Laws of Motion

• Question 465

## A helicopter of mass 1000 kg rises with a vertical acceleration of 15 m s–2. The crew and the passengers weigh 300 kg. Give the magnitude and direction of the, (a) force on the floor by the crew and passengers,(b) action of the rotor of the helicopter on the surrounding air,(c) force on the helicopter due to the surrounding air.

Solution

Given,
Mass of the helicopter, mh = 1000 kg
Mass of the crew and passengers, mp = 300 kg
Total mass of the system, m = 1300 kg
Acceleration of the helicopter, a = 15 m/s
Using Newton’s second law of motion, the reaction force R,
R – mpg = ma

= mp(g + a)
= 300 (10 + 15)
= 300 × 25
= 7500 N
The reaction force will be directed upwards, the helicopter is accelerating vertically upwards.
According to Newton’s third law of motion, the force on the floor by the crew and passengers  = 7500 N, directed downward.
(b)
Using Newton’s second law of motion, the reaction force R’ experienced by the helicopter can be calculated as,
R' - mg = ma

= m(g + a)
= 1300 (10 + 15)
= 1300 × 25

= 32500 N
The reaction force experienced by the helicopter from the surrounding air is acting upward. Hence, as per Newton’s third law of motion, the action of the rotor on the surrounding air will be 32500 N, directed downward.
(c) The force on the helicopter due to the surrounding air is 32500 N, directed upwards.

Question 466

## A stream of water flowing horizontally with a speed of 15 m s–1 gushes out of a tube of cross-sectional area 10–2m2, and hits a vertical wall nearby. What is the force exerted on the wall by the impact of water, assuming it does not rebound?

Solution
Given,
Speed of the water stream, v = 15 m/s
Cross-sectional area of the tube, A = 10–2 m2

Volume of water coming out from the pipe per second,
VAv
= 15 × 10–2 m3/s
Density of water, ρ = 103 kg/m

Mass of water flowing out through the pipe per second = ρ ×V
= 150 kg/s
The water strikes the wall and does not rebound.
Therefore, according to Newton's second law of motion,
Force exerted by the water on the wall,
F = Rate of change of momentum =
=
= 150 × 15
= 2250 N
Question 467

## Ten one-rupee coins are put on top of each other on a table. Each coin has a mass m. Give the magnitude and direction of,(a) the force on the 7th coin (counted from the bottom) due to all the coins on its top,(b) the force on the 7th coin by the eighth coin,(c) the reaction of the 6th coin on the 7th coin.

Solution
(a) Force on the seventh coin is exerted by the weight of the three coins on its top.
Weight of one coin = mg
Weight of three coins = 3mg
Hence, the force exerted on the 7th coin by the three coins on its top is 3mg.
This force acts vertically downward.
(b) Force on the seventh coin by the eighth coin is because of the weight of the eighth coin and the other two coins (ninth and tenth) on its top.
Weight of the eighth coin = mg
Weight of the ninth coin = mg
Weight of the tenth coin = mg
Total weight of these three coins = 3m
Hence, the force exerted on the 7th coin by the eighth coin is 3mg.
This force acts vertically downward.

(c) The 6th coin experiences a downward force because of the weight of the four coins (7th, 8th, 9th, and 10th) on its top.
Therefore, the total downward force experienced by the 6thcoin is 4mg.
As per Newton’s third law of motion, the 6th coin will produce an equal reaction force on the 7th coin, but in the opposite direction.
Hence, the reaction force of the 6th coin on the 7thcoin is of magnitude 4mg.
This force acts in the upward direction.
Question 468

## A train runs along an unbanked circular track of radius 30 m at a speed of 54 km/h. The mass of the train is 106 kg. What provides the centripetal force required for this purpose – The engine or the rails? What is the angle of banking required to prevent wearing out of the rail?

Solution
Given,
Radius of the circular track, r = 30 m
Speed of the train, v = 54 km/h = 15 m/s
Mass of the train, m = 106 kg
The centripetal force is provided by the lateral thrust of the rail on the wheel.
As per Newton’s third law of motion, the wheel exerts an equal and opposite force on the rail.
This reaction force is responsible for the wear and rear of the rail.
The angle of banking θ, is related to the radius (r) and speed (v) by the relation:
tan θ = v2 / rg
= 152 / (30 × 10)
= 225 / 300
θ = tan-1 (0.75) = 36.87

Therefore, the angle of banking is about 36.87°.