11Cexamples


 * Applications of Newton's 2nd Law**

**Example 1** A block of mass 5kg is pulled along a horizontal plane by a force of 20 Newtons inclined at 30º to the horizontal. The coefficient of friction between the block and the plane is 0.3. What is the acceleration of the block, correct to 1 decimal place?


 * //Resolving vertically://**

... ... Vertical forces must add to zero (or cancel each other out).

math \\ . \qquad N+20\sin30^\circ = 5g \\ .\\ . \qquad N+10 = 5g \\. \\ . \qquad N = 5g-10 \qquad \textbf{(1)} math


 * //Resolving horizontally to find R://**

math \\ . \qquad R=ma \qquad \qquad \textbf{(2)} \\. \\ . \qquad R=20\cos30^\circ - Fr \qquad \textbf{(3)} \\. \\ . \textbf{(2)}=\textbf{(3)} \\. \\ math math \\ . \qquad 5a=20\cos30^\circ-\mu N \\. \\ . \qquad 5a=20\cos30^\circ-0.3(5g-10) \qquad \text{using } \textbf{(1)} \\. \\ . \qquad a=4\cos30^\circ-0.06(5g-10) \\. \\ . \qquad a=1.1 \; m/s^2 math

**Example 2** A body of mass 10kg is on a rough plane inclined at an angle of alpha to the horizontal, where sin alpha is 3/5. A force, __P__ is acting up the plane and the object is on the point of moving up the plane. The coefficient of friction is 0.2. Find the magnitude of __P__.

math \\ . \qquad \text {Given that } \; \sin\alpha=\frac{3}{5} \\. \\ . \qquad \text {we get }\; \cos\alpha=\frac{4}{5} math

Note that if the plane is at an angle of alpha to the horizontal then the angle from the vertical down the perpendicular is also alpha {see second diagram}.

The easiest way to do these questions is to resolve parallel to the plane and perpendicular to the plane.

math \\ . \qquad N=10g \cos\alpha \\. \\ . \qquad N=8g \; \text{Newtons} math
 * //Resolve perpendicular to the plane://**


 * {8g Newtons is sometimes called 8 kilogram-weight or 8 kg wt}


 * //Resolve Parallel to the plane//**

math \\ . \qquad P=Fr+10g \sin\alpha \\ .\\ . \qquad P=\mu N + 10g \sin\alpha \\. \\ . \qquad P=0.2 \times 8g+6g \\ .\\ . \qquad P=7.6g \; \text{Newtons} \qquad \textit{(or 7.6 kg wt)} \\. \\ . \qquad P=74.48 \; \text{Newtons} math

A 35 kg girl is holding the strings of 15 helium balloons so that she is on the point of being lifted off the ground. ... ... a) Calculate the lifting force exerted by the balloons ... ... b) Hence calculate the lifting force exerted by a single balloon.
 * Example 3 **

The girl then adds another 5 helium balloons. ... ... c) Calculate the new lifting force being exerted by the balloons ... ... d) Hence find the girl's acceleration as she is lifted into the air.

She is on the point of being lifted, so
 * Solution:**
 * Normal Reaction force (from the ground) is __zero__
 * Lifting Force from the balloons __equals__ weight force of the girl

a) Lifting Force
 * LF 15 = 35g
 * LF 15 = 343 Newtons

b) Single Balloon
 * LF 1 = 343 ÷ 15
 * LF 1 = 22.87 Newtons

c) Now 20 balloons
 * LF 20 = 22.87 × 20
 * LF 20 = 457.33 Newtons

d) Resultant Force = Lifting Force - Weight Force .... .... ..... ..... .... = 457.33 - 343 .... .... ..... ..... .... = 114.33 Newton


 * Using R = ma
 * 35a = 114.33
 * a = 114.33 ÷ 35
 * a = 3.27 m/s 2.


 * FUN FACT **


 * Helium actually has a lifting force able to support 1 gram per litre (1000 cm 3 )


 * Therefore to support our 35kg girl (on the point of lifting), we would need
 * 35,000 litres of helium or 35,000, 000 cm 3.

math \\ .\qquad V = \frac{4}{3} \pi r^3 \\. \\ . \qquad \; \; = 14137 \; cm^3 \\. \\ . \qquad \; \; = 14.13 \; litres math
 * A typical party balloon with a diameter of 30 cm ( radius = 15 cm) has Volume:


 * So a helium party balloon can support about 14 grams
 * Our 35kg girl would therefore need 35,000 ÷ 14 = 2500 party balloons just to be on the point of lifting.


 * With 15 balloons, each balloon is holding 35,000 ÷ 15 = 2,333 grams.


 * So each balloon must have a volume of 2,333,000 cm 3.

math \\ . \qquad V = \frac {4}{3} \pi r^3 \\. \\ . \qquad 2,333,000 = \frac {4}{3} \pi r^3 \\. \\ . \qquad r^3 = 556963 \\. \\ . \qquad r = 82 cm math

.
 * So each balloon in the example above must have had a diameter of 1.64m