# 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:

The expression of the mechanical advantage of a lever is:

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,

Effort arm = 30 cm

As we know,

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

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Given,

The total length of a crowbar = 120 cm

Effort arm = 120 – 20 = 100 cm

We know that,

=>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,

Effort arm = 7.5 cm

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,

Effort arm = 10 cm

We also know that, M.A. = load/effort

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