M1.5 - Statics of a particle

A force has magnitude and direction, so treat it as a vector. Choose fixed perpendicular axes, resolve every force into signed components, and add corresponding components. The resultant is the single vector with the same combined effect as all the forces.

These component formulae assume that $\theta$ is measured from the positive $x$-direction. If the angle is measured from another axis, use the adjacent component with cosine and the opposite component with sine, then assign signs from the actual directions rather than from the diagram alone.

Forces of $10$ N east and $6$ N at $60^\circ$ north of west act on a particle. Taking east and north as positive,$$\mathbf R=(10-6\cos60^\circ)\mathbf i+(6\sin60^\circ)\mathbf j=7\mathbf i+3\sqrt3\mathbf j.$$Thus $|\mathbf R|=\sqrt{76}=8.72$ N. Its direction is $\tan^{-1}(3\sqrt3/7)=36.6^\circ$ north of east, equivalent to bearing $053.4^\circ$.

A particle is in equilibrium when the vector resultant of all forces acting on it is zero. It may be at rest or moving with constant velocity. Draw the force diagram for one chosen particle, choose two independent axes, and resolve the equilibrium condition along both.

Horizontal and vertical axes are often convenient, but axes parallel and perpendicular to an inclined surface can remove unnecessary components. With three non-parallel forces in equilibrium, their vectors form a closed triangle; resolving remains the safest general method. With only two forces, they must be equal, opposite and collinear.

A $4$ kg particle rests on a smooth plane inclined at $30^\circ$ to the horizontal and is held by a light string parallel to the plane. Perpendicular equilibrium gives$$R-4g\cos30^\circ=0,$$so $R=2\sqrt3g$ N. Parallel equilibrium gives$$T-4g\sin30^\circ=0,$$so $T=2g=19.6\text{ N}.$$The smooth plane contributes no friction.

After solving, verify that each unknown force has a physically possible direction and magnitude. A negative result means the assumed direction was reversed or the proposed contact cannot supply that force. A normal reaction is not automatically $mg$: it is determined by all components perpendicular to the surface.

For a particle in equilibrium on a rough surface, static friction is a response force. Its magnitude adjusts to the value needed to prevent relative motion, up to a maximum set by the coefficient of friction $\mu$ and the normal reaction $R$.

Only when the particle is on the point of sliding is friction limiting, so $F=\mu R$. Its direction opposes the impending motion. If no impending direction is stated, first solve the other equilibrium equations: the sign of the required friction reveals its direction.

A $5$ kg block rests on a horizontal rough floor. It is pulled by a force $P$ at $30^\circ$ above the horizontal and $\mu=0.4$. At impending motion to the right,$$R=5g-P\sin30^\circ,\qquad P\cos30^\circ=0.4R.$$Hence $P(\cos30^\circ+0.4\sin30^\circ)=0.4(5g)$, giving $P=18.4$ N to three significant figures. The angled pull reduces $R$, so using $R=5g$ would overestimate the limiting friction.

Questions asking for the smallest or largest applied force that preserves equilibrium usually have two limiting boundaries. At one boundary the particle is about to move in one direction; at the other it is about to move in the opposite direction, so friction reverses. Set up and solve the two cases separately, then state the permitted interval.