Gravity weight elasticity and friction

Forces in nature are conveniently classified by origin, but they are described in the same way — through Newton's second law and a free-body diagram. Gravity is the gravitational attraction of the Earth; near the surface it gives a body an acceleration $g \approx 9.8\ \text{m/s}^2$, while weight is the force with which a body acts on a support or suspension: $W = mg$. Elastic force appears when a body is deformed and tends to restore its original shape; for small deformations its magnitude obeys Hooke's law $F = kx$, where $k$ is the stiffness and $x$ is the extension. Friction opposes relative sliding: static friction varies from zero up to $\mu_s N$, whereas kinetic friction is approximately $\mu_k N$ and is directed against the motion. The normal force $N$ is always perpendicular to the surface and is found from equilibrium in that direction, not automatically equal to $mg$. First draw all forces on a free-body diagram, then project them onto the axes and write the equilibrium or motion equations.

A body of mass $m = 2.0\ \text{kg}$ rests on a horizontal table. Find the weight of the body and the normal force from the table. Take $g = 9.8\ \text{m/s}^2$.

A block of mass $5.0\ \text{kg}$ rests on a horizontal surface. The coefficient of static friction is $\mu_s = 0.40$ and the coefficient of kinetic friction is $\mu_k = 0.30$. A horizontal force $F = 10\ \text{N}$ is applied to the block. Will the block move, and what is the friction force?

A block of mass $m = 2.0\ \text{kg}$ slides down an inclined plane with angle $\theta = 30^\circ$; the coefficient of kinetic friction is $\mu_k = 0.25$. The block is attached to a spring of stiffness $k = 100\ \text{N/m}$ lying along the plane. Find the extension of the spring if the block moves with constant velocity. Take $g = 9.8\ \text{m/s}^2$.

A safe self-check in a simulation. Open the PhET simulation "Masses and Springs" or "Forces and Motion: Basics". 1) By hanging different masses, plot the spring extension against the mass and find the stiffness $k$ from the slope; verify that $mg = kx$. 2) On a horizontal surface, vary the applied force and the friction coefficient, and note the moment the object starts moving: check that static friction grows linearly up to $\mu_s N$ and then is replaced by the almost constant kinetic friction $\mu_k N$. 3) Tilt the surface and repeat the measurements: verify that $N = mg\cos\theta$ and that the component of gravity along the plane is $mg\sin\theta$. Safety: do not hang real weights above your feet, glass, or a hard floor, and do not stretch springs to their limit — use the simulation or light, safe weights on a sturdy support.

To show that gravity, elastic force, friction, and the normal force are forces of different origins acting simultaneously on one body, and how their magnitudes and directions are related.

A free-body diagram of a block on an inclined plane attached to a spring: arrows $m\vec g$, $\vec N$, $\vec F_{\text{el}}$, and $\vec F_{\text{fr}}$ with labels and the angle $\theta$. Next to it, a graph of friction force versus applied horizontal force: the linear static-friction region up to $\mu_s N$, its peak, and a horizontal kinetic-friction line at $\mu_k N$.

An official basketball has a mass of about $m=0.60\ \text{kg}$. A player holds it still in the palm for a second. Earth pulls the ball downward, while the palm pushes upward. If the ball is released, it starts falling with an acceleration of about $g=9.8\ \text{m/s}^2$ (air resistance is still small).

Here the weight law $W=mg$ applies: with $m=0.60\ \text{kg}$ and $g=9.8\ \text{N/kg}$, the weight is $W=5.88\ \text{N}$. In a free-body diagram this is the gravitational force vector directed toward Earth's center; while the ball is at rest, the net force of weight and palm force is zero.

A cyclist together with the bicycle has a mass of $m=80\ \text{kg}$. On dry asphalt the coefficient of friction between rubber and the road is about $\mu_s=0.8$. Before a traffic light, the cyclist brakes sharply but without locking the wheels; friction between the tires and the road prevents sliding forward.

Here dry friction is used: the normal force from the road equals the weight $N=mg=80\cdot9.8=784\ \text{N}$, so the maximum static friction is $F_{s,\max}=\mu_s N=0.8\cdot784\ \text{N}=627\ \text{N}$. This force points backward and during braking produces a deceleration $a=F_{s,\max}/m=7.84\ \text{m/s}^2$; if the wheel locks, the interaction switches to kinetic friction $F_k=\mu_k N$.

A bag of flour of mass $m=2.0\ \text{kg}$ rests on a kitchen spring scale. The spring stretches by $x=4.0\ \text{cm}$ from the zero position and indicates weight. If a bag of mass $m=1.0\ \text{kg}$ is placed on it, the stretch becomes about $x=2.0\ \text{cm}$.

Here Hooke's law for elastic deformation is applied: $F_{\text{el}}=kx$. For $m=2.0\ \text{kg}$ the weight is $F=mg=19.6\ \text{N}$, the stretch is $x=4.0\ \text{cm}=0.040\ \text{m}$, so the stiffness is $k=F/x=490\ \text{N/m}$. For $m=1.0\ \text{kg}$ the result is $x=9.8/490=0.020\ \text{m}=2.0\ \text{cm}$; this holds as long as the spring has not passed its elastic limit.