From Mechanics

Theory

Exercises

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Key Points

Newton's Second Law

If the resultant force \displaystyle F on a particle of mass \displaystyle m is not zero, then it will have an acceleration \displaystyle a given by,

\displaystyle F=ma

If the acceleration is not known one assumes a direction for the acceleration \displaystyle a and applies this law in that direction.

The vector form of this equation is

\displaystyle \mathbf{F}=m\mathbf{a}

If there is friction the friction equation holds.

\displaystyle F = \mu R

Newton's Second Law is sometimes referred to as the Law of Motion.

Example 10.1

A lift and its passengers have a total mass of 300 kg. Find the tension in the lift cable if;

a) it accelerates upwards at 0.2 \displaystyle \text{m}{{\text{s}}^{-2}},

b) it accelerates downwards at 0.05 \displaystyle \text{m}{{\text{s}}^{-2}}.

Solution

The lift and its passengers have been modelled as a single particle.

The diagram shows the forces acting on the lift.

Assume that the lift is accelerating upwards as indicated in the diagram.

Consider the resultant force on the particle, taking the upward direction as positive.

Resultant Force \displaystyle =T-2940

Applying Newton’s second law gives:

\displaystyle \begin{align} & 300a=T-2940 \\ & T=2940+300a \\ \end{align}

a) If \displaystyle a=0\textrm{.}2 , then \displaystyle T=2940+300\times 0\textrm{.}2=3000\text{ N} .

b) If \displaystyle a=-0\textrm{.}05 (since the lift is accelerating downwards then:

\displaystyle T=2940+300\times (-0\textrm{.}05)=2925\text{ N} .

Example 10.2

A car accelerates at 1.8 \displaystyle \text{m}{{\text{s}}^{-2}} along a straight horizontal road. The mass of the car is 1200 kg. A forward force of magnitude 3000 N acts on the car. Find the magnitude of the resistance force that also acts on the car.

Solution

The diagram shows the horizontal forces that are acting on the car, which has been modelled as a particle. The acceleration has also been included.

First find the resultant force acting on the car.

Resultant Force \displaystyle =\text{ }3000-R

Then apply Newton’s Second Law, \displaystyle F=ma , to give:

\displaystyle \begin{align} & 3000-R=1200\times 1\textrm{.}8 \\ & R=3000-2160 \\ & =840\text{ N} \end{align}

The resistance force has magnitude 840 N.

Example 10.3

A van has mass 2500 kg. A forward force of 5000 N acts on the van and a resistance force of 2000 N also acts. Find the acceleration of the van on a horizontal surface.

Solution

The diagram shows the horizontal forces that are acting on the van, which has been modelled as a particle.

First find the resultant force acting on the car.

Resultant Force \displaystyle =\text{ }5000-2000=3000\text{ N}

Then apply Newton’s Second Law, \displaystyle F=ma , to give:

\displaystyle \begin{align} & 3000=2500a \\ & a=\frac{3000}{2500}=1\textrm{.}2\text{ m}{{\text{s}}^{\text{-2}}} \end{align}

Example 10.4

A van, of mass 1200 kg, rolls down a slope, inclined at \displaystyle 3{}^\circ to the horizontal and experiences a resistance force of magnitude 400N. Find the acceleration of the car.

Solution

Model the van as a particle.

The diagram shows the forces acting on the van as it rolls down the slope.

The forces perpendicular to the slope will be in equilibrium.

The resultant force will be directed down the slope, and is found by resolving the forces parallel to the slope.

Resultant Force \displaystyle =11760\sin 3{}^\circ -400

Then apply Newton’s Second Law, \displaystyle F=ma , to give:

\displaystyle \begin{align} & 11760\sin 3{}^\circ -400=1200a \\ & a=\frac{11760\sin 3{}^\circ -400}{1200}=0\textrm{.}180\text{ m}{{\text{s}}^{\text{-2}}}\text{ (to 3 sf)} \\ \end{align}

Example 10.5

A block, of mass 3 kg, is pulled across a rough horizontal plane, by a string inclined at \displaystyle 30 ^\circ to the horizontal. The tension in the string is 20 N. The coefficient of friction between the particle and the plane is 0.2. Find the acceleration of the particle.

Solution

Model the crate as a particle.

The diagram shows the forces acting on the crate.

The vertical components of the forces will be in equilibrium.

\displaystyle \begin{align} & R+20\sin 30{}^\circ =29\textrm{.}4 \\ & R=29\textrm{.}4-20\sin 30{}^\circ \\ \end{align}

As we know R, we can calculate F using \displaystyle F=\mu R . This gives:

\displaystyle \begin{align} & F=0.2\times \left( 29\textrm{.}4-20\sin 30{}^\circ \right) \\ & =5\textrm{.}88-4\sin 30{}^\circ \end{align}

Now resolving horizontally, the resultant force can be found.

Resultant Force

\displaystyle \begin{align} & =20\cos 30{}^\circ -F \\ & =20\cos 30{}^\circ -(5\textrm{.}88-4\sin 30{}^\circ ) \\ & =20\cos 30{}^\circ -5\textrm{.}88+4\sin 30{}^\circ \end{align}

Then apply Newton’s Second Law, \displaystyle F=ma , to give:

\displaystyle \begin{align} & 20\cos 30{}^\circ -5\textrm{.}88+4\sin 30{}^\circ =3a \\ & a=\frac{20\cos 30{}^\circ -5\textrm{.}88+4\sin 30{}^\circ }{3}=4\textrm{.}48\text{ m}{{\text{s}}^{\text{-2}}}\text{ (to 3 sf)} \\ \end{align}