Distinguish mass and weight, calculate weight using $F_g = mg$, and determine the normal force on an object on a surface, including on a horizontal surface and an incline.

What this topic is asking

This dot point pins down two of the most common forces in mechanics: weight (the gravitational pull on an object) and the normal force (the support a surface provides). The Regents asks you to keep mass and weight distinct, to calculate weight with $F_g = mg$, and to find the normal force in standard setups, on a horizontal surface and on an incline. These two forces appear in nearly every free-body diagram, so getting them right is essential for the dynamics module.

Mass versus weight

The two are easy to confuse because they are proportional ($F_g = mg$), but they are different kinds of quantity. A $60$ kg student has a mass of $60$ kg whether on Earth or the Moon, but weighs about $590$ N on Earth and only about $98$ N on the Moon, because the Moon's $g$ is roughly one sixth of Earth's. On the Regents, "weight" always means a force in newtons.

Calculating weight

The weight of an object is found directly from the Reference-Table equation:

with $g = 9.81$ m/s squared near Earth's surface. This is simply Newton's second law applied to gravity, since a freely falling object has acceleration $g$. Always check that mass is in kilograms before substituting, since the Regents sometimes gives a mass in grams.

The normal force

The normal force adjusts to circumstances. On a level floor with nothing else pushing vertically, it equals the weight. If you press down on the object, the normal force increases; if something lifts part of the weight, it decreases. In an elevator accelerating upward, the normal force exceeds the weight (you feel heavier); accelerating downward, it is less.

Normal force on a horizontal surface and on an incline

On a horizontal surface with no vertical acceleration, the vertical forces balance: the upward normal force equals the downward weight, $F_N = mg$.

On an inclined plane at angle $\theta$ to the horizontal, the weight is resolved into a component along the slope ($mg\sin\theta$, which tends to slide the object down) and a component perpendicular to the slope ($mg\cos\theta$, which presses into the surface). The surface only has to support the perpendicular component, so

which is less than the full weight (since $\cos\theta < 1$ for any incline). This smaller normal force is why friction is weaker on a slope, an idea that recurs in incline problems.

Reference Tables note

The equation $F_g = mg$ is printed in the Mechanics section of the Reference Tables, and $g = 9.81$ m/s squared is in the constants list. The normal-force expressions ($F_N = mg$ on the level, $F_N = mg\cos\theta$ on an incline) are not printed; you derive them by applying Newton's second law perpendicular to the surface. The trigonometric resolution uses the right-triangle definitions also given on the tables.

Try this

Q1. State the difference between mass and weight, including their units. [2 points]

• Cue. Mass is the amount of matter in kilograms (a scalar, same everywhere); weight is the gravitational force $F_g = mg$ in newtons (a vector that varies with $g$).

Q2. A $3.0$ kg object rests on a level floor. Calculate the normal force on it ($g = 9.81$ m/s squared). [2 points]

• Cue. $F_N = mg = (3.0)(9.81) = 29.4$ N, about $29$ N.