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From Mechanics

Theory

Exercises

Key Points

Newton's First Law

A particle will move with a constant velocity or remain at rest if the resultant force on the particle is zero.

Equilibrium

If the resultant force on a particle is zero, then the forces acting on the particle are said to be in equilibrium.

The Universal Law of Gravitation

F=d 2Gm1m2 G=6.6710−11 kg-1m3s-2

Gravitational force from an heavenly body

For a particle in the neighbourhood of a planet, moon or star the distance d is measured from the centre to the particle. Thus d=R+h, where h is distance to the surface.

Gravity on Earth

The force of gravity is often called the weight.

F=mgg=9.8 ms-2

Data

Radius of Earth is 6.37106 metres

Mass of Earth is 5.981024 kg

Example 2.1

Describe whether or not the forces acting on the following objects are in equilibrium:

a) A passenger in a train that travels at a constant speed.

b) A hot air balloon rising at a constant rate.

c) A stone dropped into a very deep well full of water.

Solution

a) Yes, if it is travelling in a straight line.

b) Yes, if it is travelling in a straight line.

c) Yes, if it reaches a terminal velocity, so that it is travelling in a straight line at a constant speed.

Example 2.2

Find the magnitude of the force of gravity (weight) acting on a lorry of mass 22 tonnes.

Solution

This is calculated using the fact that the weight is given by mg.

mg=220009.8=215600 N

The diagram shows the lorry and its weight.

Note that reaction forces also act upwards on each wheel.

R1+R2+R3+R4=215600 N

Example 2.3

A box of mass 30 kg is at rest on a table.

a) Calculate the weight of the box.

b) State the magnitude of the upward force that the table exerts on the box.

Solution

a)

W=309.8=294 N

b) An upward force of 294 N must act for the box to remain in equilibrium.

Example 2.4

A satellite, of mass 400 kg, is at a height of 12 km above the surface of the earth. Find the magnitude of the gravitational attraction on the satellite.

Data: G=6.6710−11 kg−1m3s−2

Radius of earth =6.37106m

Mass of earth =5.981024kg

Solution

d 2Gm1m2=6.37106+1200026.6710−114005.981024=3917 N

Example 2.5

The sun has mass 1.991030kg . The mean distance of the earth from the sun is approximately 1.51011m .

(a) Calculate the force that the sun exerts on the earth.

(b) State the force that the earth exerts on the sun.

(c) Explain why this force varies.

Solution

(a)

d 2Gm1m2=1.5101126.6710−111.9910305.981024=3.531022 N

(b) 3.531022 N

(c) The distance between the earth and the sun varies.