Machine Numericals Class 10 ICSE
Last updated on June 14th, 2023 at 03:19 pm
In this post, we will solve a set of selected Numericals based on the Machines chapter from the class 10 physics syllabus of ICSE.
Formulas Used
The mechanical advantage of a machine is equal to the product of its efficiency and velocity ratio.
M.A. = V.R. x η
The expression of the mechanical advantage of a lever is:
M.A.=Effort arm/Load arm
The expression of the mechanical advantage of a lever is:
M.A.=Load/Effort
The principle of moments states that when a system is balanced the sum of the anti-clockwise turning
moments equals the sum of the clockwise turning moments.
Machine Numericals for Class 10 ICSE
Question 1)
The figure below shows a uniform meter rule of weight W supported on a fulcrum at the 60 cm mark by applying the effort E at the 90 cm mark.
(a) State with reasons whether the weight W of the rule is greater than, less than or equal to the effort E.
(b) Find the mechanical advantage in an ideal case.
Solution
(a) The weight W of the scale is greater than E.
The effort arm is 30 cm and the load arm is 10 cm. In order to balance the scale — the weight (W) of the scale has to be more than the effort (E).
(b) Given,
Load arm = 10 cm
Effort arm = 30 cm
As we know,
M.A.=Effort arm/Load arm
Substituting the values in the formula we get,
M.A.=30/10
⇒M.A.=3
Question 2) 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.
Solution
Given,
The total length of a crowbar = 120 cm
Effort arm = 120 – 20 = 100 cm
Load arm = 20 cm
We know that,
Mechanical advantage M.A.=Effort arm/Load arm
=>M.A.=100/20
⇒M.A.=5
Question 3) A pair of scissors has its blades 15 cm long, while its handles are 7.5 cm long. What is its mechanical advantage?
Solution
Given,
Load arm = 15 cm
Effort arm = 7.5 cm
Mechanical advantage M.A.=Effort arm/Load arm.=7.5/15=0.5
Question 4) A force of 5kgf is required to cut a metal sheet. A pair of shears used for cutting the metal sheet has its blades 5 cm long, while its handles are 10 cm long. What effort is needed to cut the sheet?
Solution
Given,
Load arm = 5 cm
Effort arm = 10 cm
Mechanical advantage M.A.=Effort arm/Load arm=10/5=2
We also know that, M.A. = load/effort
Effort=Load/M.A.
Substituting the values in the formula we get,
Effort=5/2
⇒Effort=2.5 kgf
Question 5) In the Figure below the man applies a downward force of 150 N at a big distance of 2.0 m from the pivot. A rock is placed a small distance of 0.4 m from the pivot.
How much force does the lever exert on the rock?
Solution
Applying the Principle of moments,
turning moment exerted by the man = turning moment exerted on the rock
So, 150 × 2 = F × 0.4
F = 750 N
The lever exert 750 N on the rock
Question 6) A small crane is being used to load a boat. A 5000 N counterbalance weight on the left-hand side is used to stabilize the crane and stop it from toppling over.
Where should the counterbalance weight be placed when the crew uses the crane to lift a 2000 N crate into the boat?
Solution:
The anticlockwise moment from the counterbalance = The clockwise moment from the crate
5000 × d = 2000 × 3.0
d = 6000/5000
d = 1.2 m (answer)
Question 7) In the Figure below the teeth of gear wheels exert a force of 250 N on each other. The radius of the smaller wheel is 10 cm, and the larger wheel has a radius of 20 cm.
a) Calculate the turning moment produced by each wheel.
b) Explain how gears can be used in machines.
c) Which gear wheel is likely to wear out first?
Solution Link: you get the solution to this numerical on gear wheels here
