Question 477

(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 478

(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.

Question 479

Solution

We have,

Mass of the box,*m* = 40 kg

Coefficient of friction, Mass of the box,

Initial velocity,

Acceleration,

Distance of the box from the end of the truck,

According to Newton's second law of motion,

Force on the box caused by the accelerated motion,

= 80 N

According to Newton’s third law of motion, a reaction force of 80 N is acting on the box in the backward direction.

The backward motion of the box is opposed by the force of friction

This force is given by,

= 0.15 × 40 × 10

= 60 N

Therefore,

Net force acting on the block,

The backward acceleration produced in the box is,

Using the second equation of motion,

Time t is,

s' = ut + (1/2) a

5 = 0 + (1/2) × 0.5 × t

∴ t = √

Hence, the box will fall from the truck after s from the start.

The distance

s = ut + (1/2)at

= 0 + (1/2) × 2 × (

= 20 m

Therefore, at a distance of 20 m, the box will fall off the truck.

Question 480

Solution

Coin placed at 4 cm from the centre

Given,

Mass of each coin = Given,

Radius of the disc,

Frequency of revolution,

=

= rev/s

Coefficient of friction,

In the given situation, the coin having a force of friction greater than or equal to the centripetal force provided by the rotation of the disc will revolve with the disc. If this is not the case, then the coin will slip from the disc.

Case 1: Coin placed at 4 cm

Radius of revolution,

Angular frequency,

= 2 x

= 3.49 s

Frictional force,

= 0.15 ×

= 1.5

Centripetal force on the coin is,

= m × 0.04 × (3.49)

= 0.49 m N

Since

Case 2: Coin placed at 14 cm

Radius, r" = 14 cm = 0.14 m

Angular frequency,

Frictional force,

Centripetal force is given as,

=

= 1.7

Since