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Physics Laws of Motion

5.37 A disc revolves with a speed of $33 \frac{1}{3}$ rev/min, and has a radius of 15 cm. Two coins are placed at 4 cm and 14 cm away from the centre of the record. If the co-efficient of friction between the coins and the record is 0.15, which of the coins will revolve with the record ?

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Vishal Baghel

Contributor-Level 10

Speed of revolution of the disc, n = $33 \frac{1}{3}$ rev/min = 100/3 rpm = 100/ (3 $* 60) = 0.56 r p s$

Angular acceleration $ω$ = 2 $π n$ = 2 $* \frac{22}{7} * 0.56$ = 3.492 rad/s

The coins revolve with the disc. The centripetal force is provided by the frictional force $m v^{2} / r \leq μ m g$ …. (1)

As v = r $ω$ , the above equation becomes mr $ω^{2} / r \leq μ m g$

r $\leq μ g / ω^{2}$

r $\leq$ (0.15 $* 10) / {3.492}^{2}$ = 12 cm

For coin A, r = 4 cm

The condition (r $\leq$ 12 ) is satisfied for the coin placed at r = 4 cm, so coin A will revolve with the disc.

The condition (r $\leq$ 12 ) is not satisfied for the coin placed at r = 14 cm, so coin B will not revolve with the disc.

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Physics Laws of Motion

5.36 The rear side of a truck is open and a box of 40 kg mass is placed 5 m away from the open end as shown in Fig. 5.22. The coefficient of friction between the box and the surface below it is 0.15. On a straight road, the truck starts from rest and accelerates with 2 m s^-2. At what distance from the starting point does the box fall off the truck? (Ignore the size of the box).

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Vishal Baghel

Contributor-Level 10

Force on the box, F = MA = 40 $*$ 2 N = 80 N

Frictional force, Ff = $? m g$ = 0.15 $* 40 * 10 =$ 60 N

Net force = F – Ff = 80 – 60 = 20 N

From the equation F = ma, we get the backward acceleration produced in the box

a = 20/40 = 0.5 m/s2

From the equation s = ut + $\frac{1}{2} a t^{2}$ , to travel s = 5 m by the box to fall off from the truck, we get

5 = 0 $* t$ + $\frac{1}{2}$ $*$ 0.5 $* t^{2}$

5 = 0.25 $t^{2}$ , t = 4.47 s

The travel of truck during t = 4.47 s is

= 0 $* t$ + $\frac{1}{2}$ $*$ 0.5 $* t^{2}$ = 0.5 $* 2 * {4.47}^{2}$ = 19.98 m

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Physics Laws of Motion

5.35 A block of mass 15 kg is placed on a long trolley. The coefficient of static friction between the block and the trolley is 0.18. The trolley accelerates from rest with 0.5 m s^-2 for 20 s and then moves with uniform velocity.

Discuss the motion of the block as viewed by

(a) A stationary observer on the ground

(b) An observer moving with the trolley

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Vishal Baghel

Contributor-Level 10

Mass of the block = 15 kg

Coefficient of static friction between the block and the trolley $μ$ = 0.18

Acceleration of the trolley = 0.5 m/s2

(a) Force experienced by block, F = MA = 15 $*$ 0.5 = 7.5 N. This fore acts in the direction of motion of the trolley

Force of friction, Ff = $μ m g$ = 0.18 $* 15 * 10$ N = 27 N

Force experienced by the block is less than the friction, hence for a stationary observer on the ground, the block will be stationary

(b) When an observer moves with the trolley, the trolley will appear to be at rest

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Physics Laws of Motion

5.34 Two bodies A and B of masses 5 kg and 10 kg in contact with each other rest on a table against a rigid wall (Fig. 5.21). The coefficient of friction between the bodies and the table is

0.15. A force of 200 N is applied horizontally to A. What are

(a) The reaction of the partition

(b) The action-reaction forces between A and B ? What happens when the wall is removed?

Does the answer to (b) change, when the bodies are in motion?

Ignore the difference between μs and μk

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Vishal Baghel

Contributor-Level 10

Mass of the body A, $m_{A}$ = 5 kg

Mass of the body B, $m_{B}$ = 10 kg

Applied force = 200 N

Coefficient of friction between the bodies and the table, μs = 0.15

(a) The frictional force is given by the relation

fs = $μ$ ( $m_{A}$ + $m_{B}$ )g = 0.15 $* ((5 + 10)) =$ 22.5 N. towards left

Hence, the force on the partition = 200 – 22.5 N = 177.5 N, acting rightwards

According to Newton's 3rd law, the reaction of the partition will be 177.5 N, acting towards left

(b) Force of friction on mass A is given by

Fa = $μ$ mAg = 0.15 $* 5 * 10 =$ 7.5 N, acting leftward

The net force exerted by mass A on mass B = 200 -7.5 N =

...more

Mass of the body A, $m_{A}$ = 5 kg

Mass of the body B, $m_{B}$ = 10 kg

Applied force = 200 N

Coefficient of friction between the bodies and the table, μs = 0.15

(a) The frictional force is given by the relation

fs = $μ$ ( $m_{A}$ + $m_{B}$ )g = 0.15 $* ((5 + 10)) =$ 22.5 N. towards left

Hence, the force on the partition = 200 – 22.5 N = 177.5 N, acting rightwards

According to Newton's 3rd law, the reaction of the partition will be 177.5 N, acting towards left

(b) Force of friction on mass A is given by

Fa = $μ$ mAg = 0.15 $* 5 * 10 =$ 7.5 N, acting leftward

The net force exerted by mass A on mass B = 200 -7.5 N = 192.5 N, acting rightward

An equal amount of force will be exerted by mass B on A = 192.5 N, acting leftward

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

The net force acting on the moving system = 177.5 N

The equation of motion for the system of acceleration a, can be written as

Net force = ( $m_{A}$ + $m_{B}$ )a

a = Net force / ( $m_{A}$ + $m_{B}$ )= 177.5 / (5+10) = 11.833 m/s²

Net force causing mass A to move

Fa = $m_{A}$ a = 5 $* 11.833 = 59.17 N$

Net force exerted by mass A on mass B = 192.5 – 59.17 N = 133.33 N

This force acts in the direction of motion. According to Newton's 3rd law, force exerted by mass B on mass A will be 133.33 N.

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a year ago

Physics Laws of Motion

5.33 A monkey of mass 40 kg climbs on a rope (Fig. 5.20) which can stand a maximum tension of 600 N. In which of the following cases will the rope break: the monkey

(a) Climbs up with an acceleration of 6 m s^-2

(b) Climbs down with an acceleration of 4 m s^-2

(c) Climbs up with a uniform speed of 5 m s^-1

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

(Ignore the mass of the rope)

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Vishal Baghel

Contributor-Level 10

Mass of the monkey = 40 kg

Maximum tension of the rope Tmax = 600 N

(a) When the monkey climbs up with an acceleration of 6m / s²

Tension in the rope, T – mg = ma

T = m (g+a) = 40 (10+6)= 640 N

Since T > Tmax, the rope will break

(b) When the monkey climbs down with an acceleration of 4m/s²

Tension in the rope T is given by mg – T = ma

T = m (g-a) = 40 (10-4) N = 240 N

Since T < T max, the rope will not break

(c) Climbs up with an uniform speed of 5 m/s

In this case, the acceleration = 0

The tension in the rope is given by

T –mg = ma

T = mg = 40 $* 10 = 400 N$

Since T < Tmax, the rope will not break

(d) When the monkey falls down freely under gravity

The tension in the rope is given by the equat

...more

Mass of the monkey = 40 kg

Maximum tension of the rope Tmax = 600 N

(a) When the monkey climbs up with an acceleration of 6m / s²

Tension in the rope, T – mg = ma

T = m (g+a) = 40 (10+6)= 640 N

Since T > Tmax, the rope will break

(b) When the monkey climbs down with an acceleration of 4m/s²

Tension in the rope T is given by mg – T = ma

T = m (g-a) = 40 (10-4) N = 240 N

Since T < T max, the rope will not break

(c) Climbs up with an uniform speed of 5 m/s

In this case, the acceleration = 0

The tension in the rope is given by

T –mg = ma

T = mg = 40 $* 10 = 400 N$

Since T < Tmax, the rope will not break

(d) When the monkey falls down freely under gravity

The tension in the rope is given by the equation

T + mg = ma, since a = g, T = 0

Since T < Tmax, the rope will not break

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New answer posted

a year ago

Physics Laws of Motion

5.32 A block of mass 25 kg is raised by a 50 kg man in two different ways as shown in Fig. 5.19. What is the action on the floor by the man in the two cases ? If the floor yields to a normal force of 700 N, which mode should the man adopt to lift the block without the floor yielding ?

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Vishal Baghel

Contributor-Level 10

Mass of the block = 25 kg

Mass of the man = 50 kg

Acceleration due to gravity = 10 m/s2

Weight of the block = 250 N

Weight of the man = 500 N

In the 1st case, the man lifts the block directly, applying an upward force, same as the block = 250 N

Due to Newton's 3rd law of motion, the downward reaction of the man on the floor = 500 N + 250 N = 750 N

In the 2nd case, the man applies a downward force of 25 kg wt. According to Newton's 3rd law, the action on the floor by the man 500N-250N = 250N

The man should adopt the second case

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New answer posted

a year ago

Physics Laws of Motion

5.31 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 ?

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Vishal Baghel

Contributor-Level 10

Given

Radius, r = 30m

Speed, v = 54 km/h = 15 m/s

The mass of the train, m = 106 kg

(a) The required centripetal force is provided by the rails to the wheels of the train

(b) The angle of banking required to prevent wearing out of the rails

$\tan ? θ$ = $v^{2}$ / rg = $15^{2}$ / (30 $* 10)$

$θ$ = 37 $°$

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New answer posted

a year ago

Physics Laws of Motion

5.30 An aircraft executes a horizontal loop at a speed of 720 km/h with its wings banked at 15°. What is the radius of the loop ?

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Vishal Baghel

Contributor-Level 10

The speed of the aircraft, v = 720 km/h = 200 m/s

The angle of banking = 15 $°$

From the relation $\tan ? θ$ = $v^{2}$ /rg we get r = $v^{2}$ / ( g $* \tan ? θ)$ = 14928 m

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New answer posted

a year ago

Physics Laws of Motion

5.29 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

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Vishal Baghel

Contributor-Level 10

(a) The force on the 7th coin is due to the 3 coins kept above it

The weight of the 3 coins = 3m

Force exerted on the 7th coin is (3mg) N and the force is acting vertically downwards

(b) The force on the 7th coin by the 8th coin will be the force exerted by the 3 coins above it = (3mg) N, acting downwards

(c) The 6th coin will experience the force of 4 coins above it, acting downwards = 4mg

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New answer posted

a year ago

Physics Laws of Motion

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

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Vishal Baghel

Contributor-Level 10

Speed of water, v = 15 m/s

Cross-sectional area of the tube, A = 10^-2 m²

Density of water $ρ$ = 1000 kg/m3

Mass of the water hitting the wall = $ρ * A * v$ = 1000 $*$ 10^-2 $* 15$ = 150 kg/s

Force exerted on the wall because of impact of water = mass $*$ velocity = 150 $* 15$ N = 2250 N

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