Question 473

(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 474

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°.

Question 475

Solution

Given,

Mass of the block, *m* = 25 kg

Mass of the man, *M* = 50 kg

Acceleration due to gravity, g = 10 m/s^{2 }

Force applied on the block, *F* = 25 × 10

= 250 N

Weight of the man, *W* = 50 × 10 = 500 N

Case (a): When the man lifts the block directly

In this case, the man applies a force in the upward direction.

This increases his apparent weight.

Therefore,

Action on the floor by the man = 250 + 500 = 750 N

Case (b): When the man lifts the block using a pulley

In this case, the man applies a force in the downward direction. This decreases his apparent weight.

Therefore,

Action on the floor by the man = 500 – 250

= 250 N

If the floor can yield to a normal force of 700 N, then the man should adopt the second method to easily lift the block by applying lesser force.

Question 476

(a) climbs up with an acceleration of 6 m s

(b) climbs down with an acceleration of 4 m s

(c) climbs up with a uniform speed of 5 m s

(d) falls down the rope nearly freely under gravity?

(Ignore the mass of the rope).

Solution

Case (a):

Mass of the monkey, Acceleration due to gravity,

Maximum tension that the rope can bear,

Acceleration of the monkey,

Using Newton’s second law of motion, equation of motion is

∴

= 40 (10 + 6)

= 640 N

Since

Case (b)

Acceleration of the monkey,

Using Newton’s second law of motion, the equation of motion is,

∴

= 40(10 – 4)

= 240 N

Since

Case (c)

The monkey is climbing with a uniform speed of 5 m/s.

Therefore, its acceleration is zero, i.e.,

Using Newton’s second law of motion, equation of motion is,

∴

= 40 × 10

= 400 N

Since

Case (d)

When the monkey falls freely under gravity, acceleration will become equal to the acceleration due to gravity, i.e.,

Using Newton’s second law of motion, we can write the equation of motion as:

∴

Since