Weight, elastic force and friction: formulas and exercises

October, Year 1 of high school: if your school already has Physics in the first year, as is normally the case in the technological sector of the Italian technical institutes, in these weeks vectors arrive and straight afterwards the concrete forces: weight, elastic force, friction. The typical calendar is “vectors, forces (weight, Hooke, friction, tension), equilibrium of the particle”, and the test falls in six to nine weeks, so between mid-November and early December. A clarification: for Year 1 classes alone in 2026/27 the new curriculum with “Scienze sperimentali” (experimental sciences) in place of separate subjects is also planned; if and how Physics changes depends on your school’s planning, so look at the work plan the teacher has given you.

This post is the piece that is usually missing between “I can resolve a vector” and “I can solve a problem”: which are the three forces that keep appearing, what formula each one has, and where mistakes are made. There are three exercises, with round numbers, and gravity is always $g=9.8{\text{m/s}}^2$ (if your teacher uses $9.81$ or $10$, the decimal digits change, not the method).

Mass and weight: two different quantities

Mass $m$ measures how much matter there is in a body, or rather how much the body resists changes of velocity. It is measured in kilograms (kg) and never changes: a body has the same mass on Earth, on the Moon and in space.

Weight $P$ is a force: the attraction that the planet exerts on the body. It is measured in newtons (N), has a vertical direction pointing downwards, and depends on where you are. The link with mass is the second law applied to free fall:

with $g$ the acceleration of gravity: about $9.8{\text{m/s}}^2$ on the Earth’s surface, about $1.6{\text{m/s}}^2$ on the Moon. An object of 1 kg therefore weighs about $9.8\text{N}$ on Earth. The newton is the force that gives a mass of 1 kg an acceleration of $1{\text{m/s}}^2$:

This is your first dimensional check: any force, if you have done the steps right, must come out in newtons.

F = ma and P = mg: why they are not the same formula

The second law says that the resultant of the forces applied to a body equals mass times acceleration:

$P=mg$ is a special case of it: if the only force acting is the weight, the resulting acceleration is exactly $g$. If instead the body rests on a table, the weight is still there but the acceleration is zero, because the reaction of the table balances it. Weight does not disappear when the body is at rest: other forces compensate it. How to find the constraint reaction you will find in the post on constraint reactions in statics; here it is enough to remember that on a horizontal plane the normal force is $N=mg$ as long as there are no other vertical forces.

Elastic force and Hooke’s law

A spring, when you stretch or compress it by an amount $x$ relative to its rest length, pulls or pushes to return. As long as the deformation is small (within the “elastic limit”) the force is proportional to the deformation:

$k$ is the spring constant and is measured in N/m: a spring with a large $k$ is stiff, one with a small $k$ is soft. The minus sign does not mean that the force is “negative” in the sense of small: it says that the force has a direction opposite to the displacement. If you stretch it to the right ($x>0$) the spring pulls to the left ($F<0$). When in exercises you only need magnitudes write $F=kx$ and reason about the direction yourself with the drawing.

How $k$ is measured in the lab. You hang a body of known mass on the spring, wait for it to stop and measure the extension. At equilibrium the weight is balanced by the elastic force, so $mg=kx$ and

You repeat with several masses and get several $(x,F)$ pairs: on the force-extension graph the points lie on a straight line through the origin, and $k$ is its slope. If the lab relation is asked in the test, this is the reasoning to report. Watch the unit: extensions are measured in centimetres, but to get N/m you have to convert them to metres.

Static and kinetic friction

Friction is the force that opposes sliding between two surfaces in contact. It has two versions, and the distinction is what loses the most marks in tests.

Static friction. It acts when the body is at rest relative to the surface. It does not have a fixed value: it adapts to the force you try to apply, equal and opposite, until you reach a maximum threshold:

If you push with 30 N and static friction can reach 50 N, the friction is exactly 30 N: the body stays still. Only when the push exceeds ${\mu }_{s}N$ does the body start moving.

Kinetic friction. It acts when the body slides, has a direction opposite to the velocity and a precise value, not a maximum:

In general ${\mu }_d<{\mu }_s$: that is why getting a heavy crate moving takes more effort than keeping it moving. The coefficients $\mu$ are pure numbers, with no units, and depend on the materials of the two surfaces and on their state (smooth, rough, wet). $N$ is the normal reaction, the component of the reaction of the plane perpendicular to the surface: it is not always equal to the weight, it is only on a horizontal plane with no other vertical forces.

Equilibrium with friction: the method

A body is in equilibrium if the vector sum of the forces is zero. With friction the procedure is always the same.

1. Draw the body and all the forces: weight, normal force, applied force, friction.
2. Choose two axes, one along the surface and one perpendicular, and write the sum of the forces along each.
3. Along the perpendicular axis you find $N$.
4. Along the axis of the surface you compare the force that tries to make the body slide with ${\mu }_{s}N$: if it is less than or equal, the body stays still and the friction equals that force; if it is greater, the body moves and you switch to ${\mu }_{d}N$.

On the inclined plane the force that tries to make the body slide is the component of the weight along the plane, and the limit equilibrium condition becomes $\tan \theta ={\mu }_s$: you will find it developed in dynamics and the inclined plane in Year 2, while the resolution of forces with angles is explained in the post on resolving forces without trigonometry. Here we stay on the horizontal plane, where the threshold yes/no logic is easier to see.

Typical mistakes

Confusing kg and N. “The weight of the crate is 20 kg” is a mistake that gets penalised: 20 kg is the mass, the weight is $20\cdot 9.8=196\text{N}$. In everyday life people say “weight” for mass and the scale is calibrated in kg, but in a physics problem the difference matters. Always check the units: if a force comes out in kg, you have forgotten a $g$.

Thinking friction depends on the contact area. In the model studied at school the friction force depends only on $\mu$ and $N$, not on how large the supporting surface is. A brick resting on its long side or its short side has the same friction force: the pressure changes, not the force.

Getting the sign of Hooke wrong. If you write $F=kx$ and then use that value as a “restoring” force in the direction of the displacement, the direction is wrong. Always draw the arrow: the elastic force points towards the rest position.

Using ${\mu }_s$ when the body is moving (or vice versa). If the body starts, the friction is ${\mu }_{d}N$, not ${\mu }_{s}N$. And if the body stays still, the friction is not ${\mu }_{s}N$, it equals the applied force.

Swapping the normal force and the weight. $N=mg$ only if the surface is horizontal and there are no other vertical forces. If you pull the crate upwards with an inclined rope, $N$ decreases.

Forgetting the conversions. Centimetres to metres for the spring, grams to kilograms for the mass: half the numerical mistakes are there.

Three worked examples

For all: $g=9.8{\text{m/s}}^2$.

Exercise 1 — mass and weight. An object has mass $m=5\text{kg}$. Calculate its weight on Earth and on the Moon ($g_M=1.6{\text{m/s}}^2$). What is its mass on the Moon?

On Earth:

On the Moon:

The mass stays $5\text{kg}$ everywhere: only the weight changes, because $g$ changes. The ratio between the two weights is $49/8\approx 6$, that is, on the Moon the same object weighs about a sixth.

Exercise 2 — measuring $k$ and predicting. A spring hung vertically stretches by $5\text{cm}$ when you attach a mass of $0.5\text{kg}$ to it. Calculate $k$. By how much would it stretch with a mass of $1.5\text{kg}$?

The force that deforms the spring is the weight of the hanging mass:

The extension in metres is $x=0.05\text{m}$, so

With $m=1.5\text{kg}$ the force is $F=1.5\cdot 9.8=14.7\text{N}$ and the extension becomes

Check: the mass has tripled and the extension has tripled too (5 cm $\to$ 15 cm), as the direct proportionality of Hooke predicts.

Exercise 3 — friction: does it stay still or move? A crate of mass $m=20\text{kg}$ rests on a horizontal floor, with ${\mu }_s=0.5$ and ${\mu }_d=0.4$. You push it with a horizontal force. a) What happens with $80\text{N}$? b) And with $120\text{N}$? c) How much force is needed to keep it at constant speed once it has started?

First the normal force. In the vertical there is no acceleration, so

The threshold of static friction is

a) $80\text{N}<98\text{N}$: the crate stays still. Static friction is exactly $80\text{N}$ (not $98$) and the resultant is zero.

b) $120\text{N}>98\text{N}$: the crate starts. Now the friction is kinetic:

The resultant force is $120-78.4=41.6\text{N}$ and the acceleration

c) At constant speed the acceleration is zero, so the push equals the kinetic friction: $F=78.4\text{N}$. It is less than the starting threshold ($98\text{N}$), which is why a crate “breaks away” with effort and then slides more easily.

The check to make every time is on the units: every force result is in N, the acceleration in ${\text{m/s}}^2$, and $\mu$ has no units.

Where to find more

On this topic there is not yet a dedicated simulation: to train with the format of physics tests you can look at the category of simulations for high school, where there are tests on other topics. For a more theoretical revision on vectors, which are the basis of everything else, there is vectors and forces in Year 3, and to understand how the first serious test of a physics course is structured, physics in Year 3 and the first test: they are written for later years, but the method (drawing, axes, equations, units) is the same. The general picture of the subject is on the Physics page.

If in a few weeks you find you know the formulas but cannot get started on the problems, usually all that is missing is the force diagram done properly. We can look at it together in a free introductory call.