Explore topic-wise InterviewSolutions in .

Inclined plane acts as a machine since less effort is needed in lifting a load to a higher level by moving over an inclined plane as compared to that in lifting the load directly.

Four examples of inclined plane acting as machine : 

1. Ramp 

2. Railway Bridge 

3. Stair case 

4. Mountainous foot path or roads. 

All these act as machine as less effort is needed in lifting the load to higher level.

(b) A crowbar

Number of teeth in first wheel = 10

Number of teeth in second wheel = 50

For gain in speed, the second wheel of 50 teeth (NA = 50) is used as driving wheel and the first wheel of 10 teeth (NB = 10) is used as driven wheel.

Gear ratio = N_A/N_B = 10/50 = 1/5 = 1:5

Gain in speed = N_A/N_B = 50/10 = 5/1 = 5

For gain in torque, the second wheel of 50 teeth (N_B = 50) is used as driven wheel and the first wheel of 10 teeth (NA = 10) is used as driving wheel.

Gear ratio = N_A/N_B = 50/10 = 5/1 = 5 : 1

Gain in torque = N_A/N_B = 10/50 = 1/5 = 1:5

Gain in torque = N_B/N_A = 50/10 = 5

Gradient of inclination : “The ratio between the vertical distance moved by a body and the horizontal distance travelled along the inclined plane is called grade of inclination” or gradient of inclined plane. e.g. If a body riser 10m vertically when it moves along the inclined plane by 

200m or sinθ 10/200 = 1/20 

= then inclination grade is said to be 1 in 20. 

Higher is the magnitude of gradient the more difficult is the slope to climb and vice-versa. 

While laying hill roads, the gradient is kept allow as possible.

(a) Gear : “A gear is a wheel with teeth around its rim.” Or “Is a precise device to transfer the rotatory motion from one point to the other”. 

(b) Driven wheel “A wheel which receives motion from driver wheel and is connected to the load and rotates in opposite direction to driver wheel.” 

(c) Driving wheel : “A gear wheel which is closer to the source of power and effort is applied to this wheel is called driving wheel”.

The radius of driving wheel r_A=2cm

The radius of driven wheel r_B=20cm

(a) the gear ratio  r_A/r_B = 2/20 = 1 : 10

(b) The number of rotations made per minute by the driving wheel is na= 100

The number of rotations made per minute by the driven wheel nb = r_A/r_B x na = 2/20 x 100 = 10

(a) Number of teeth in driven wheel nb = 40.

Number of teeth in driven wheel na = ra/rb x nb

⇒ 2/20 x 20 = 4

Given, gear ratio = N_A/N_B = 10/a

It is possible to obtain a change in direction,

Using wheel C, if the number of teeth in

Wheel C is equal to the number of teeth in wheel A.

∴ NC = NA = 10

Hence, the gear ratio of wheels A and C:

Gear ratio = N_A/N_B = 10/10

∴ The required gear ration is 1:1

Given, radius of driving wheel ra = 1 cm

No. of teeth in the driving wheel Na= 8

Speed of rotation of driving wheel na = 12 rpm

Speed of rotation of driven wheel nb = 1 rpm

(a) Radius of driven wheel rb = ra x na/nb = 1 x 12/1 = 12 cm

(b) No. of teeth in the driven wheel Nb = Na x na/nb = 8 x 12/1 = 96

Total length of crowbar = 120 cm

Load arm = 20 cm

Effort arm = 120 − 20 = 100 cm

Mechanical advantage M.A =  effort arm/Load arm

M.A = 100/20=5

(c) changes the direction of rotation

Velocity ratio (gear ratio) of a car gaining speed is : more than 1.

Total length of rod=4 m = 400 cm

(a) 18kgf load is placed at 60 cm from the support.

W kgf weight is placed at 250 cm from the support.

By the principle of moments

18 x 60 = W x 250

W = 4.32 kgf

(b) Given W=5 kgf

18kgf load is placed at 60 cm from the support.

Let 5 kgf of weight is placed at d cm from the support.

By the principle of moments

18 x 60 = 5 x d

d = 216 cm from the support on the longer arm

(c) It belongs to class I lever.

(a) This is a class II lever.

(b) Given: FA=80 cm, AB = 20 cm, BF= FA+AB=100cm

Mechanical advantage M.A = BF/AF = 100/80 = 1.25

(c) Effort (E) = Load (L)/M.A = 5/1.25 = 4 Kgf

(a) The principle of moments: Moment of the load about the fulcrum=moment of the effort about the fulcrum

FB × Load = FA × Effort

(b) Sugar tongs the example of this class of lever.

(c) Given: FA = 10 cm, AB = 500 cm, BF = 500 + 10 = 510 cm.

The mechanical advantage

M.A = AF/BF = 10/510 = 1/51

The minimum effort required to lift the load

Effort = Load/M.A = 50/1/51 = 2550 N

The expression for the mechanical advantage of an inclined plane in terms of its length l and vertical height h is:

M.A = I/h

Gear system: A gear system is a device to transfer precisely the rotator motion from one point to the other. A gear is a wheel with teeth around its rim. The teeth act as the components of a machine and they transmit rotational motion to the wheel by successively engaging the teeth of the other rotating gear.

Working: Each tooth of a gear acts like a small lever of class I. A gear when in operation, can be considered as a lever with an additional property that it can be continuously rotated instead of moving back and forth as is the case with an ordinary lever. Each gear wheel is mounted on an axle which rotates at a speed depending upon the motion transmitted to it. The gear wheel closer to the source of power is called the driver, while the gear wheel which receives motion from the driver is called the driven gear. The driven gear rotates in a direction opposite to the driving gear when the two gears make an external contact. On the other hand, if the gears make an internal contact, both gears rotate in the same direction.

It is the ratio of the useful work done by the machine to the work put into the machine by the effort.

In actual machine there is always some loss of energy due to friction and weight of moving parts, thus the output energy is always less than the input energy.

Effort arm = 7.5 cm

Load arm = 15 cm

Mechanical advantage M.A = Effort arm/Load arm = 7.5/15 = 0.5

(a) This is a class I lever.

(b) Given AB=1m, AF=0.4m and BF=0.6 m

Mechanical advantage M.A = BF/AF = 0.6/0.4 = 1.5

(c) Load = 15kfg

Effort = Load/M.A = 15/1.5 = 10 kgf

Fire tongs has its arms =20 cm

Effort arm = 15 cm

Load arm =20 cm

(i) Mechanical advantage M.A = Effort arm/Load arm = 15/20 = 0.75

(ii) Effort = Load/M.A = 1.5/0.75 = 2.0 kgf

Mechanical advantage of an inclined plane is always greater than 1.

(a) A machine acts as a force multiplier when the effort arm is longer than the load arm. The mechanical advantage of such machines is greater than 1.

(b) A machine acts a speed multiplier when the effort arm is shorter than the load arm. The mechanical advantage of such machines is less than 1.

It is not possible for a machine to act as a force multiplier and speed multiplier simultaneously. This is because machines which are force multipliers cannot gain in speed and vice-versa.

Effort arm = 10 cm

Load arm = 5 cm

Mechanical advantage = M.A = Effort arm/Load arm = 10/5 = 2

Load = 5kgf

Effort = Load/M.A = 5/2 = 2.5 kgf

Load to be raised L = mg = 30 x 10 = 300N

Effort applied E = 350N

M.A. = L/E = 300/350 = 6/7 = 0.86

(a) The fixed pulley P₂ only changes the direction of the force i.e. force can be applied downwards which is more convenient. 

(b) The load will be raised by a distance x/2. 

(c) Distance moved by load dL = x/2 

Distance moved by the effort dE = x 

Load (L) = 20 kgf 

Effort = ? 

Load × Load arm = effort × effort arm

20 x x/2 = x x E or E = 10kgf.

The force applied by the women is = 70 N

The mass of bucket and water together is = 6 kg

Total load = 6 × 10 = 60 N

Mechanical advantage M.A = Load/Effort = 60/70 = 0.857

(a) The velocity ratio of a single fixed pulley is always more than 1.(false)

(b) The velocity ratio of a single movable pulley is always 2.(true)

(c) The velocity ratio of a combination of n movable pulleys with a fixed pulley is always 2ⁿ.(true)

(d) The velocity ratio of a block and tackle system is always equal to the number of strands of the tackle supporting the load. (true)

Pulley : “is a flat circular disc, having a groove in its edge and capable of rotating about a fixed point passing through its centre commonly called axle.” 

(a) More than one: shears used for cutting the thin metal sheets.

(b) Less than one: a pair of scissors whose blades are longer than its handles.

When the mechanical advantage is less than 1, the levers are used to obtain gain in speed. This implies that the displacement of load is more as compared to the displacement of effort.

A pulley system has 5 pulley in all. Its velocity ratio is 5.

To increase the load arm the blades of scissor are longer so that blades move over longer when the effort arm is moved a little. 

The effort arms of shears made longer as compared to blades so that it becomes force multiplier and helps in cutting the metal sheet. 

The ideal mechanical advantage of a single fixed pulley is 1. It cannot be used as force multiplier.

It helps in applying effort in a convenient direction.

Explanation: A single fixed pulley though does not reduce the effort but helps in changing the direction of effort applied. As it is far easier to apply effort in downward direction, the single fixed pulley is widely used.

Force multiplier

Explanation: The mechanical advantage of movable pulley is greater than 1. Thus, using a single movable pulley, the load can be lifted by applying an effort equal to half the load (in ideal situation), i.e. the single movable pulley acts as a force multiplier.

The effort can be applied in a more convenient direction with the single fixed pulley. One can conveniently make use of his own weight also for the effort.

The three classes of levers are:

(i) Class I levers: In these types of levers, the fulcrum F is in between the effort E and the load L. Example: a seesaw, a pair of scissors, crowbar.

(ii) Class II levers: In these types of levers, the load L is in between the effort E and the fulcrum F. The effort arm is thus always longer than the load arm. Example: a nut cracker, a bottle opener.

(iii) Class III levers: In these types of levers, the effort E is in between the fulcrum F and the load L and the effort arm is always smaller than the load arm. Example: sugar tongs, forearm used for lifting a load.

1. Its parts are less than frictionless. 

2. Its string is not perfectly elastic. 

3. Weight of movable pulley. 

The efficiency of a single movable pulley system is not 100% this is because (i)The friction of the pulley bearing is not zero , (ii)The weight of the pulley and string is not zero.

Single movable pulley: A pulley, whose axis of rotation is not fixed in position, is called a single movable pulley. Mechanical advantage in the ideal case is 2.

(b) velocity ratio 2, and actual mechanical advantage less than 2

M.A = Effortarm/Loadarm

This is the expression of the mechanical advantage of a lever.

(c) both (a) and (b) 

(c) less than 2, but more than 1

This is because the output work is always less than the input work, so the efficiency is always less than 1 because of energy loss due to friction.

M.A = V.R × n

The mechanical advantage of second order levers is always more than 1 because the effort arm is always longer than the load arm. 

(M.A.) = (η x V.R)

M.A. of second order of lever is more than one as effort arm is longer than load arm M.A = Effort arm/Load arm.

(c) η × V.R. = M.A.