (such as a ball striking and bouncing off a wall) or over an extended period (such as a person rolling a tire up a hill).

In most contexts, the contact force exerted on a moving object is the vector sum of normal and frictional forces. The frictional force acts exactly opposite the directions of motion, while the normal force acts perpendicular to this direction if the object is moving horizontally with respect to gravity.

This force is equal to the ​coefficient of friction​ μ between the object and the surface multiplied by the object's weight, which is its mass multiplied by gravity. Thus:

\(F_f=\mu mg\)

Find the value of μ by looking it up in an online chart such as the one at Engineer's Edge. ​Note:​ Sometimes you will need to use the coefficient of kinetic friction and at other times you will need to know the coefficient of static friction.

Assume for this problem that F_f = 5 Newtons.

This force, F_N, is simply the object's mass times the acceleration due to gravity times the sine of the angle between the direction of motion and the vertical gravity vector g, which has a value of 9.8 m/s². For this problem, assume that the object is moving horizontally, so the angle between the direction of motion and gravity is 90 degrees, which has a sine of 1. Thus F_N = mg for present purposes. If the object were sliding down a ramp oriented at 30 degrees to the horizontal, the normal force would be:

\(F_N=mg\times\sin{(90-30)}=mg\times \sin{60}=mg\times 0.866\)

For this problem, however, assume a mass of 10 kg. F_N is therefore 98 Newtons.

If you picture the normal force F_N acting downward and the frictional force F_f acting horizontally, the vector sum is the hypotenuse the completes a right triangle joining these force vectors. Its magnitude is thus:

\(\sqrt{F_N^2+F_f^2}\)

which for this problem is

\(\sqrt{15^2+98^2}=\sqrt{225+9604}=99.14\text{ N}\)