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Force and Pressure Class 8 Numerical

By Anupam M

Last updated on June 25th, 2023 at 04:41 pm

In this post, we will solve Numerical problems based on the Force and Pressure chapter of class 8 science. Here, first, we will go through the formula of pressure & other related data, and then attempt the numerical questions. Solution & answer is given for all the numerical questions.

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Formulas Used

Formula of pressure is as follows:
Pressure = Force / [Area on which the force in applied]

The air pressure at the sea level is called 1 atmosphere pressure. Air pressure decreases as one goes up in height from the sea level.

1 Atmosphere = 101 kPa = 101 x 103 Pa = 1 bar = 103 mbar

Force and Pressure Class 8 Numerical Questions and Answers [solved]

Example 1. The area of the bottom of a tiffin box is 0.25 m2 and weight is 50 N, Calculate the pressure exerted by the box on the shelf.

Solution:

Given: Area = 0.25 m2, weight of the box = 50 N, Pressure = ?

Pressure =force/area = 50 N/0.25 m2 = 200 N/ m2

Example 2. The area of the tip of a screw is 0.5 mm2 and its weight is 0.5 N. Calculate the pressure (in Pa) exerted by the screw on a wooden plank.

Solution:

Given: Area = 0.5 x10-6 m2

Weight of the screw = 0.5 N, Pressure =?

Pressure =weight/area =0.5N / (0.5×10-6 m2) = 106 N/m2 = 106 Pa

Example 3. Mass of a block of metal is 10 kg and its dimensions are length 50 cm, breadth 10 cm, height 20 cm. (See fig.) If the metal block is placed on the following surface on the table, find out on which of the surface ABCD, CDEF and BCFG will the pressure exerted on the table be maximum.

Solution:

See also  A smooth wooden 40.0 N block is placed on a smooth wooden table. A force of 14.0 N is required to keep the block moving at constant velocity. (a) What is the coefficient of sliding friction between the block and the tabletop? (b) If a 20.0 N brick is placed on top of the wooden block, what force will be required to keep the block and brick moving at constant velocity?

Given: Weight of the block = mg =10 x 9.8N =98 N

For the surface ABCD, length = 50 cm, height = 20 cm.

Area = 50 cm x 20 cm= 1000 cm2 = 0.1 m2

Pressure = weight/area= 98/(0.1) = 980 Pa

For the surface CDEF, length = 50 cm, breadth = 10 cm

Area = 50 cm x 10 cm = 500 cm2= 0.05 m2

Pressure =weight/area= 98/0.05 = 9800/5 = 1960 Pa

For the surface BCFG, height = 20 cm, breadth = 10 cm

Area = 20 cm x 10 cm = 200 cm2= 0.02 m2

Pressure =weight/area= 98 N/0.02 m2 = 4900 Pa

From all 3 cases, we found that the Maximum Pressure is on the surface BCFG.

Hence, the smaller the area of contact, the larger the pressure.

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