Free-body diagrams
Draw only the forces acting on the object itself, and most mechanics problems solve themselves.
How to draw one
Shrink the object to a dot and draw one arrow for each external force acting on it, with tails starting at the dot and lengths reflecting relative sizes. Include only forces on the object, never forces the object exerts on other things, and never add a separate arrow for ma, which is the result of the forces, not one of them.
\sum \vec{F}_{\text{on object}} = m\vec{a}Normal force and tension
The normal force is a surface pushing perpendicular to itself, and it adjusts itself to whatever makes the perpendicular motion come out right, so it is not always equal to mg. In an elevator accelerating up at 2 m/s^2, the floor must push with m(g + 2). Tension in a rope pulls along the rope toward both ends, and a rope around an ideal pulley has the same tension throughout.
N - mg = ma \implies N = m(g+a)Inclines and components
On a slope, tilt your axes so x runs along the incline and y is perpendicular to it. The weight then decomposes into mg sin(theta) down the slope and mg cos(theta) into the surface. For a 10 kg block on a 30 degree incline, those components are 98 sin(30) = 49 N along the slope and 98 cos(30) = 84.9 N into the surface, which is what sets the normal force on a frictionless incline.
mg\sin\theta \ \text{(along slope)}, \quad mg\cos\theta \ \text{(into surface)}Flashcard
Normal force in an upward-accelerating elevator
Check your understanding
A 50 kg person stands in an elevator accelerating upward at 2.0 m/s^2. What does the scale read? Use g = 9.8 m/s^2.