Learning Lab

Newtons 1st law of motion

Every object continues to be at rest, or continues with constant velocity, unless it experiences an unbalanced force.

Seatbelts are fitted in cars to take into account Newtons 1st Law of Motion. If, for instance, you had to brake suddenly and you were not wearing your seatbelt, then by Newtons 1st Law, you would continue to travel at a uniform speed – the speed of the car just before braking – until you made contact with the windscreen. That is, the sum of the forces acting on you was zero at the instant you braked even though the car was being acted upon by a braking force. The windscreen would then exert a force on you as you hit it. As it did to the author of this worksheet when he was 13 years old.

Newtons 2nd law of motion

This law relates to the (sum of \(\Sigma\) ) the (forces F) written as \(\Sigma F\) s acting on an object and the acceleration produced as a consequence of this resultant force. It is given by:

Sum of the Forces = \(\Sigma F=ma\)

where \(\Sigma F\) is the sum of the forces (in Newtons), ‘m’ is the mass of the object (in\(kg\)), and ‘a’ is the acceleration (in \(ms^{-2}\)). Note: \(\Sigma F\) must have the same direction as acceleration as force is also a vector.

When doing Newtons 2nd law problems, draw all forces acting on the object in question.

Example 1

A toy car of mass \(2kg\) has a driving force \(F_{D}\) of \(20N\). The frictional force \(F_{F}\) acting on the car is \(10Newton\). Find the cars acceleration

Solution:

The net force acting on the car is the driving force minus the frictional force: \[\begin{align*} \Sigma F & =F_{D}-F_{F}\\ & =20-10\\ & =10N. \end{align*}\] Using Newton’s 2nd law11 The second law is: \(\Sigma F=ma\). we get \[\begin{align*} ma & =10\\ 2a & =10\\ a & =5ms^{-2}. \end{align*}\] The magnitude of the car’s acceleration is \(5ms^{-2}\).

Newton’s 3rd law of motion

When two objects interact, the forces they exert on each other are equal in size, but opposite in direction. These pairs of forces are called action-reaction forces.

You need to appreciate that each force in the action-reaction pair acts on different objects. This is different to Newtons 2nd law where you analyze forces that are acting on the same object. Note that each force is equal in size but opposite in direction to its paired force.

Objects in Lifts

• If accelerating down (as shown): \[\begin{align*} \Sigma F & =ma\\ mg-F & =ma \end{align*}\]

Open imageAccelerating down then gravity is winning or larger. Accelerating up then Normal force is winning or larger. Acceleration has magnitude and is directional , it is a vector. where \(g\) is the acceleration of gravity (see diagram)

• If accelerating up: \[\begin{align*} \Sigma F & =ma\\ F_{N}-mg & =ma \end{align*}\]

• If not accelerating at all:22 It is stationary or moving at a constant velocity. \[\begin{align*} \Sigma F & =0\\ F_{N}-mg & =0\\ F_{N} & =mg \end{align*}\]

Objects Bouncing

During the bounce (the instant the ball is in contact with the surface):

• \(F_{N}\) > mg, because a > 0 and is upwards. Therefore there is a resultant force acting which is upwards (as the ball will now head up).
• \(F_{N}\) can often be much larger than mg (for the ball to bounce).
• At the instant the object is stationary

Open image(just before it changes direction and heads up)

\[\begin{align*} \Sigma F & =0\\ F_{N}-mg & =0\\ F_{N} & =mg \end{align*}\]

Object being pulled along a surface

\(T\) = tension in rope , \(a\) = acceleration of object , \(m\) = mass of object, \(F_{N}\) is the normal force of the table pushing on the block, \(mg\) is the force of the block pushing on the table. Remember that \(g\) is acceleration of gravity, \(F_{f}\) is the force of friction the table exerts on the object

Horizontally: \(\Sigma F\)= \(T\) \(=\) \(ma\)

Vertically: \(\Sigma F=F_{N}-mg=0\)

Therefore \(F_{N}=\) \(mg\)

Horizontally: \(\sum F\)= T - \(F_{f}\)=\(ma\)

Vertically: \(\sum F=F_{N}-mg=0\)

Therefore \(F_{N}=\) \(mg\)

Example:

A car of mass 800kg accelerates from rest to \(20\) \(ms^{-1}\) in \(\text{$8.0\sec$ }\). The resistance forces acting on the car total \(1000N\).

Find (a) the acceleration of the car, and (b) the driving force of the car.

1. Acceleration is given by the change in velocity divided by the time taken, or

Open imageremember “from rest” means \(u=0\), final velocity is \(v=20\) and time \(t=8\)

\[\begin{align*} a & =\frac{v-u}{t}\\ & =\frac{20-0}{8}\\ & =2.5ms^{-2} \end{align*}\]

2. Net Force \(\Sigma F=F_{D}-F_{F}=ma\)

Therefore \[\begin{align*} F_{D}-F_{F} & =m\times a\\ F_{D}-1000 & =800\times2.5\\ F_{D} & =800\times2.5+1000\\ F_{D} & =2000+1000\\ F_{D} & =3000N \end{align*}\]

The driving force of the car is therefore \(3000\) Newton.