Selina Concise Class 10 Physics Solution Chapter No. 3- ‘Machines’ For ICSE Board Students.
Selina Concise Class 10 Physics Chapter 3 Machines Exercise All Questions and Answers by Physics Teacher here in this post.
Exercise 3 (A)
Question: 1
(a) What do you understand by a simple machine?
(b) State the principle of an ideal machine.
Solutions: Machine is a device by which we can either overcome a large resistive force at some point by applying a small force at a convenient point and in a desired direction or by which we can obtain a gain in speed.
An ideal machine is that is which there is no loss of energy is any manner. Here the work output is equal to the work input.
Question: 2
State four ways in which machines are useful to us?
Solutions: Speed gain can be achieved using machine.
The point of application of efforts can be changed to a convenient point.
The direction of efforts can be changed to convenient direction.
Machine can be used to lift heavy load easily.
Question: 3
Name a machine for each of the following use:
(a) to multiply the force,
(b) to change the point of application of force,
(c) to change the direction of force,
(d) to obtain the gain in speed.
Solutions:
(a) Jack used to lift heavy objects
(b) Big spanner used to turn nuts using minimum efforts.
(c) Pulley to bring water out of well where force is applied in opposite direction
(d) Vehicles are example machine is used to gain speed.
Question: 4
What is the purpose of a jack in lifting a car by it?
Solutions: The jack is used to make the efforts highly efficient, jack is used to increase the force applied to lift the heavy objects. It is machine used to for lifting heavy devices by applying minimum forces.
Question: 5
What do you understand by an ideal machine? How does it differ from a practical machine?
Solutions: An ideal machine is that is which there is no loss of energy is any manner. Here the work output is equal to the work input. The ideal machine has 100 % efficiency. On the contrary output of a practical machine is always less than that of ideal as friction plays big role in the consumption of energy.
Question: 6
Explain the term mechanical advantage. State its unit.
Solutions: The term mechanical advantage is a ratio of load to efforts. As the mechanical advantage is ratio it has no unit.
Question: 7
Define the term velocity ratio. State its unit.
Solutions: It is ratio of velocity of efforts to the velocity of load. As the term is a ratio it has no unit.
Question: 8
How is mechanical advantage related to the velocity ratio for
(i) an ideal machine,
(ii) a practical machine?
Solutions: (i) for ideal machine mechanical advantage is equal to velocity ratio
(ii) For practical machine the mechanical advantage is always less than velocity ratio.
Question: 9
Define the term efficiency of a machine. Give two reasons for a machine not to be 100% efficient?
Solutions: Efficiency of machine is the ratio of machine’s work done on the load to the work done on the machine by the efforts.
Due to friction and weight of moving parts machine can never be 100 % efficient.
Question: 10
When does a machine act as (a) a force multiplier and (b) a speed multiplier? Can a machine act as a force multiplier and speed multiplier simultaneously?
Solutions: When value of mechanical advantage is greater than 1 the machine become force multiplier. When value of mechanical advantage becomes less than 1 the machine becomes speed multiplier. Machine can never act as a force multiplier and speed multiplier simultaneously.
Question: 11
A machine works as a (i) force multiplier, (ii) speed multiplier. In each case state whether the velocity ratio is more than or less than 1.
Solutions: When value of velocity ratio is greater than 1 the machine become force multiplier. When value of velocity ratio becomes less than 1 the machine becomes speed multiplier.
Question: 12
(a) State the relationship between mechanical advantage, velocity ratio and efficiency.
(b) Name the term that will not change for a machine of a given design.
Solutions:
(a) Work output = load * displacement of load
Work input = effort * displacement of effort
As we know,
Efficiency = work output / work input
Efficiency = load * displacement of effort / effort * displacement of effort
Mechanical advantage (M.A.) = load / effort
Velocity ratio (V.R.) = displacement of effort / displacement of load
Hence, Efficiency = M.A. / V.R.
M.A. = efficiency * V.R.
(b) velocity ratio is the term that will not change for a machine of given design.
Question: 13
Derive a relationship between mechanical advantage, velocity ratio and efficiency of a machine.
Solutions: Work output = load * displacement of load
Work input = effort * displacement of effort
As we know,
Efficiency = work output / work input
Efficiency = load * displacement of effort / effort * displacement of effort
Mechanical advantage (M.A.) = load / effort
Velocity ratio (V.R.) = displacement of effort / displacement of load
Hence,
Efficiency = M.A. / V.R.
M.A. = efficiency * V.R.
Question: 14
How is the mechanical advantage related to the velocity ratio for an actual machine? State whether the efficiency of such a machine is equal to 1, less than 1 or more than 1.
Solutions: The mechanical advantage of actual machine is product of efficiency and velocity ratio.
Efficiency = M.A. / V.R.
M.A. = efficiency * V.R.
Efficiency of machine is always less than 1 as the loss of energy is unavoidable due to friction or in any other form of unwanted energy.
Question: 15
State one reason why is mechanical advantage less than the velocity ratio for an actual machine.
Solutions: Mechanical advantage is always less than velocity ratio as the mechanical advantage get reduced because of friction and weight of parts on the other hand velocity ratio remains constant.
Question: 16
What is a lever? State its principle.
Solutions: Lever is a rigid, straight or bent bar which is capable of turning about the fixed axis.
When two equal forces are acting in opposite directions applied on a uniform lever at equal distance from the fulcrum, the force establish state of equilibrium in the lever.
Question: 17
Write down a relation expressing the mechanical advantage of a lever.
Solutions: A lever works on the principle of moments according to which at the equilibrium position of the lever, moment of load about the fulcrum must be equal to the moment of effort about the fulcrum and the two moments must always be in opposite directions. Moment of load L about the fulcrum F is clockwise, while the moment of effort E about the fulcrum F is anticlockwise. Thus,
Clockwise moment of load about the fulcrum = anticlockwise moment of efforts about the load
Load * load arm = efforts * effort arm
L * FB = E * FA
L /E = FA / FB
Mechanical advantage = load / efforts
Hence, Mechanical advantage = effort arm / load arm
Question: 18
Name the three classes of levers and state how are they distinguished. Give two examples of each class.
Solutions:
(a) Class I lever, when the fulcrum is between efforts and load like seesaw, crowbar.
(b) Class II lever, when the load is between fulcrum and effort like nail cutter, wheel barrow.
(c) Class III lever, when the effort is between fulcrum and load like foot treadle, knife etc.
Question: 19
Give one example each of a class I lever where the mechanical advantage is (a) more than 1, and (b) less than 1.
Solutions: Class I lever with mechanical advantage more than one is physical balance with equal arm length.
Class I lever with mechanical advantage less than one is pair of scissors whose blades are longer than its handle.
Question: 20
What is the use of lever if its mechanical advantage is (a) more than 1, (b) equal to 1, and (c) less than 1?
Solutions:
(a) When mechanical advantage is more than 1 the lever acts as force multiplier used to overcome a large resistive force by small efforts.
(b) Mechanical advantage is equal to one when effort arm and load arm of equal lengths.
(c) When the mechanical advantage is less than 1, it is used to gain speed.
Question: 21
Both a pair of scissors and a pair of pliers belong to the same class of levers. Name the class of lever. Which one has the mechanical advantage less than 1?
Solutions: Both a pair of scissors and a pair of a plier belong to the class I lever.
The mechanical advantage pair of scissors is less than 1 as blades of the scissors are longer than the handles. Hence, the effort arm is shorter than the load arm.
Question: 22
Explain why scissors for cutting cloth may have blades longer than the handles, but shears for cutting metals have short blades and long handles.
Solutions: For pair of scissors cutting cloth the effort arm is shorter than load arm that is blade so blades move longer with short movement of handle. While shears have short blades hence the efforts arm is longer than load arm so small effort is used to cut the metal.
Question: 24
Which type of lever has a mechanical advantage always more than 1? Give reason with one example. What change can be made in this lever to increase its mechanical advantage?
Solutions: Lever having effort arm larger than load arm have always mechanical advantage more than 1. For example, shears used to cut the metal where blades are larger than the handle. The effect arm is larger than the load arm. Hence, we can use it to minimise efforts to cut the metal. To increase the mechanical advantage of lever we have to make sure length of effort arm must be greater than load arm.
Question: 25
Draw a diagram of a lever which is always used as a force multiplier. How is the effort arm related to the load arm in such a lever?
Solutions:
In such alever, the effect arm is longer than the load arm.
Question: 26
Explain why the mechanical advantage of a class II type of lever is always more than 1.
Solutions: In class II levers the fulcrum and the effort are placed at the two ends of lever and load is placed between the two of them. Hence, the length of load arm is always less than effort arm. Therefore, the mechanical advantage of class II lever is always more than 1.
Questions: 27
Draw a labelled diagram of a Class II lever. Give one example of such a lever.
Solutions:
This is an example of class II lever like a door.
Question: 28
Shows a lemon crusher.
(a) In the diagram, mark the position of the directions of load L and effort E.
(b) Name the class of lever.
Solutions:
It is a class two lever.
Question: 30
State the kind of lever which always has the mechanical advantage less than 1. Draw a labelled diagram of such a lever.
Solutions: Class III lever is the kind of lever which always has mechanical advantage less than 1.
Question: 31
Explain why the mechanical advantage of the class III lever is always less than 1.
Solutions: In class III levers, the load and fulcrum are placed and the ends of lever and the effort placed in between them hence length of load arm is aways greater than effort arm. Hence in class III levers the mechanical advantage is always less than 1.
Question: 32
Classes III levers have mechanical advantage less than 1. Why are they then used?
Solutions: In class III levers, the load and fulcrum are placed and the ends of lever and the effort placed in between them hence length of load arm is aways greater than effort arm. Hence in class III levers the mechanical advantage is always less than 1. They are used to gain speed.
Question: 34
State the class of levers and the relative positions of load (L), effort (E) and fulcrum (F) in
(a) a bottle opener, and
(b) sugar tongs.
Solutions:
(a) Bottle opener is class II lever, where load and fulcrum is placed at same point and effort is placed at opposite end.
(b) Sugar tongs are class III lever, where effort is placed between load and the fulcrum at the opposite end.
Question: 36
Classify the following into levers as class I, class II or class III:
(a) A door
(b) A catapult
(c) Claw hammer
(d) A wheelbarrow
(e) A fishing rod.
(f) sugar tongs.
Solutions:
(a) Class II lever
(b) Class I lever
(c) Class I lever
(d) Class II lever
(e) Class III lever
(f) Class III lever.
Question: 37
What type of lever is formed by the human body while (a) raising a load on the palm, and (b) raising the weight of body on toes?
Solutions: Human body forms class III lever while raising a load on the palm
While raising the weight of body on the toe human body form class II lever.
Question: 39
Give an example of each class of lever in a human body.
Solution: Class one lever = doing bicep curls
Class II lever – raising body weight on toes
Class III lever = raising load on the palm.
Question: 40
Complete the following sentences:
(a) Mechanical advantage = ___________ × velocity ratio
(b) In class II lever, effort arm is __________ than the load arm.
(c) A scissors is a ___________ multiplier.
Solutions:
(a) Efficiency
(b) Longer
(c) speed
Multiple Choice Type:
Question: 1
Mechanical advantage (M.A.), load (L) and effort (E) are related as:
a.) M.A. = L x E
b.) M.A. x E = L
c.) E = M.A. x L
d.) None of these
Solutions: (b) M.A. x E = L
Question: 2
The correct relationship between the mechanical advantage (M.A.), velocity ratio (V.R.) and efficiency (η) is:
a.) M.A. = η x V.R.
b.) V.R. = η x M.A.
c.) η = M.A. x V.R.
d.) None of these
Solutions: (a) M.A. = η x V.R.
Question: 3
Select the incorrect statement:
(a) A machine always has the efficiency less than 100%.
(b) The mechanical advantage of a machine can be less than 1.
(c) A machine can be used as a speed multiplier.
(d) A machine can have a mechanical advantage greater than the velocity ratio.
Solutions: (d) A machine can have a mechanical advantage greater than the velocity ratio.
Question: 4
The lever for which the mechanical advantage is less than 1 has the:
(a) Fulcrum at mid-point between load and effort.
(b) Load between effort and fulcrum.
(c) Effort between fulcrum and load.
(d) Load and effort acting at the same point.
Solutions: (c) Effort between fulcrum and load.
Question: 5
Class II levers are designed to have:
a.) M.A. = V.R.
b.) M.A. > V.R.
c.) M.A. > 1
d.) M.A. < 1
Solutions: (c) M.A. > 1
Numerical
Question: 1
A crowbar of length 120 cm has its fulcrum situated at a distance of 20 cm from the load. Calculate the mechanical advantage of the crowbar.
Solutions: Given
Length of crowbar = 120
Load arm = 20 cm
Effort arm = 120 – 20 = 100 cm
Mechanical advantage = effort arm / load arm
A. = 100 / 20
M.A. = 5
Question: 2
A pair of scissors has its blades 15 cm long, while its handles are 7.5 cm long. What is its mechanical advantage?
Solutions: Given
For pair of scissors effort arm = 7.5 cm
Load arm = 15 cm
Mechanical advantage = effort arm / load arm
A. = 7.5 / 15
A. = 0.5
Question: 3
A force of 5kgf is required to cut a metal sheet. A shears used for cutting the metal sheet has its blades 5 cm long, while its handle is 10 cm long. What effort is needed to cut the sheet?
Solutions: Given
Load =5kg
Load arm = 5 cm
Effort arm = 10 cm
According to principle of lever,
Effort * effort arm = load * load arm
Effort = 5 * 5 / 10
Effort = 2.5 kg
Question: 4
The diagram below shows a lever in use.
(a) To which class of lever does it belong?
(b) If AB =1 m, AF= 0.4 m, find its mechanical advantage.
(c) Calculate the value of E.
Solutions:
(a) This is class I liver
(b) Given
AB = 1m
AF = 0.4m
BF = 0.6m
Mechanical advantage = BF /AF
M.A. = 0.6/ 0.4
M.A. = 1.5
(c) load = 15 kg
Effort = load / M.A.
= 15 / 1.5
= 10 kg
Question: 5
A man uses a crowbar of length 1.5 m to raise a load of 75kgf by putting a sharp edge below the bar at a distance 1 m from his hand. (a) Draw a diagram of the arrangement showing the fulcrum (F), load (L) and effort (E) with their directions. (b) State the kind of lever. (c) Calculate: (i) load arm, (ii) effort arm, (iii) mechanical advantage, and (iv) the effort needed.
Solutions:
(a) Diagram
(b) Crowbar is a class I lever
(c) (i) total length of bar = 1.5 m
As the bar is 1m from hand
Effort arm = 1m
Hence, load arm = 1.5 – 1 = 0.5 m
(ii) load arm = `1m
(iii) mechanical advantage = effort arm / load arm
M.A. = 1/ 0.5 = 2
(iv) effort needed = load / M.A. = 75 / 2 = 37.5 kg
Question: 6
A pair of scissors is used to cut a piece of a cloth by keeping it at a distance 8.0 cm from its rivet and applying an effort of 10 kgf by fingers at a distance 2.0 cm from the rivet.
(a) Find: (i) the mechanical advantage of scissors and (ii) the load offered by the cloth.
(b) How does the pair of scissors act: as a force multiplier or as a speed multiplier?
Solutions: Given
Finger at a distance from rivet = effort arm = 2 cm
Distance from cloth = load arm = 8 cm
Effort = 10 kg
(a) (i) mechanical advantage = effort arm / load arm
M.A. = 2/8 = 0.25
(ii) load offered = effort needed * mechanical advantage
Load = 10 * 0.25
Load = 2.5 kg
(b) Scissors acts as speed multiplier because mechanical advantage of scissors is less than 1.
Question: 7
A 4 m long rod of negligible weight is to be balanced about a point 125 cm from one end and a load of 18 kgf is suspended at a point 60 cm from the support on the shorter arm.
(a) If a weight W is placed at a distance of 250 cm from the support on the long arm, Find W.
(b) If a weight 5 kgf is kept to balance the rod, find its position.
(c) To which class of lever does it belong?
Solutions: Given
(a) Load = 18 kg
Load arm = 60 cm
Effort arm = 250 cm
According to principle of moment
Load * load arm = effort * effort arm
18 * 60 = W * 250
W = 4.32 kg
(b) W = 5kg
We have to find its position
According to principle of moment
Load * load arm = effort * effort arm
18 * 60 = 5 * effort arm
Effort arm = 1080 / 5
Effort arm = 216 cm on the longer arm
(c) Class I lever
Question: 8
A lever of length 9 cm has its load arm 5 cm long and the effort arm is 9 cm long. (a) To which class does it belong? (b) Draw diagram of the lever showing the position of fulcrum F and directions of both the load L and effort E. (c) What is the mechanical advantage and velocity ratio if the efficiency is 100%? (d) What will be the mechanical advantage and velocity ratio if the efficiency becomes 50%?
Solutions: Given
Length of lever = 9 cm
Load arm = 5 cm
Effort arm = 9cm
(a) Lever belongs to class II
(b) Mechanical advantage = effort arm / load rm
M.A. = 9 / 5 = 1.8
Velocity ratio = 1.8 (for 100% efficiency)
(c) As for 50 % efficiency
We know that, the value of velocity ratio remains constant
M.A. = efficiency * V.R.
M.A. = 0.5 * 1.8
M.A. = 0.9
Question: 9
The diagram below shows a lever in use.
(a) To which class of lever does it belong?
(b) Without changing the dimensions of the lever, if the load is shifted towards the fulcrum what happens to the mechanical advantage of the lever.
Solutions:
(a) The lever belongs to class II
(b) Mechanical advantage off the lever increases as the length of load arm decreases when we move load towards the fulcrum.
Question: 10
The figure shows a wheelbarrow of mass 15 kg carrying a load of 30 kgf with its centre of gravity at A. The points B and C are the centre of wheel and tip of the handle such that the horizontal distance AB = 20 cm and AC = 40 cm.
Find: (a) the load arm, (b) the effort arm, (c) the mechanical advantage, and (d) the
minimum effort required to keep the leg just off the ground.
Solutions:
(a) Load arm = 20 cm, as A is the point of centre of gravity and load
(b) Effort arm = 60 cm, as the handle is placed at point C
(c) Mechanical advantage = effort arm / load arm
M.A. = 60 / 20 = 3
(d) total load = 45 kg
Load arm = 20 cm
Effort arm = 60 cm
According to principle of lever
Load * load arm = effort * effort arm
45 * 20 = effort * 60
Effort = 900 / 60
Effort = 15 kg
Minimum effort required are 15 kg.
Question: 11
The diagram below shows the use of a lever.
(a) State the principle of moments as applied to the above lever.
(b) To which class of lever does it belong? Give an example of this class of lever.
(c) If FA = 10cm, AB = 490cm, calculate: (i) the mechanical advantage, and (ii) the minimum effort required to lift the load (= 50N).
Solutions:
(a) According to principle of levers
Load * load arm = effort * effort arm
Here effort arm = FA
Load arm = FB
Hence,
Load * FB = effort * FA
(b) This lever belongs to class III, for example foot treadle.
(c) Effort arm = 10 cm
Load arm = 490 + 10 = 500 cm
Mechanical advantage = effort arm / load arm
M.A. = 10/ 500
M.A. = 1/50
Efforts = load * FB/ FA
Efforts = 50 * 500 / 10
Efforts = 2500 N
Question: 12
A fire tongs has its arms 20 cm long. It is used to lift a coal of weight 1.5kgf by applying an effort at a distance 15 cm from the fulcrum. Find: (i) the mechanical advantage of fire tongs and (ii) the effort needed.
Solutions: Given
Load arm = 20 cm
Load = 1.5 kg
Effort arm = 15 cm
(i) Mechanical advantage = effort arm / load arm
M.A. = 15 / 20
M.A. = 0.75
(ii) according to principle of levers
Load * load arm = effort * effort arm
1.5 * 20 = effort * 15
Effort = 30 /15
Effort = 2 kg
Exercise 3 (b)
Question: 1
What is a fixed pulley? State its one use.
Solutions: Fixed pulley is a pulley which has axis of rotation in stationary position. Pulley is used for lifting of small load.
Question: 2
What is the ideal mechanical advantage of a single fixed pulley? Can it be used as a force multiplier?
Solutions: Ideal mechanical advantage of single fixed pulley is 1. No it can not used as force multiplier.
Question: 3
Name the pulley which has no gain in mechanical advantage. Explain, why is such a pulley is then used?
Solutions: Single fixed pulley is the pulley which has no gain in mechanical advantage. Fixed pulley is used to change the direction of efforts to be applied in more convenient direction.
Question: 4
What is the velocity ratio of a single fixed pulley?
Solutions: the velocity ratio of single fixed pulley is one and the effort moved downward by certain distance then the load moves upward by came distance.
Question: 5
In a single fixed pulley, if the effort moves by a distance x downwards, by what height is the load raised upwards?
Solutions: As the velocity ratio is one, if the effort moves by distance x upwards then load moves by distance x downwards.
Question: 6
What is a single movable pulley? What is its mechanical advantage in the ideal case?
Solutions: single movable pulley is the pulley whose axis of rotation is movable of not fixed. Mechanical advantage of single movable pulley is 2 in the ideal case.
Question: 7
Name the type of single pulley that has an ideal mechanical advantage equal to 2. Draw a labelled diagram of the pulley mentioned by you.
Solutions:
It is a single moveable pulley.
Question: 8
Give two reasons why the efficiency of a single movable pulley system is not 100%.
Solutions: The efficiency of single movable pulley will always be less than 100% as the bearing in pulley are not frictionless and the weight of pulley and string can never be zero.
Question: 9
In which direction the force need be applied, when a single pulley is used with a mechanical advantage greater than 1? How can you change the direction of force applied without altering its mechanical advantage? Draw a labelled diagram of the system.
Solutions: When a single pulley is used with mechanical advantage of 2 forced is used in upward direction. To change the direction of force applied without changing the mechanical advantage we have to add single fixed pulley along with single movable pulley to change the direction of efforts.
Question: 10
What is the velocity ratio of a single movable pulley? How does the friction in the pulley bearing affect it?
Solutions: Velocity ratio for single movable pulley is 2. Velocity ratio for a pulley never changes it remains constant.
Question: 11
In a single movable pulley, if the effort moves by a distance x upwards, by what height is the load raised?
Solutions: If the effort move distance x upwards the load moves 2x upwards.
Question: 12
Draw a labelled diagram of an arrangement of two pulleys, one fixed and other moveable. In the diagram, mark the directions of all forces acting on it. What is the ideal mechanical advantage of the system? How can it be achieved?
Solutions: The ideal mechanical advantage of system is two. When we assume mass of string and pulley’s friction zero. We can achieve the desired mechanism advantage.
Question: 13
The diagram alongside shows a pulley arrangement.
(a) Name the pulleys A and B.
(b) In the diagram, mark the direction of tension on each strand of string.
(c) What is the purpose of the pulley B?
(d) If the tension is T, deduce the relation between (i) T and E, and (ii) E and L.
(e) What is the velocity ratio of the arrangement?
(f) Assuming that the efficiency of the system is 100%, what is the mechanical advantage?
Solutions:
(a) A is single movable pulley and B is single fixed pulley.
(b)
(c) Purpose of pulley B is to change the direction of effort.
(d) The effort E balance the tension T. hence, E = T, and the load is twice to the effort. Hence, E = 2L
(e) Velocity ratio of this arrangement is 2.
(f) Assuming that the efficiency of the system is 100%. The mechanical advantage is 2.
Question: 14
State four differences between a single fixed pulley and a single movable pulley.
Solutions:
Single fixed pulley | Single movable pulley |
This pulley is fixed to rigid point. | This pulley is not fixed at any rigid point. |
Mechanical advantage for ideal pulley is 1. | Mechanical advantage of this pulley is 2. |
The velocity ratio is 1 | Velocity ratio is 2 |
It doesn’t act as force multiplier. | It act as force multiplier. |
Question: 15
The diagram alongside shows and arrangement of three pulleys A, B and C. The load is marked as L and the effort as E.
(a) Name the pulleys A, B and C.
(b) Mark in the diagram the direction of load (L), effort (E) and tension T1 and T2 in the two strings.
(c) How are the magnitudes of L and E related to the tension T1?
(d) Calculate the mechanical advantage and velocity ratio of the arrangement.
(e) What assumptions have you made in parts (c) and (d)?
Solutions:
(a) A and B are movable pulleys and C is fixed pulley.
(b)
(c) The effort E= T1 and the value of load L = 4 T1
(d) Mechanical advantage is 4 as well as velocity ratio is also 2.
(e) Assumptions made are pulley A and B are weightless and frictionless.
Question: 16
Draw a diagram of combination of three movable pulleys and one fixed pulley to lift up a load. in the diagram, show the directions of load, effort and tension in each strand. Find: (i) the mechanical advantage, (ii) velocity ratio and (iii) the efficiency of the combination in ideal situation.
Solutions:
As mechanical advantage of one moveable pulley is two. The system contains three moveable pulleys in equilibrium so mechanical advantage is 23 .
Let the distance load moves up by be d.
The efforts needed to move the load by distance d be 23 d
Velocity ratio = efforts needed / distance moved
V.R. = 23d /d
V.R = 23
efficiency = M.A/ V.R.
efficiency = 23 /23
efficiency = 100 %
Question: 17
Draw a diagram of a block and tackle system of pulleys having a velocity ratio of 5. In your diagram indicate clearly the points of application and the directions of the load L and effort E. Also mark the tension T in each strand.
Solutions:
Question: 18
Give reasons for the following:
(a) In a single fixed pulley, the velocity ratio is always more than the mechanical advantage.
(b) The efficiency of a movable pulley is always less than 100%.
(c) In the case of a block and tackle system, the mechanical advantage increases with the increase in the number of pulleys.
(d) The lower block of a block and tackle pulley system must be of negligible weight.
Solutions:
(a) Velocity ratio is always more than mechanical advantage as the some of the efforts are wasted to overcome the friction of the pulley.
(b) Efficiency of movable pulley is always less than 100 % as the weight of pulley and string is not zero and efforts are wasted to overcome friction.
(c) The mechanical advantage increases as the number of pulleys increases because mechanical advantage is equal to the total number of pulleys.
(d) Efficiency of pulley is related to mass of lower block. Therefore, efficiency decreases with increase in the weight of lower block of pulleys.
Question: 19
Name a machine which is used to:
(a) Multiply force
(b) Multiply speed, and
(c) Change the direction of force applied.
Solutions:
(a) A movable pulley is used to multiply the force.
(b) Gear system is used to multiply speed.
(c) Single fixed pulley used to change in the direction of force applied.
Question: 20
State whether the following statements are true or false.
(a) The velocity ratio of a single fixed pulley is always more than 1.
(b) The velocity ratio of a single movable pulley is always 2.
(c) The velocity ratio of a combination of n movable pulleys with a fixed pulley is always 2n.
(d) The velocity ratio of a block and tackle system is always equal to the number of strands of the tackle supporting the load.
Solutions:
(a) False
(b) True
(c) True
(d) True
Multiple Choice Type:
Question: 1
A single fixed pulley is used because it:
(a) Has a mechanical advantage greater than 1
(b) Has a velocity ratio less than 1
(c) Gives 100% efficiency
(d) Helps to apply the effort in a convenient direction.
Solutions: (d) Helps to apply the effort in a convenient direction.
Question: 2
The mechanical advantage of an ideal single movable pulley is:
(a) 1
(b) 2
(c) less than 2
(d) less than 1.
Solutions: (b) 2
Question: 3
A movable pulley is used as:
(a) a force multiplier
(b) a speed multiplier
(c) a device to change the direction of effort
(d) an energy multiplier
Solutions: (a) a force multiplier
Numericals
Question: 1
A woman draws water from a well using a fixed pulley. The mass of the bucket and water together is 6 kg. The force applied by the women is 70 N. calculate the mechanical advantage. (Take g = 10 m s-2).
Solutions: Given
Force = 70 N
Mass of bucket or load = 6 kg
Total load = mass * g = 6 * 10 = 60 N
Mechanical advantage = load / effort
M.A. = 60/70 = 0.857
Question: 2
A fixed pulley is driven by a 100 kg mass falling at a rate of 8.0 m in 4.0 s. It lifts a load of 75.0 kgf. Calculate:
(a) The power input to the pulley taking the force of gravity on 1 kg as 10 N.
(b) The efficiency of the pulley, and
(c) The height to which the load is raised in 4.0 s.
Solutions:
(a) Effort = mass * g
Effort = 100 * 10
Effort = 1000 N
Therefore, input power = effort * distance / time
Input power = 1000 * 8/ 4
Input power = 2000 W
(b) Load pulled by pulley
load = load * g
load = 75 * 10
load = 750 N
mechanical advantage = load / effort
M.A. = 750 / 1000
M.A. = 0.75
Velocity ratio = 1
Efficiency = M.A. / V.R. * 100
Efficiency = 0.75 / 1 * 100
Efficiency = 75%
(c) Load moves by the same distance upwards when the effort moves downwards. Hence, load moves by 8m.
Question: 3
A single fixed pulley and a movable pulley both are separately used to lift a load of 50 kgf to the same height. Compare the efforts applied.
Solutions: For single fixed pulley load is equal to effort applied. That is 50 kg
For single movable pulley efforts applied are half of the load, that is 25 kg.
Hence ratio of efforts applied = 50/25 = 2/1
Ratio of effort by movable pulley to fixed pulley is 2:1.
Question: 4
In a block and tackle system consisting of 3 pulleys, a load of 75 kgf is raised with an effort of 25 kgf. Find: (i) the mechanical advantage, (ii) the velocity ratio, and (iii) the efficiency.
Solutions: Given
Pulleys used = 3
Load = 75 kg
Effort = 25 kg
(i) Mechanical advantage = load / effort
M.A. = 75 /25
M.A. = 3
(ii) Velocity ratio = 3
(iii) Efficiency = M.A. / V.R.
= 3/3
= 100%
Question: 5
A block and tackle system has 5 pulleys. If an effort of 1000 N is needed in the downward direction to raise a load of 4500 N, calculate:
(a)The mechanical advantage,
(b)The velocity ratio and,
(c)The efficiency of the system.
Solutions: Given
Pulleys used = 5
Effort = 1000 N
Load = 4500 N
Mechanical advantage = load / effort
M.A. = 4500/1000
M.A. = 4.5
(b) The velocity ratio = 5
(c) The efficiency = M.A. / V.R.
= 4.5 / 5
= 0.9
= 90 %
Question: 6
In figure draw a tackle to lift the load by applying the force in the download direction.
(a) Mark in the diagram the direction of load L and effort E.
(b) If the load is raised by 1 m, through what distance will the effort move?
(c) State how many strands of tackle are supporting the load.
(d) What is the mechanical advantage of the system?
Solutions:
(b) If the load is raised by 1 m, effort = 1 * 5 = 5
(c) 5 strands of tackle are supporting the load.
(d) Mechanical advantage = load / effort
M.A. = 5T / T
= 5
Question: 7
A pulley system has a velocity ratio 3. Draw a diagram showing the point of application and direction of load (L), effort (E) and tension (T). If lifts a load of 150 N by an effort of 60 N. Calculate its mechanical advantage. Is the pulley system ideal? Give reason.
Solutions: Given
Load = 150 N
Effort = 60 N
Mechanical advantage = load / effort
M.A. = 150/60
M.A. = 2.5
Pulley system is not ideal. The mechanical advantage is less than velocity ratio.
Question: 8
Figure shows a system of four pulleys. The upper two pulleys are fixed and the lower two are movable.
(a) Draw a string around the pulleys. Also show the point of application and direction in which the effort E is applied.
(b) What is the velocity ratio of the system?
(c) How are load and effort of the pulley system related?
(d) What assumption do you make in arriving at your answer in part (c)?
Solutions:
(a)
(b) Velocity of ratio of system = number of pulleys = 4
(c) The mechanical advantage = load / effort. As velocity ratio is four
Load = 4 * effort.
(d) Assumptions are there is no friction in the pulley bearings, weight of pulley can be neglected and effort is applied downwards.
Question: 9
Figure shows a block and tackle system of pulleys used to lift a load.
(a) How many strands of tackle are supporting the load?
(b) Draw arrows to represent tension T in each strand.
(c) What is the mechanical advantage of the system?
(d) When load is pulled up by a distance 1 m, how far does the effort end move?
(e) How much effort is needed to lift a load of 100 N?
Solutions:
(a) 4 strands of tackle are supporting the load.
(b)
(c) Mechanical advantage of system = load / effort
M.A. = 4T / T
M.A, = 4
(d) When load is moved by 1m then the effort moves 4 times that is 4m.
(e) Mechanical advantage = load / effort
effort = load / M.A.
effort = 100 / 4
effort = 25 N
Question: 10
A block and tackle system has the velocity ratio 3. Draw a labelled diagram of the system indicating the points of application and the directions of load L and effort E. A man can exert a pull of 200 kgf. (a) What is the maximum load he can raise with this pulley system if its efficiency is 60%? (b) If the effort end moves a distance 60 cm, what distance does the load move?
Solutions:
Velocity ratio = number of pulleys = 3
If efficiency = 60%
For block and tackle system
Efficiency = 1 – w / nE
0.6 = 1 – w / 3 * 200
w/ 600 = (1- 0.6)
w = 240 kg
hence, load is
L = n E – w
L = 600 – 240
L = 360 kg
The velocity ratio of system = 3
V.R. = distance moved by effort/ distance moved by load
Distance moved by load = 60 / 3
Distance moved by load = 20 cm
Question: 11
You are given four pulleys and three strings. Draw a neat and labelled diagram to use them so as to obtain a maximum mechanical advantage equal to 8. In your diagram mark the directions of load L, effort E and tension in each strand.
What assumptions have you made to obtain the required mechanical advantage?
Solutions:
The assumptions made are pulley and string are mass less and there is no friction in the pulley.