Question 469

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 470

(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

Question 471

(a) the reaction of the partition

(b) the action-reaction forces between

What happens when the wall is removed? Does the answer to (b) change, when the bodies are in motion? Ignore the difference between μ

Solution

(a)

Mass of body A,*m*_{A} = 5 kg

Mass of body B,*m*_{B} = 10 kg

Applied force, Mass of body A,

Mass of body B,

Coefficient of friction,

The force of friction is given by the relation,

f

= 0.15 (5 + 10) × 10

= 1.5 × 15

= 22.5 N, leftward

Net force acting on the partition = 200 – 22.5

= 177.5 N, rightward

According to Newton’s third law of motion, the reaction force of the partition will be in the direction opposite to the net applied force.

Hence, the reaction of the partition will be 177.5 N,directed leftwards.

(b)

= 0.15 × 5 × 10

= 7.5 N leftward

Net force exerted by mass A on mass B = 200 – 7.5 = 192.5 N, rightward

Now, according to Newton’s third law of motion, an equal amount of reaction force will be exerted by mass B on mass A.

i.e., Net force acting on mass A by mass B= 192.5 N acting leftward.

When the wall is removed, the two bodies will move in the direction of the applied force.

Net force acting on the moving system = 177.5 N

The equation of motion for the system of acceleration

Net force = (m

∴ Acceleration, a =

=

Net force causing mass A to move,

= 5 × 11.83

= 59.15 N

Net force exerted by mass A on mass B = 192.5 – 59.15 = 133.35 N

This force will act in the direction of motion.

According to Newton’s third law of motion, an equal amount of force will be exerted by mass B on mass A, i.e., 133.3 N, acting opposite to the direction of motion.

Question 472

(a) a stationary observer on the ground,

(b) an observer moving with the trolley.

Solution

Given,

Mass of the block,*m* = 15 kg

Coefficient of static friction,*μ* = 0.18

Acceleration of the trolley,*a* = 0.5 m/s^{2 }

According to Newton’s second law of motion,

Force (*F*) on the block caused by the motion of the trolley is given by the relation,

* F* = *ma *

= 15 × 0.5

= 7.5 N

This force is acted in the direction of motion of the trolley.

Force of static friction between the block and the trolley,

* f* = *μ**m*g

= 0.18 × 15 × 10

= 27 N

The applied external force is less than the force of static friction between the block and the trolley. Therefore, the block will appear to be at rest, for an observer on the ground.

When the trolley moves with uniform velocity, only the force of friction will act in this situation and there will be no applied external force.

(b) The person who is moving with the trolley has some acceleration.

The frictional force is acting backwards on the trolley and is opposed by a pseudo force of the same magnitude. But, this force is acting in the backward direction. Therefore, for an observer moving with the trolley, the trolley will appear to be at rest.

Mass of the block,

Coefficient of static friction,

Acceleration of the trolley,

According to Newton’s second law of motion,

Force (

= 7.5 N

This force is acted in the direction of motion of the trolley.

Force of static friction between the block and the trolley,

= 0.18 × 15 × 10

= 27 N

The applied external force is less than the force of static friction between the block and the trolley. Therefore, the block will appear to be at rest, for an observer on the ground.

When the trolley moves with uniform velocity, only the force of friction will act in this situation and there will be no applied external force.

(b) The person who is moving with the trolley has some acceleration.

The frictional force is acting backwards on the trolley and is opposed by a pseudo force of the same magnitude. But, this force is acting in the backward direction. Therefore, for an observer moving with the trolley, the trolley will appear to be at rest.