From Mechanics

Theory

Exercises

Key Points

Newton's First Law

If the resultant force on a particle is zero, then it will move with constant velocity or remain at rest.

Note that the reverse is also true, that if a particle moves with constant velocity or remains at rest the resultant force on the particle must be zero.

Friction exists where rough surfaces move relative to one another and satisfies the equation

\displaystyle F = \mu R

Here \displaystyle F is the friction force.

The coefficient of friction is \displaystyle \mu.

The normal reaction force is \displaystyle R.

Example 9.1

A helicopter of mass 880 kg is rising vertically at a constant rate. Find the magnitude of the lift force acting on the helicopter. How would your answer change if the helicopter was descending at a constant rate?

Solution

As the helicopter is rising at a constant rate the forces on it must be in equilibrium.

The lift force is 8624 N.

If the helicopter is descending at a constant rate the lift force will be 8624 N.

Example 9.2

A cyclist, of mass 75 kg, freewheels down a slope inclined at to the horizontal at a constant speed. Find the magnitude of the resistance force acting on the cyclist.

Solution

The cyclist and cycle is modelled as a particle.

The diagram shows the forces acting, where the resistance force has magnitude P.

Resolving parallel to the slope gives:

\displaystyle P=735\sin 6{}^\circ =76\textrm{.}8\text{ N}

Example 9.3

A skier, of mass 60 kg, skis down a slope inclined at to the horizontal at a constant speed. If the coefficient of friction between her skis and the slope is 0.1, find the magnitude of the air resistance force acting on her.

Solution

The skier is modelled as a particle.

The diagram shows the forces acting, where the resistance force has magnitude .

Resolving perpendicular to the slope gives:

\displaystyle R=588\cos 30{}^\circ

Then using \displaystyle F=\mu R , because the skier is sliding gives:

\displaystyle

Resolving parallel to the slope gives:

\displaystyle \begin{align} & F+P=588\sin 30{}^\circ \\ & P=588\sin 30{}^\circ -F \\ & =588\sin 30{}^\circ -58\textrm{.}8\cos 30{}^\circ \\ & =243\text{ N} \end{align}