Laws of Motion

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New answer posted•a year ago
New answer posted•a year ago
S
a year ago

Let θ be the angle made by the radius vector joining the bead and the centre of the wire with the downward direction. Let, N be the normal reaction.

mg = N cos?θ …….(1)

mr ω2 = N sin?θ ……(2)

m(R sin?θ ) ω2 = N sin?θ

Hence N = m(R) ω2

Substituting the value on N

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New answer posted•a year ago
S
a year ago

Mass of the man, m = 70 kg

Radius of the drum, r = 3 m

Coefficient of friction between the wall and his clothing,  μ = 0.15

Number of revs of hollow cylindrical drum = 200 rev/min = 200/60 rev/s = 3.33 rev/s

The centripetal force required is provided by the normal N of the wall on the man

N = m&

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New answer posted•a year ago
S
a year ago

When the motorcyclist is at the uppermost point of the death well, the normal reaction R on the motorcyclist by the ceiling of the chamber acts downwards. His weight mg also acts downwards. The outward centrifugal force acting on the motorcyclist is balanced by two forces.

R + mg = m v2/r , whe

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New answer posted•a year ago
S
a year ago

Speed of revolution of the disc, n = 3313 rev/min = 100/3 rpm = 100/ (3 ×60)=0.56rps

Angular acceleration ω = 2 πn = 2 ×227×0.56 = 3.492 rad/s

The coins revolve with the disc. The centripetal force is provided by the frictional force mv2/r≤μmg …. (1)

As v = r ω , the above eq

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New answer posted•a year ago
S
a year ago

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

Frictional force, Ff = ? mg = 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 + 12at2 , to travel s = 5 m by

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New answer posted•a year ago
S
a year ago

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 = μmg 

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New answer posted•a year ago
New answer posted•a year ago
S
a year ago

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

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New answer posted•a year ago
S
a year ago

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 t

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