Introduction to simple machines - class-X

Graphite powder is used in some machines to reduce friction. Ball bearing coverts sliding friction into rolling friction.

The mechanical advantage increases with decrease in height of inclined plane.

The mechanical advantage of an inclined plane is given by length/height. As the height decreases, the ratio length/height increases, thus increasing the mechanical advantage.

In an ideal machine there will be zero energy loss hence output will be equal to input

The mechanical advantage (MA) of any machine, including a lever, is defined as the ratio of the load force overcome to the effort force applied. Therefore, it is expressed as load divided by effort.

A wedge is a simple machine consisting of two inclined planes joined back to back. It is used to split or cut objects by converting force applied to its blunt end into forces perpendicular to its inclined surfaces.

Mechanical advantage (MA) is the ratio of the load to its effort.
efficiency of machine (n) = MA/VR
Velocity Ratio (VR) is the ratio of the velocity of effort to velocity of load.

For ideal machine, efficiency = 1

For a practical machine, velocity ratio is greater than mechanical advantage
V.R > M.A.

Ratio of load to the effort is k/a mechanical advantage of the machine. The ratio of the velocity of effort to its velocity of the load is called its velocity ratio of machine.

Ratio of the velocity of effort to the velocity of load is called velocity ratio of machine. Ratio of the load to its effort is called mechanical advantage of machine.

A steeper incline (greater inclination) effectively acts as a shorter inclined plane for a given horizontal distance, which reduces the mechanical advantage. Consequently, more effort is required to climb it.

VR = Velocity of effort / velocity of load

Mass given $= 400\  kg$

Load = $w = 200N$
Distance moved by the load = $5cm = .05m$
$Effort (P) = 10N$
$n(Efficiency) = 50%$
$MA = Load/Effort = W/P=200N/10N=20$

Given:  $VR=2$


From the relation $VR=\dfrac{\text{Distance moved by effort}}{\text{distance moved by load}}$


Using formula we get distance moved by the load is $2.5m$

Velocity ratio = $\displaystyle \frac { Distance\quad moved\quad by\quad the\quad effort }{ Corresponding\quad distane\quad moved\quad by\quad the\quad load } $
It only depends on the design of machine and is always same for a particular machine.

As no machine is perfectly efficient.
So, we define efficiency as efficiency = $\displaystyle \frac { Energy\quad input }{ Energy\quad output } \times 100$
Also Efficiency = $\displaystyle \frac { mechanical\quad advantage }{ velocity\quad ratio } \times 100$
A petrol engine is about 30% efficient, a diesel engine is 40% efficient.

A machine in which displacement of load is more than the displacement of effort, will have the velocity ratio more than 1.

Friction has effect on Mechanical Advantage and efficiency but not Voltage Ratio.

$\eta=\frac{M.A}{V.R}=\frac{l}{h(l/h)}=1$
$M.A=\frac{l}{h}, V.R=\frac{l}{h}$

An inclined plane is a sloping surface that behaves like a simple machine whose mechanical advantage is always greater than $1$.

The vertical force is the gravitational force means weight of the load. 

Mechanical advantage is the ratio of the load lifted to the effort applied. It is expressed as load divided by effort.

A crowbar is a classic example of a first-class lever, where the fulcrum is between the effort and the load. Therefore, stating it is a second-class lever is incorrect.

Mechanical advantage is the amplification in force due to the use of a machine . The mechanical advantage of wheel and axle is given by 

Efficiency is the (often measurable) ability to avoid wasting materials, energy, efforts, money, and time in doing something or in producing a desired result. In a more general sense, it is the ability to do things well, successfully, and without waste. In more mathematical or scientific terms, it is a measure of the extent to which input is well used for an intended task or function (output).

A wheelbarrow is example 3 class third lever rest all are example of simple machine. hence option D is incorrect

C lever. The edge of the can serves as the fulcrum and the lid is the load.

The velocity ratio is define as $VR=\dfrac{{Effort}{} _{distance}}{{load}{} _{distance}}$
hence option A is correct.

the distance between fulcrum and effort is called effort distance.
Option A is correct

The mechanical advantage of each machine is determined by  multiplying the mechanical advantage for each simple machine.
Option D is correct

The $effort=\dfrac{MA}{load}=\dfrac{35lb}{3}=11.7lb$

For an ideal machine the Velocity ratio is always equal to Machanical advantage.

$VR=MA$

Hence option D is correct

$\eta=\dfrac{load}{effortVR}100$

$load=\etaeffortVR=0.61218=130N$

Let us take first case,

In an ideal pulley system, the mechanical advantage is equal to the number of segments or strands of rope actively supporting the resistance force (load). Counting these strands gives the direct multiplier for the input force.

The VR is define as ratio of distance travelled by effort:distance travelled by load

The VR of second order lever is always grater than 1 because effort arm is always longer than load arm.

Work input = 500 N * 20 m = 10,000 J. Work output = 10,000 N * 0.8 m = 8,000 J. Work lost to friction = 10,000 - 8,000 = 2,000 J. Effort lost = 2,000 J / 20 m = 100 N.

$effiency=\eta=\dfrac{Machanical-advantage}{velocity-ratio}*100$

In a first-class lever, the fulcrum position can be adjusted. If the effort arm is longer than the load arm, the mechanical advantage (and velocity ratio) is greater than 1; if shorter, it is less than 1; if equal, it is 1.

$\eta=\dfrac{Load}{effort}*100=\dfrac{1000}{1200}*100=83.33%$

$\eta=\dfrac{Load}{effort}*100=\dfrac{1200}{1260}*100=95.23%$

The vertical plane $W=mgh$
for inclined plane $W'=mghsin(Q)$
$sinQ<1$
$W>W'$
hence only one part of weight is to overcome

Mechanical advantage of an inclined plane at an angle $\theta$ is  $ \dfrac{Length }{height}$$ = \dfrac{l}{h} = \dfrac{1}{sin \theta}$

For a wheel and axle, the mechanical advantage is the ratio of the radius of the wheel (R) to the radius of the axle (r).

Force required to roll the load up the ramp (F) $=\displaystyle \frac{load \times height}{distance}$
$= 800 \displaystyle \times \frac{2}{4} = 400 N$

The mechanical advantage of a wheel and axle $= \displaystyle \frac{Radius \ of \ wheel}{Radius \ of \ the \ axle}$
$MA _A = \displaystyle \frac{R _1}{r _1} $ and  $MA _B = \displaystyle \frac{R _2}{r _2}$
Given $ r _1  :  r _2  = 5 : 4$
$R _1  :  R _2  = 3  :  7$
$ \displaystyle \frac{MA _A}{MA _B} = \frac{R _1}{r _1} \times \frac{r _2}{R _2}$
$\displaystyle = \frac{3}{5} \times \frac{4}{7} = \frac{12}{35}$
$MA _B >  MA _A$

The mechanical advantage of a wheel and axle $= \displaystyle \frac{Radius  of  wheel}{Radius  of  the  axle}$
Given radius of wheel $=$ 25 $\times$ radius of axle
$MA = \displaystyle \frac{25 \times radius  of  axle}{radius  of  axle} = 25$

Velocity ratio $= \displaystyle \frac{distance \ of \ the \ effort}{distance \ moved \ by \ the \ load}$
$= \displaystyle \frac{80  cm}{1  cm}= 80$

The work done by Akhil in lifting the load vertically up $=$ Force $\times$ height
$= 800 \times 2 = 1600 \ J$

Using the principle of moments: F1 * L1 = F2 * L2. 80 N * 0.25 m = 32 N * L2. L2 = (80 * 0.25) / 32 = 20 / 32 = 0.625 m.

Load $ = 75 kgf$
Length of crowbar $= 1.5 m$
Effort arm $= 1 m $
So, Load Arm $= \text{Length of crowbar - Effort arm}$

Load $ = 75 kgf$
Length of crowbar $= 1.5 m$
Effort arm $= 1 m $
So, Load Arm $= \text{Length of crowbar - Effort arm}$
$\Rightarrow 1.5 - 1 = 0.5 m$
Mechanical Advantage $= \dfrac{\text{Effort Arm}}{\text{Load Arm}} = \dfrac{1}{0.5} =2$

Length of the sloping wooden plank , $l = 2 m$
Vertical height , $h = 1 m$
We, know that for a inclined plane,
Mechanical Advantage $= \dfrac{Length \ of \ the \ incline}{vertical \ height}$
$\Rightarrow \dfrac{l}{h} = \dfrac{2}{1} = 2$

Effort Arm $ = 15 cm$
Load Arm $ = 20 - 15 = 5 cm$
Mechanical Advantage $= \dfrac{\text{Effort Arm}}{\text{Load Arm}} = \dfrac{15}{5} = 3$

$Load= 1.5kgf$
$Load \ arm= 20cm$
$Effort \ arm=15 cm$
$Effort \ = 20 \times 1.5/15= 2.0kgf$
The fulcrum in tongs is the joint end of the arms

A butter knife has a blunt edge (high wedge angle), which results in a low mechanical advantage, making it harder to cut compared to a sharp knife.

As steps become more inclined, the vertical height gained per step increases, requiring more work against gravity. This reduces the mechanical advantage because a greater portion of the applied force goes into lifting vertically rather than moving forward.

In a third-class lever, the effort is applied between the fulcrum and the load. Since the effort arm is always shorter than the load arm, the mechanical advantage is always less than 1.

In a second-class lever, the load is between the fulcrum and the effort. This means the effort arm is always longer than the load arm, resulting in a mechanical advantage always greater than 1.

Definition of effort $\left( \varepsilon \right)$. The resistive or opposing force to be overcome by a machine is called the load $\left(L\right)$.

Definition of load. The force applied on the machine to overcome the load is called effort $\left(\varepsilon \right)$.

A screw is an inclined plane wrapped around a cylinder. When force is applied, the screw must rotate multiple times to penetrate the wood, resulting in a longer travel distance compared to a nail, which is driven straight in.

$sin \theta=\dfrac{1}{M.A}=\dfrac{Effort}{Load}=\dfrac{10}{20}=\dfrac{1}{2}$
Thus, by placing a wooden plank at $\theta=30^o$ an angle equal to $30^o$ with the horizontal group can push the load of mass $20$Kg at any height by exerting a force(or effort) of $10$kg.

The velocity ratio of a machine is equal to the displacement of effort such a machine changes the direction of effort.

Increasing the closeness of threading (decreasing the pitch) increases the mechanical advantage of the screw. A higher mechanical advantage means that less effort is required to drive the screw into the material.

This appears to be a duplicate of question 481409. A butter knife has a thick cutting edge with a large wedge angle, which distributes force over a larger area. This reduces mechanical advantage, requiring more force to cut compared to a sharp knife.

Efficiency = (Mechanical Advantage) / (Velocity Ratio). Given efficiency = 0.60 and Velocity Ratio = 18, the Mechanical Advantage (MA) = 0.60 * 18 = 10.8. Since MA = W/P, then P = W/10.8. The law of machine is P = mW + c. Using the provided option A, P = W/18 + 4.8, this does not match the calculated slope. However, assuming the question implies a standard linear form, the provided answer is likely based on specific machine parameters not fully defined.

The law of machine is P = mW + c. We have two points: (2400, 150) and (3000, 180). Slope m = (180 - 150) / (3000 - 2400) = 30 / 600 = 0.05. Substituting into P = 0.05W + c: 150 = 0.05(2400) + c => 150 = 120 + c => c = 30. Thus, P = 0.05W + 30.