Question 465

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, m_{h} = 1000 kg

Mass of the crew and passengers, m_{p} = 300 kg

Total mass of the system, m = 1300 kg

Acceleration of the helicopter, *a* = 15 m/s^{2 }Using Newton’s second law of motion, the reaction force *R*,* R* – *m*_{p}g* = ma*

=

= 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

=

= 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

Solution

Given,

Speed of the water stream,*v* = 15 m/s

Cross-sectional area of the tube, Speed of the water stream,

Volume of water coming out from the pipe per second,

= 15 × 10

Density of water,

Mass of water flowing out through the pipe per second =

= 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

(a) the force on the 7

(b) the force on the 7

(c) the reaction of the 6th coin on the 7

Solution

(a)** **Force on the seventh coin is exerted by the weight of the three coins on its top.

Weight of one coin = Weight of three coins = 3

Hence, the force exerted on the 7

This force acts vertically downward.

(b)

Weight of the eighth coin =

Weight of the ninth coin =

Weight of the tenth coin =

Total weight of these three coins = 3

Hence, the force exerted on the 7

This force acts vertically downward.

Therefore, the total downward force experienced by the 6

As per Newton’s third law of motion, the 6

Hence, the reaction force of the 6

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 10^{6} 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, Radius of the circular track,

Mass of the train,

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

= 15

= 225 / 300

θ =

Therefore, the angle of banking is about 36.87°.