Forces and equilibrium

4.1 Forces and equilibrium

Forces to look for

Start every mechanics question with a clear diagram showing each force on the object.

Components and resultants

A force $F$ at angle $\theta$ to a direction has a component $F\cos\theta$ along it and $F\sin\theta$ at right angles to it.

To combine forces, find the total $X$ and the total $Y$:

$X = 10\cos 30^\circ + 6 = 14.66$ and $Y = 10\sin 30^\circ = 5$.

The resultant is $\sqrt{14.66^2 + 5^2} = 15.5$ N, at $\tan^{-1}\dfrac{5}{14.66} = 18.8^\circ$ above the horizontal.

Equilibrium of a particle

Resolve in two directions at right angles. Choose directions that leave the fewest unknowns: along and perpendicular to a slope, or horizontally and vertically on level ground.

Resolving horizontally: $T_1\sin 30^\circ = T_2\sin 60^\circ$.

Resolving vertically: $T_1\cos 30^\circ + T_2\cos 60^\circ = 20$.

The first gives $T_1 = \sqrt{3}\,T_2$. Then $\tfrac{3}{2}T_2 + \tfrac{1}{2}T_2 = 20$, so $T_2 = 10$ N and $T_1 = 17.3$ N.

Contact forces and friction

A contact force between two surfaces has two components: the normal force $R$ and the friction force $F$.

A smooth contact has no friction. It is a model: real surfaces are never exactly smooth, but friction may be small enough to ignore.

For a rough contact, friction opposes the direction the object would otherwise move, and it grows only as large as it needs to be, up to a limit:

Limiting equilibrium means the object is on the point of moving, so $F = \mu R$.

Inclined planes

On a slope at angle $\theta$, resolve along and perpendicular to the slope:

$\text{down the slope: } mg\sin\theta \qquad \text{into the slope: } mg\cos\theta$

Perpendicular: $R = mg\cos 30^\circ$. Along: $F = mg\sin 30^\circ$. With $F = \mu R$:

$\mu = \frac{\sin 30^\circ}{\cos 30^\circ} = \tan 30^\circ = 0.577$

A force at an angle

A pull at an angle lifts part of the weight off the surface, which reduces $R$ and so reduces the friction. A push downwards at an angle does the opposite.

Vertically: $R + P\sin 30^\circ = 40$, so $R = 40 - 0.5P$.

Horizontally: $P\cos 30^\circ = F = 0.5R = 20 - 0.25P$.

So $1.116P = 20$, and $P = 17.9$ N.

Friction either way

An object on a slope held by a force $P$ may be about to slide down or about to be pushed up. Friction reverses between the two, so these give the least and greatest values of $P$:

Newton's third law

When two objects touch, the force each exerts on the other is equal in size and opposite in direction. These two forces act on different objects, so they never cancel in one object's equation.

The weight is the Earth's pull on the book. Its partner is the book's pull on the Earth. The normal reaction on the book is a different force, and it is paired with the book's push on the table.