# Simple machine question and answer 1

. A machine has a velocity ratio 5 and 80% efficient. What effort should be needed to lift a load of 200N with the aid of the machine? $Efficiency\quad =\frac { M.A }{ V.R }$

V.R = 5, Efficiency = 80% $and\quad M.A\quad =\frac { Load }{ effort }$ $so\quad \quad \frac { 80 }{ 100 } =\frac { M.A }{ 5 } \quad \quad =\frac { 80\times 5 }{ 100 } \quad =M.A$ $Therefore\quad \quad M.A\quad =\quad 4,\quad but\quad M.A\quad =\frac { load }{ effort } \quad =\frac { 200 }{ effort }$ $\frac { 80\times 5 }{ 100 } =\frac { 200 }{ effort } ,\quad effort\quad =\frac { 200\times 100 }{ 80\times 5 } =\quad 50N$

Effort needed = 50N

# Simple machine question and answers 2

What does it mean that the M.A is 5? $it\quad means\quad that\quad \frac { load }{ effort\quad } =5,\quad or\quad ratio\quad of\quad M.A\quad to\quad V.R\quad =5:1$

# Simple machine question and answer 3

What does it mean that the V.R  is 4? $it\quad means\quad that\quad \frac { distance\quad moved\quad by\quad the\quad effort }{ distance\quad moved\quad by\quad the\quad load\quad } \quad =4$

Or it means that the distance moved by the effort is four times greater than the distance moved by the load.

# Simple machine question and answers 4

A machine with V.R 5 require  10000J of work to raise a load of 500N through a vertical distance of 1.5m. find the efficiency and the mechanical advantage M.A of the machine. $Efficiency\quad =\frac { work\quad output }{ work\quad input } \quad \times 100%$ $Eficiency\quad =\quad \frac { load\quad \times distance\quad moved\quad by\quad the\quad load }{ work\quad inpu } \times 100%$ $therefore\quad efficiency\quad =\quad \frac { 500\times 1.5 }{ 1000 } \times 100%\quad \quad = 75%$

So efficiency = 75%

To solve M.A $Use\quad efficiency\quad =\frac { M.A }{ V.R }$

Where  V.R = 5 , efficiency = 75% $\therefore \quad M.A\quad =\frac { 5 }{ 75 } \times 100 =3.75$

M.A = 3.75

# Simple machine question and answers 5

Simple machine are classified in to three. Name them

They are First order, second Order and third order levers.

# Simple machine question and answer 6

A block and tackle consisting of 5 pulleys is used to raise a load of 400N through height of 10m. if the work done against friction is 1000J, calculate

(a). the work done by the effort

(b). the efficiency of the machine

(c). the effort applied

(a). the work done by the effort is equal to work done against friction + work done in raising the load

ie   1000 + 400 x 10 = 5000N

(b). $the\quad efficiency\quad =\frac { work\quad output }{ work\quad input } \times 100%\quad =\frac { 400\times 10 }{ 5000 } \times 100%$ $=\quad \frac { 4000 }{ 5000 } \times 100%=80%$

Efficiency = 80%

(c). Note that for block and tackle that has 5 pulleys the V.R = 5 (velocity ratio) $efficiency\quad =\frac { M.A }{ V.R } \times 100\quad %$

Where V.R = 5, efficiency = 80% $therefore\quad M.A\quad \frac { 5\quad \times \quad 80 }{ 100 } \quad =\quad 4,\quad but\quad M.A\quad =\frac { load }{ effort }$ $\therefore \quad effort\quad =\frac { 400 }{ 4 } \quad =\quad 100N$

Effort = 100N

# Simple machine question and answer 7

(7). What is the velocity ratio of an inclined plane of length 6m, if the higher end is 2m above the ground ? if the efficiency of this machine used is 60%. Calculate the mechanical advantage M.A. $V.R\quad =\quad \frac { lenght\quad of\quad the\quad plane }{ hight\quad ofthe\quad plane } =\quad \frac { 6 }{ 2 } =\quad 3$ $Using\quad \varepsilon =\frac { M.A }{ V.R } \times 100\quad$

Where Ɛ=60% and V.R = 3 $\therefore \quad 60\quad =\quad \frac { M.A }{ 3 } \times 100\quad$ $\therefore \quad M.A\quad =\frac { 60\times 3 }{ 100 } =\quad \frac { 180 }{ 100 } \quad =1.8$

M.A =1.8. $Note\quad that\quad the\quad velocity\quad ratio\quad of\quad inclined\quad plane\quad =\frac { 1 }{ sin\theta } \quad if\quad there\quad is\quad no\quad friction.\quad \theta \quad =\quad angle\quad of\quad inclination.\\$

# Simple machine question and answers 8

(8). The pitch of the screw of a screw of 0.5cm. When used to raise a load, the handle turns through a circle of radius 40cm. What is the mechanical advantage of the jack if its efficiency is 25%? (π =3.14)

First let find the V.R $V.R\quad =\frac { 2\pi a }{ Pitch }$

Where Pitch is the distance between consecutive threads =0.5cm, and a = the radius of the thread=40cm. $So\quad V.R\quad =\frac { 2\times 3.14\times 40 }{ 0.5 } =502.4$ $Using\quad Efficiency=\frac { M.A }{ V.R } \times 100\\ \\ \\ 100M.A\quad =Efficiency\times V.R\\ \\ \\ M.A=\frac { Eficiency\times V.R }{ 100 } =\frac { 25\times 502.4 }{ 100 } \quad =125.6$

# Simple machine question and answers 9

. A wheel and axel is used to raise a man of weight 700N by application of effort 200N. If the radii of the wheel and axle are 400mm and 100mm respectively, determine the efficiency of machine. $V.R\quad =\frac { a }{ b } =\frac { 400 }{ 100 } =4\\ \\ M.A\quad =\frac { load }{ effort } =\frac { 700 }{ 200 } =\frac { 7 }{ 2 } =3.5\\ \\ Efficiency\quad =\frac { M.A }{ V.R } \times 100\quad =\frac { 3.5\times 100 }{ 4 } =\frac { 350 }{ 4 } =87.5$

Efficiency =87.5%

# Exercise

(1). An inclined plane of 300 is used as simple machine . What is it Velocity Ratio. V.R?

(2). The pitch of a screw Jack is 0.5cm. The arm is 50cm long and its mechanical advantage is 250. What is its efficiency? ( take π = 3.142)

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